$\displaystyle\varphi_{Y}(r)$
$\displaystyle=\exp\left(ir^{\prime}\mu\right)\prod_{k=1}^{K}\frac{K_{\frac{\nu_{k}}{2}}(\sqrt{\nu_{k}r^{\prime}\Omega_{k}r})\left(\nu_{k}r^{\prime}\Omega_{k}r\right)^{\frac{\nu_{k}}{4}}}{\Gamma\left(\frac{\nu_{k}}{2}\right)2^{\frac{\nu_{k}}{2}-1}},\quad{\rm
where}\quad r\in\mathbb{R}^{n},$

where $\Omega_{k}=\Xi_{k}\Xi_{k}^{\prime}$ for $k=1,\ldots,K$. It is evident
that only $\mu$ and the pairs, $\\{\nu_{k},\Omega_{k}\\}$, $k=1,\ldots,K$,
matter for the distribution. Moreover, because the multivariate-$t$
distribution is not a stable distribution, except for $\nu=1$ (Cauchy) and
$\nu=\infty$ (Gaussian), we can identify $\\{n_{k},\nu_{k},\Omega_{k}\\}$,
$k=1,\ldots,K$, along with $\mu$.

$\square$

######  Lemma A.2.

If $P$ is orthonormal, $P^{\prime}P=I$, and both $B$ and $BP$ are symmetric
and positive definite, then $B=BP$.

###### Proof.

By the symmetry we have $P^{\prime}B=BP$ and

$B=PBP=P^{\prime}BP^{\prime}.$

So it follows that

$(B)^{2}=P^{\prime}BP^{\prime}PBP=P^{\prime}BBP=(BP)^{2},$

From the eigendecomposition of the two symmetric matrices, $B=Q\Lambda
Q^{\prime}$ and $BP=N\varUpsilon N^{\prime}$, we now have

$Q\Lambda Q^{\prime}Q\Lambda
Q^{\prime}=Q\Lambda^{2}Q^{\prime}=N\varUpsilon^{2}N^{\prime},$

such that$\Lambda^{2}=Q^{\prime}N\varUpsilon^{2}N^{\prime}Q$, from which we
can conclude that $\Lambda$ and $\varUpsilon$ are the same diagonal matrices,
aside from a reordering of the elements. Hence $Q^{\prime}N$ is a permutation
matrix.

$B=Q\Lambda Q^{\prime}=Q[Q^{\prime}N\varUpsilon
N^{\prime}Q]Q^{\prime}=Q[Q^{\prime}BPQ]Q^{\prime}=BP.$ $B=BP\Leftrightarrow
B(I-P)=0,$

which implies $P=I$ because $B$ is pd. ∎

Proof of Theorem 4. Since $\Omega_{k}$ is psd with rank $n_{k}$, we can
express it as $\Omega_{k}=\tilde{\Xi}_{k}\tilde{\Xi}_{k}^{\prime}$ for some
matrix $\tilde{\Xi}_{k}\in\mathbb{R}^{n\times n_{k}}$ with full column rank,
$n_{k}$. Next, let $P_{kk}\in\mathbb{R}^{n_{k}\times n_{k}}$ be orthonormal,
$k=1,\ldots,K$, and define

$\Xi=\tilde{\Xi}P,\qquad\text{where}\qquad
P=\left(\begin{array}[]{cccc}P_{11}&0&\cdots&0\\\ 0&P_{22}&\ddots&\vdots\\\
\vdots&\ddots&\ddots&0\\\
0&\cdots&0&P_{KK}\end{array}\right)\in\mathbb{R}^{n\times n}.$ (A.1)

It follow that
$\Xi_{k}\Xi_{k}^{\prime}=\tilde{\Xi}_{k}P_{kk}P_{kk}^{\prime}\tilde{\Xi}_{k}^{\prime}=\Omega_{k}$.
If $\tilde{\Xi}_{kk}$ is invertible, we can set

$P_{kk}=\tilde{\Xi}_{kk}^{\prime}(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}},$

and it is easy to verify that $P_{kk}^{\prime}P_{kk}=I_{n_{k}}$. Moreover, it
follows that
$\Xi_{kk}=\tilde{\Xi}_{kk}P_{kk}=(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{\frac{1}{2}}$
is a symmetric positive definite matrix. More generally, if $\tilde{\Xi}_{kk}$
is not invertible, but has (reduced) rank $r$, then we the singular value
decomposition gives us,

$\tilde{\Xi}_{kk}=V\Lambda_{k}U^{\prime},\qquad\text{for some}\quad
V,U\in\mathbb{R}^{n_{k}\times r},\quad\text{with}\quad
V^{\prime}V=U^{\prime}U=I_{r},$

where $\Lambda_{k}$ is an $r\times r$ diagonal matrix with the positive
eigenvalues of $\tilde{\Xi}_{kk}$ along the diagonal. For this case, we define
the auxiliary full-rank matrix,

$\check{\Xi}_{kk}=\tilde{\Xi}_{kk}+V_{\bot}U_{\bot}^{\prime},$

where $V_{\bot}\in\mathbb{R}^{n_{k}\times(n_{k}-r)}$ is orthogonal to $V$,
$V_{\bot}^{\prime}V=0$ and normalized, $V_{\bot}^{\prime}V_{\bot}=I_{n_{k}-r}$
, and $U_{\bot}$ is defined similarly. We now set

$\displaystyle P_{kk}$ $\displaystyle=$
$\displaystyle\check{\Xi}_{kk}^{\prime}(\check{\Xi}_{kk}\check{\Xi}_{kk}^{\prime})^{-\frac{1}{2}},$
$\displaystyle=$
$\displaystyle(V\Lambda_{k}U^{\prime}+V_{\bot}U_{\bot}^{\prime})^{\prime}[(V\Lambda_{k}U^{\prime}+V_{\bot}U_{\bot}^{\prime})(V\Lambda_{k}U^{\prime}+V_{\bot}U_{\bot}^{\prime})^{\prime}]^{-1/2}$
$\displaystyle=$
$\displaystyle(V\Lambda_{k}U^{\prime}+V_{\bot}U_{\bot}^{\prime})^{\prime}[(V\Lambda_{k}U^{\prime}U\Lambda_{k}V^{\prime}+V_{\bot}V_{\bot}^{\prime}]^{-1/2}$
$\displaystyle=$
$\displaystyle(U\Lambda_{k}V^{\prime}+U_{\bot}V_{\bot}^{\prime})(V\Lambda_{k}^{-1}V^{\prime}+V_{\bot}V_{\bot}^{\prime})$
$\displaystyle=$ $\displaystyle UV^{\prime}+U_{\bot}V_{\bot}^{\prime},$

and find that

$\Xi_{kk}=\tilde{\Xi}_{kk}P_{kk}=V\Lambda_{k}U^{\prime}(UV^{\prime}+U_{\bot}V_{\bot}^{\prime})=V\Lambda_{k}V^{\prime}=V\Lambda_{k}^{1/2}U^{\prime}U\Lambda_{k}^{1/2}V^{\prime}=(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{\frac{1}{2}},$

is symmetric and positive semidefinite with the same rank as
$\tilde{\Xi}_{kk}$. This proves the existence of a $\Xi$ matrix with symmetric
positive semidefinite diagonal blocks, $\Xi_{11},\ldots,\Xi_{KK}$. If
$\Xi_{kk}$ is also invertible (implying it is symmetric and positive definite)
then it follows that $\Xi_{kk}$ is unique by Lemma A.2. To see this, consider
$\Xi P$ where $P$ has the structure in (A.1), its $k$-th diagonal block is
given by $\Xi_{kk}P_{kk}$ and for it to be symmetric and pd, it follows Lemma
A.2 that $P_{kk}=I_{n_{k}}$. Thus, if $\Xi_{11},\ldots,\Xi_{KK}$ are all
invertible, then $P$ must be $I_{n}$, which implies that $\Xi$ is unique.

$\square$

Proof of Theorem 5 (Score vectors). We have
$X_{k}=e_{k}^{\prime}X=e_{k}^{\prime}A\left(Y-\mu\right),$where $X_{k}\sim
t_{\nu_{k}}\left(0,I_{n_{k}}\right)$, $k=1,\ldots,K$ are independent, and the
log-likelihood function is therefore given by

$\displaystyle\ell\left(Y\right)$
$\displaystyle=-\tfrac{1}{2}\log|\tilde{\Xi}\tilde{\Xi}^{\prime}|+\sum_{k=1}^{K}c_{k}-\tfrac{\nu_{k}+n_{k}}{2}\log\left(1+\tfrac{1}{\nu_{k}}X_{k}^{\prime}X_{k}\right),$

where
$c_{k}=\log\Gamma\left(\frac{\nu_{k}+n_{k}}{2}\right)-\log\Gamma\left(\frac{\nu_{k}}{2}\right)-\frac{n_{k}}{2}\log\left(\nu_{k}\pi\right)$.
We find that

$\displaystyle\frac{\partial\left(X_{k}^{\prime}X_{k}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}$
$\displaystyle=\frac{\partial\left(X_{k}^{\prime}X_{k}\right)}{\partial
X_{k}^{\prime}}\frac{\partial{\rm
vec}(e_{k}^{\prime}A\left(Y-\mu\right))}{\partial{\rm
vec}(A)^{\prime}}\frac{\partial{\rm vec}(A)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}$
$\displaystyle=-2X_{k}^{\prime}\left(\left(Y-\mu\right)^{\prime}\otimes
e_{k}^{\prime}\right)\left(A^{\prime}\otimes A\right)$
$\displaystyle=-2X_{k}^{\prime}\left(X^{\prime}\otimes e_{k}^{\prime}A\right)$
$\displaystyle=-2{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)^{\prime},$

and using the formulae

$\frac{\partial{\rm
vec}\left(\tilde{\Xi}\tilde{\Xi}^{\prime}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}=\left(I_{n^{2}}+K_{n}\right)\left(\Xi^{\prime}\otimes
I_{n}\right),\quad\text{and}\quad\frac{\partial\log|M|}{\partial{\rm
vec}(M)^{\prime}}={\rm vec}\left(M^{-1}\right)^{\prime},$

for any symmetric matrix $M$ with positive determinant, we obtain

$\displaystyle-\tfrac{1}{2}\frac{\partial\log|\tilde{\Xi}\tilde{\Xi}^{\prime}|}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}$ $\displaystyle=-\tfrac{1}{2}{\rm
vec}\left(AA^{\prime}\right)^{\prime}\left(I_{n^{2}}+K_{n}\right)\left(\tilde{\Xi}^{\prime}\otimes
I_{n}\right)=-{\rm vec}\left(A^{\prime}\right)^{\prime}.$

Define
$W_{k}=\left(v_{k}+n_{k}\right)/\left(v_{k}+X_{k}^{\prime}X_{k}\right)$, then
we have the form of the score given by

$\displaystyle\nabla_{\tilde{\Xi}}^{\prime}=\frac{\partial\ell}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}$ $\displaystyle=\sum_{k=1}^{K}W_{k}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)^{\prime}-{\rm
vec}\left(A^{\prime}\right)^{\prime}.$

The expressions for $\nabla_{\nu_{k}}=\frac{\partial\ell}{\partial\nu_{k}}$
and $\nabla_{\mu}^{\prime}=\frac{\partial\ell}{\partial\mu^{\prime}}$ follows
directly.

$\square$

Proof of Theorem 5 (Hessian matrix). First we note that

$\frac{\partial
X_{k}}{\partial\mu^{\prime}}=-e_{k}^{\prime}A,\quad\frac{\partial
X_{k}}{\partial{\rm vec}(\tilde{\Xi})^{\prime}}=-X^{\prime}\otimes
e_{k}^{\prime}A,\quad\frac{\partial X_{k}}{\partial\nu_{l}}=0,$

and

$\frac{\partial W_{k}}{\partial X_{k}^{\prime}}=\frac{\partial
W_{k}}{\partial\left(X_{k}^{\prime}X_{k}\right)}\frac{\partial\left(X_{k}^{\prime}X_{k}\right)}{\partial
X_{k}^{\prime}}=-\tfrac{\nu_{k}+n_{k}}{\left(\nu_{k}+Z_{k}^{\prime}Z_{k}\right)^{2}}2X_{k}^{\prime}=-\tfrac{2}{\nu_{k}+n_{k}}W_{k}^{2}X_{k}^{\prime},$

such that

$\displaystyle\frac{\partial W_{k}}{\partial\mu^{\prime}}$
$\displaystyle=\tfrac{2}{\nu_{k}+n_{k}}W_{k}^{2}X_{k}^{\prime}e_{k}^{\prime}A,$
$\displaystyle\frac{\partial W_{k}}{\partial{\rm vec}(\tilde{\Xi})^{\prime}}$
$\displaystyle=\tfrac{2}{\nu_{k}+n_{k}}W_{k}^{2}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)^{\prime},$
$\displaystyle\frac{\partial W_{k}}{\partial\nu_{k}}$
$\displaystyle=\tfrac{X_{k}^{\prime}X_{k}-n_{k}}{\left(v_{k}+X_{k}^{\prime}X_{k}\right)^{2}}=\tfrac{1}{\nu_{k}+n_{k}}\left(W_{k}-W_{k}^{2}\right).$

We proceed to derive the expressions for the three terms,
$\nabla_{\mu\mu^{\prime}}$, $\nabla_{\mu\Xi^{\prime}}$, and
$\nabla_{\mu\nu_{k}}$, derived from
$\nabla_{\mu}=\sum_{k=1}^{K}W_{k}A^{\prime}e_{k}X_{k}$. The first is given by

$\displaystyle\nabla_{\mu\mu^{\prime}}$
$\displaystyle=\sum_{k=1}^{K}A^{\prime}e_{k}X_{k}\frac{\partial
W_{k}}{\partial\mu^{\prime}}+W_{k}A^{\prime}e_{k}\frac{\partial
Z_{k}}{\partial\mu^{\prime}}$
$\displaystyle=\sum_{k=1}^{K}\tfrac{2W_{k}^{2}}{\nu_{k}+n_{k}}A^{\prime}e_{k}X_{k}X_{k}^{\prime}e_{k}^{\prime}A-W_{k}A^{\prime}J_{k}A.$

For the next term we make use of an alternative expression for $\nabla_{\mu}$
given by:

$\nabla_{\mu}=\sum_{k=1}^{K}W_{k}A^{\prime}e_{k}X_{k}=\sum_{k=1}^{K}W_{k}\left(X_{k}^{\prime}e_{k}^{\prime}\otimes
I_{n}\right){\rm vec}\left(A^{\prime}\right),$

and find that

$\displaystyle\nabla_{\mu\tilde{\Xi}^{\prime}}$
$\displaystyle=\sum_{k=1}^{K}W_{k}\left(X_{k}^{\prime}e_{k}^{\prime}\otimes
I_{n}\right)\frac{\partial{\rm vec}\left(A^{\prime}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}+A^{\prime}e_{k}X_{k}\frac{\partial
W_{k}}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}+W_{k}A^{\prime}e_{k}\frac{\partial
X_{k}}{\partial{\rm vec}(\tilde{\Xi})^{\prime}}$
$\displaystyle=\sum_{k=1}^{K}W_{k}\left(X_{k}^{\prime}e_{k}^{\prime}\otimes
I_{n}\right)K_{n}\left(A^{\prime}\otimes
A\right)+\tfrac{2W_{k}^{2}}{\nu_{k}+n_{k}}A^{\prime}e_{k}X_{k}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)^{\prime}-W_{k}A^{\prime}e_{k}\left(X^{\prime}\otimes
e_{k}^{\prime}A\right)$
$\displaystyle=\sum_{k=1}^{K}W_{k}\left(AX_{k}^{\prime}e_{k}^{\prime}\otimes
A^{\prime}\right)K_{n}+\tfrac{2W_{k}^{2}}{\nu_{k}+n_{k}}A^{\prime}e_{k}X_{k}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)^{\prime}-W_{k}A^{\prime}\left(X^{\prime}\otimes
J_{k}A\right).$

For the third term we have

$\nabla_{\mu\nu_{k}}=\sum_{k=1}^{K}A^{\prime}e_{k}X_{k}\frac{\partial
W_{k}}{\partial\nu_{k}}=\sum_{k=1}^{K}\tfrac{1}{\nu_{k}+n_{k}}A^{\prime}e_{k}X_{k}\left(W_{k}-W_{k}^{2}\right).$

Next we obtain the expressions for $\nabla_{\tilde{\Xi}\nu_{k}}$ and
$\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}$, starting from
$\nabla_{\tilde{\Xi}}=\sum_{k=1}^{K}W_{k}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)-{\rm
vec}\left(A^{\prime}\right)$. The first expression is simply

$\nabla_{\tilde{\Xi}\nu_{k}}=\tfrac{1}{\nu_{k}+n_{k}}\left(W_{k}-W_{k}^{2}\right){\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right),$

and the second takes the form,

$\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}=\sum_{k=1}^{K}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)\frac{\partial
W_{k}}{\partial{\rm vec}(\tilde{\Xi})^{\prime}}+W_{k}\frac{\partial{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}-\frac{\partial{\rm
vec}\left(A^{\prime}\right)}{\partial{\rm vec}(\tilde{\Xi})^{\prime}}$

where

$\frac{\partial{\rm vec}\left(A^{\prime}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}=K_{n}\left(A^{\prime}\otimes
A\right)=\left(A\otimes A^{\prime}\right)K_{n}=\left(I_{n}\otimes
A^{\prime}\right)K_{n}\left(I_{n}\otimes A\right),$ $\displaystyle{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)\frac{\partial
W_{k}}{\partial{\rm vec}(\tilde{\Xi})^{\prime}}$
$\displaystyle=\tfrac{2W_{k}^{2}}{\nu_{k}+n_{k}}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right){\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)^{\prime},$

and

$\frac{\partial{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}=\left(XX_{k}^{\prime}\otimes
I_{n}\right)\frac{\partial{\rm vec}\left(A^{\prime}e_{k}\right)}{\partial{\rm
vec}\left(\Xi\right)^{\prime}}+\left(I_{n}\otimes
A^{\prime}e_{k}\right)\frac{\partial{\rm
vec}\left(X_{k}X^{\prime}\right)}{\partial{\rm
vec}\left(\Xi\right)^{\prime}}.$ (A.2)

Next,

$\frac{\partial{\rm vec}\left(A^{\prime}e_{k}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}=\left(e_{k}^{\prime}\otimes
I_{n}\right)\frac{\partial{\rm vec}\left(A^{\prime}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}=\left(e_{k}^{\prime}\otimes
I_{n}\right)\left(A\otimes A^{\prime}\right)K_{n}=\left(e_{k}^{\prime}A\otimes
A^{\prime}\right)K_{n},$

shows that the first term in (A.2) can be expressed as

$\left(XX_{k}^{\prime}\otimes I_{n}\right)\frac{\partial{\rm
vec}\left(A^{\prime}e_{k}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}=\left(XX_{k}^{\prime}\otimes
I_{n}\right)\left(e_{k}^{\prime}A\otimes
A^{\prime}\right)K_{n}=\left(XX_{k}^{\prime}e_{k}^{\prime}A\otimes
A^{\prime}\right)K_{n},$

and for the second term in (A.2) we have

$\frac{\partial{\rm vec}\left(X_{k}X^{\prime}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}=\left(X\otimes I_{n_{k}}\right)\frac{\partial{\rm
vec}\left(X_{k}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}+\left(I_{n}\otimes X_{k}\right)\frac{\partial{\rm
vec}\left(X^{\prime}\right)}{\partial{\rm vec}(\tilde{\Xi})^{\prime}},$

where

$\frac{\partial{\rm vec}\left(X_{k}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}=-X^{\prime}\otimes e_{k}^{\prime}A,$

and

$\displaystyle\frac{\partial{\rm vec}\left(X^{\prime}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}$ $\displaystyle=\frac{\partial{\rm
vec}\left(\sum_{k=1}^{K}e_{k}X_{k}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}=\sum_{k=1}^{K}\frac{\partial{\rm
vec}\left(e_{k}X_{k}\right)}{\partial{\rm vec}(\tilde{\Xi})^{\prime}}$
$\displaystyle=\sum_{k=1}^{K}\frac{\partial\left(e_{k}X_{k}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}=\sum_{k=1}^{K}e_{k}\frac{\partial{\rm
vec}\left(X_{k}\right)}{\partial{\rm vec}(\tilde{\Xi})^{\prime}}$
$\displaystyle=-\sum_{k=1}^{K}e_{k}\left(X^{\prime}\otimes
e_{k}^{\prime}A\right)=-\sum_{k=1}^{K}\left(X^{\prime}\otimes J_{k}A\right)$
$\displaystyle=-\left(X^{\prime}\otimes A\right).$

So we have

$\displaystyle\frac{\partial{\rm
vec}\left(X_{k}X^{\prime}\right)}{\partial{\rm vec}(\tilde{\Xi})^{\prime}}$
$\displaystyle=-\left(X\otimes I_{n_{k}}\right)\left(X^{\prime}\otimes
e_{k}^{\prime}A\right)-\left(I_{n}\otimes X_{k}\right)\left(X^{\prime}\otimes
A\right)$ $\displaystyle=-\left(XX^{\prime}\otimes
e_{k}^{\prime}A\right)-\left(I_{n}\otimes X_{k}\right)\left(A\otimes
X^{\prime}\right)K_{n}$ $\displaystyle=-\left(XX^{\prime}\otimes
e_{k}^{\prime}A\right)-\left(A\otimes X_{k}X^{\prime}\right)K_{n}$

where we use that $\left(Y\otimes x^{\prime}\right)K_{sm}=x^{\prime}\otimes Y$
when $x$ is a vector.

Combined, we have

$\displaystyle\frac{\partial{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)}{\partial{\rm
vec}(\tilde{\Xi})^{\prime}}$
$\displaystyle=\left(XX_{k}^{\prime}e_{k}^{\prime}A\otimes
A^{\prime}\right)K_{n}-\left(I_{n}\otimes
A^{\prime}e_{k}\right)\left(\left(XX^{\prime}\otimes
e_{k}^{\prime}A\right)+\left(A\otimes X_{k}X^{\prime}\right)K_{n}\right)$
$\displaystyle=\left(XX_{k}^{\prime}e_{k}^{\prime}A\otimes
A^{\prime}\right)K_{n}-\left(XX^{\prime}\otimes
A^{\prime}J_{k}A\right)-\left(A\otimes
A^{\prime}e_{k}X_{k}X^{\prime}\right)K_{n},$

such that

$\displaystyle\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}$
$\displaystyle=-\left(A\otimes
A^{\prime}\right)K_{n}+\sum_{k=1}^{K}\tfrac{2W_{k}^{2}}{\nu_{k}+n_{k}}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right){\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)^{\prime}$
$\displaystyle\quad+\sum_{k=1}^{K}W_{k}\left[\left(XX_{k}^{\prime}e_{k}^{\prime}A\otimes
A^{\prime}\right)K_{n}-\left(XX^{\prime}\otimes
A^{\prime}J_{k}A\right)-\left(A\otimes
A^{\prime}e_{k}X_{k}X^{\prime}\right)K_{n}\right].$

Finally, we derive $\nabla_{\nu\nu^{\prime}}$ starting from

$\nabla_{\nu_{k}}=\tfrac{1}{2}\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)+1-W_{k}-\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)\right].$

For $k\neq l$ we have $\nabla_{\nu_{k}\nu_{l}}=0$, and for $k=l$ we have

$\displaystyle\nabla_{\nu_{k}\nu_{k}}$
$\displaystyle=\tfrac{1}{2}\left[\tfrac{1}{2}\psi^{\prime}\left(\tfrac{v_{k}+n_{k}}{2}\right)-\tfrac{1}{2}\psi^{\prime}\left(\tfrac{v_{k}}{2}\right)+\tfrac{1}{\nu_{k}+n_{k}}\left(W_{k}^{2}-W_{k}\right)+\tfrac{1}{\nu_{k}}\left[1-\tfrac{\nu_{k}}{\nu_{k}+X_{k}^{\prime}X_{k}}\right]\right]$
$\displaystyle=\tfrac{1}{4}\psi^{\prime}\left(\tfrac{v_{k}+n_{k}}{2}\right)-\tfrac{1}{4}\psi^{\prime}\left(\tfrac{v_{k}}{2}\right)+\tfrac{1}{2\nu_{k}}+\tfrac{1}{2}\tfrac{1}{\nu_{k}+n_{k}}W_{k}^{2}-\tfrac{1}{\left(\nu_{k}+n_{k}\right)}W_{k}.$

$\square$

Next, we derive some expectations for quantities that involve a multivariate
$t$-distribution.

######  Lemma A.3.

Suppose that $X\sim t_{\nu}(0,I_{n})$ and let
$W=\frac{\nu+n}{\nu+X^{\prime}X}$ and

$\zeta_{p,q}=\left(\tfrac{\nu+n}{\nu}\right)^{\frac{p}{2}}\left(\tfrac{\nu}{2}\right)^{\frac{q}{2}}\frac{\Gamma(\tfrac{\nu+n}{2})}{\Gamma(\tfrac{\nu}{2})}\frac{\Gamma(\tfrac{\nu+p-q}{2})}{\Gamma(\tfrac{\nu+p+n}{2})}\in\mathbb{R}.$

(i) For any integrable function $g$ and any $p>-\nu$, it holds that

$\mathbb{E}\left[W^{\frac{p}{2}}g(X)\right]=\zeta_{p,0}\mathbb{E}\left[g(Y)\right],\quad
Y\sim t_{\nu+p}\left(0,\tfrac{v}{v+p}I_{n}\right).$

(ii) Moreover, if $g$ is homogeneous of degree $q<\nu+p$, then

$\mathbb{E}\left[W^{\frac{p}{2}}g(X)\right]=\zeta_{p,q}\mathbb{E}\left[g(Z)\right],\quad
Z\sim N\left(0,I_{n}\right).$

(iii) The following expectations hold

$\displaystyle\mathbb{E}\left[W\log\left(1+\tfrac{X^{\prime}X}{\nu}\right)\right]$
$\displaystyle=\psi\left(\tfrac{\nu+n}{2}+1\right)-\psi\left(\tfrac{\nu}{2}+1\right),$
$\displaystyle\mathbb{E}\left[W\log\left(1+\tfrac{Q}{\nu}\right){\rm
vec}\left(XX^{\prime}\right)\right]$
$\displaystyle=\left[\psi\left(\tfrac{\nu+n}{2}+1\right)-\psi\left(\tfrac{\nu}{2}\right)\right]{\rm
vec}\left(I_{n}\right),$
$\displaystyle\mathbb{E}\left[\log\left(1+\tfrac{X^{\prime}X}{\nu}\right)\right]$
$\displaystyle=\psi\left(\tfrac{\nu+n}{2}\right)-\psi\left(\tfrac{\nu}{2}\right),$
$\displaystyle\mathbb{E}\left[\log^{2}\left(1+\tfrac{X^{\prime}X}{\nu}\right)\right]$
$\displaystyle=\psi^{\prime}\left(\tfrac{\nu}{2}\right)-\psi^{\prime}\left(\tfrac{\nu+n}{2}\right)+\left[\psi\left(\tfrac{\nu+n}{2}\right)-\psi\left(\tfrac{\nu}{2}\right)\right]^{2},$

where $\psi\left(\cdot\right)$ and $\psi^{\prime}\left(\cdot\right)$ are the
digamma and trigamma functions, respectively.

Note that $p$ in part (iii) may be negative, since $-\nu<0$. If $p/2$ is a
positive integer, then

$\displaystyle\zeta_{p,q}$ $\displaystyle=$
$\displaystyle\left(\tfrac{\nu+n}{\nu}\right)^{\frac{p}{2}}\left(\tfrac{\nu}{2}\right)^{\frac{q}{2}}\prod_{k=0}^{\tfrac{p}{2}-1}\frac{\nu+q+2k}{\nu+n+2k}.$

where we used $\Gamma(x+1)=x\Gamma(x)$, repeatedly, and the following
simplifications follow:

$\begin{array}[]{rclcrcl}\zeta_{0,2}&=&\tfrac{\nu}{\nu-2},&&\zeta_{0,4}&=&\frac{\nu^{2}}{\left(\nu-2\right)\left(\nu-4\right)},\\\
\zeta_{2,0}&=&1,&&\zeta_{2,2}&=&1,\\\
\zeta_{2,1}&=&\left(\tfrac{\nu+1}{\nu}\right)\left(\tfrac{\nu}{2}\right)^{\frac{1}{2}},&&\zeta_{4,2}&=&\frac{\nu+n}{\nu+n+2},\\\
\zeta_{4,0}&=&\frac{(\nu+2)(\nu+n)}{\nu(\nu+n+2)},&&\zeta_{4,4}&=&\frac{\nu+n}{\nu+n+2}.\end{array}$

Proof of Lemma A.3. Let
$\kappa_{\nu,n}=\Gamma(\tfrac{\nu+n}{2})/\Gamma(\tfrac{\nu}{2})$, and the
density for $X\sim t_{\nu}(0,I_{n})$ is

$f_{x}(x)=\kappa_{\nu,n}[\nu\pi]^{-\frac{n}{2}}\left(1+\tfrac{x^{\prime}x}{\nu}\right)^{-\frac{\nu+n}{2}},$

whereas the density for $Y\sim t_{\nu+p}(0,\frac{v}{v+p}I_{n})$ is

$\displaystyle f_{y}(y)$ $\displaystyle=$
$\displaystyle\kappa_{\nu+p,n}[(\nu+p)\pi]^{-\frac{n}{2}}\left(\tfrac{\nu}{\nu+p}\right)^{-\frac{n}{2}}\left(1+\tfrac{1}{\nu+p}x^{\prime}\left[\tfrac{\nu}{\nu+p}I_{n}\right]^{-1}x\right)^{-\frac{\nu+p+n}{2}}$
$\displaystyle=$
$\displaystyle\kappa_{\nu+p,n}[\nu\pi]^{-\frac{n}{2}}\left(1+\tfrac{x^{\prime}x}{\nu}\right)^{-\frac{\nu+p+n}{2}}.$

The expected value we seek is given by

$\displaystyle\mathbb{E}\left[W^{\frac{p}{2}}g(X)\right]$ $\displaystyle=$
$\displaystyle\int\left(\tfrac{\nu+n}{\nu+x^{\prime}x}\right)^{\frac{p}{2}}g(x)\kappa_{\nu,n}[\nu\pi]^{-\frac{n}{2}}\left(1+\tfrac{x^{\prime}x}{\nu}\right)^{-\frac{\nu+n}{2}}\mathrm{d}x$
$\displaystyle=$
$\displaystyle\left(\tfrac{\nu+n}{\nu}\right)^{\frac{p}{2}}\int
g(x)\kappa_{\nu,n}[\nu\pi]^{-\frac{n}{2}}\left(1+\tfrac{x^{\prime}x}{\nu}\right)^{-\frac{\nu+p+n}{2}}\mathrm{d}x$
$\displaystyle=$
$\displaystyle\left(\tfrac{\nu+n}{\nu}\right)^{\frac{p}{2}}\frac{\kappa_{\nu,n}}{\kappa_{\nu+p,n}}\int
g(x)f_{y}(x)\mathrm{d}x,$ $\displaystyle=$
$\displaystyle\zeta_{p,0}\mathbb{E}\left[g(Y)\right],\quad Y\sim
t_{\nu+p}(0,\tfrac{v}{v+p}I_{n}),$

and the results for part (i) follows, since

$\zeta_{p,0}=\left(\tfrac{\nu+n}{\nu}\right)^{\frac{p}{2}}\frac{\Gamma(\tfrac{\nu+n}{2})/\Gamma(\tfrac{\nu}{2})}{\Gamma(\tfrac{\nu+p+n}{2})/\Gamma(\tfrac{\nu+p}{2})}=\left(\tfrac{\nu+n}{\nu}\right)^{\frac{p}{2}}\frac{\kappa_{\nu,n}}{\kappa_{\nu+p,n}}.$

This completes the proof of (i).

To prove (ii) we use that $Y\sim t_{\nu+p}\left(0,\frac{v}{v+p}I_{n}\right)$
can be expressed as $Y=Z/\sqrt{\xi/\nu}$ where $Z\sim N(0,I_{n})$ and $\xi$ is
an independent $\chi^{2}$-distributed random variable with $\nu+p$ degrees of
freedom. Hence, $Y=\psi Z$, with $\psi=1/\sqrt{\xi/\nu}$, such that

$\psi^{q}=\left(\tfrac{\nu}{\xi}\right)^{\frac{q}{2}}.$

Now using part (i) and that $g$ is homogeneous, we find

$\displaystyle\mathbb{E}\left[W^{\frac{p}{2}}g(X)\right]$ $\displaystyle=$
$\displaystyle\zeta_{p,0}\mathbb{E}\left[g(Y)\right]=\zeta_{p,0}\mathbb{E}\left[\psi^{q}g(Z)\right]$
$\displaystyle=$
$\displaystyle\zeta_{p,0}\nu^{\frac{q}{2}}\mathbb{E}[\xi^{-\frac{q}{2}}]\mathbb{E}\left[g(Z)\right],$
$\displaystyle=$
$\displaystyle\zeta_{p,0}\nu^{\frac{q}{2}}\frac{\Gamma(\tfrac{\nu+p-q}{2})}{\Gamma(\tfrac{\nu+p}{2})}\left(\tfrac{1}{2}\right)^{\frac{q}{2}}\mathbb{E}\left[g(Z)\right]$
$\displaystyle=$
$\displaystyle\left(\tfrac{\nu+n}{\nu}\right)^{\frac{p}{2}}\frac{\Gamma(\tfrac{\nu+n}{2})/\Gamma(\tfrac{\nu}{2})}{\Gamma(\tfrac{\nu+p+n}{2})/\Gamma(\tfrac{\nu+p}{2})}\nu^{\frac{q}{2}}\frac{\Gamma(\tfrac{\nu+p-q}{2})}{\Gamma(\tfrac{\nu+p}{2})}\left(\tfrac{1}{2}\right)^{\frac{q}{2}}\mathbb{E}\left[g(Z)\right]$
$\displaystyle=$ $\displaystyle\zeta_{p,q}\mathbb{E}\left[g(Z)\right],$

where we used that $\xi\sim\chi_{\nu+p}^{2}$ and $Z$ are independent, and that

$\mathbb{E}\left(\xi^{-\frac{q}{2}}\right)=\frac{\Gamma(\tfrac{\nu+p-q}{2})}{\Gamma(\tfrac{\nu+p}{2})}\left(\tfrac{1}{2}\right)^{\frac{q}{2}},\qquad\text{for}\quad
q<\nu+p.$

This completes the proof of (ii).

For the first expression in (iii), we find that for $X\sim
t_{\nu}\left(0,I_{n}\right)$, we have $X^{\prime}X/n\sim F_{n,v}$, and
$S=(1+X^{\prime}X/\nu)^{-1}\sim{\rm
Beta}\left(\tfrac{\nu}{2},\tfrac{n}{2}\right)$. So we have

$\displaystyle\mathbb{E}\left(W\log\left(1+\tfrac{X^{\prime}X}{\nu}\right)\right)$
$\displaystyle=\tfrac{\nu+n}{\nu}\mathbb{E}\left(S\log\left(S\right)\right)=\tfrac{\nu+n}{\nu}\frac{1}{\mathrm{B}(\tfrac{\nu}{2},\tfrac{n}{2})}\int_{0}^{1}\log\left(x\right)x^{\frac{\nu}{2}}(1-x)^{\frac{n}{2}-1}\mathrm{d}x.$

By using the following integral from Gradshteyn and Ryzhik, (2007, p. 540),

$\int_{0}^{1}x^{\mu-1}\left(1-x^{r}\right)^{s-1}\log x\ {\rm
d}x=\tfrac{1}{r^{2}}\mathrm{B}\left(\tfrac{\mu}{r},s\right)\left[\psi\left(\tfrac{\mu}{r}\right)-\psi\left(\tfrac{\mu}{r}+s\right)\right],$
(A.3)

where $\psi\left(\cdot\right)$ is the digamma function, and let
$r=1,s=\tfrac{n}{2},\mu=\frac{\nu}{2}+1$, we have

$\mathbb{E}\left(W\log\left(1+\tfrac{X^{\prime}X}{\nu}\right)\right)=\tfrac{\nu+n}{\nu}\tfrac{\mathrm{B}\left(\frac{\nu}{2}+1,\frac{n}{2}\right)}{\mathrm{B}(\tfrac{\nu}{2},\tfrac{n}{2})}\left[\psi\left(\tfrac{\nu}{2}+1\right)-\psi\left(\tfrac{\nu+n}{2}+1\right)\right]=\psi\left(\tfrac{\nu}{2}+1\right)-\psi\left(\tfrac{\nu+n}{2}+1\right).$

For the first expression in (iii), it is well known that for an elliptic
distribution $X\sim EL\left(0,I_{n}\right)$, the two random variables,
$Q\equiv X^{\prime}X$ and $U\equiv Q^{-\frac{1}{2}}X$, are independent, and
$U$ has a uniform distribution on a unit sphere, see e.g. Mitchell, (1989,
section 2). This result applies to the multivariate $t$-distribution, $X\sim
t_{\nu}\left(0,I_{n}\right)$, such that

$\displaystyle\mathbb{E}\left[W\log\left(1+\tfrac{Q}{\nu}\right){\rm
vec}\left(XX^{\prime}\right)\right]$
$\displaystyle=\mathbb{E}\left[QW\log\left(1+\tfrac{Q}{\nu}\right){\rm
vec}\left(XX^{\prime}/Q\right)\right]$
$\displaystyle=\mathbb{E}\left[QW\log\left(1+\tfrac{Q}{\nu}\right){\rm
vec}\left(UU^{\prime}\right)\right]$
$\displaystyle=\mathbb{E}\left[QW\log\left(1+\tfrac{Q}{\nu}\right)\right]\mathbb{E}\left[{\rm
vec}\left(UU^{\prime}\right)\right]$
$\displaystyle=\tfrac{\nu+n}{n}\mathbb{E}\left[\tfrac{Q/v}{1+Q/v}\log\left(1+\tfrac{Q}{\nu}\right)\right]{\rm
vec}\left(I_{n}\right),$

where the last equality uses that $\mathbb{E}\left[{\rm
vec}\left(UU^{\prime}\right)\right]=\tfrac{1}{n}{\rm vec}\left(I_{n}\right)$.
Then we have

$\displaystyle\mathbb{E}\left[\tfrac{Q/v}{1+Q/v}\log\left(1+\tfrac{Q}{\nu}\right)\right]$
$\displaystyle=-\mathbb{E}\left[\left(1-S\right)\log\left(S\right)\right]$
$\displaystyle=\frac{-1}{\mathrm{B}(\tfrac{\nu}{2},\tfrac{n}{2})}\int_{0}^{1}\log\left(x\right)x^{\frac{\nu}{2}-1}(1-x)^{\frac{n}{2}}dx$
$\displaystyle=\frac{\mathrm{B}\left(\frac{\nu}{2},\tfrac{n}{2}+1\right)}{\mathrm{B}(\tfrac{\nu}{2},\tfrac{n}{2})}\left[\psi\left(\tfrac{\nu+n}{2}+1\right)-\psi\left(\tfrac{\nu}{2}\right)\right]$
$\displaystyle=\tfrac{n}{\nu+n}\left[\psi\left(\tfrac{\nu+n}{2}+1\right)-\psi\left(\tfrac{\nu}{2}\right)\right],$

where we let $r=1,s=\tfrac{n}{2}+1,\mu=\frac{\nu}{2}$ in the integral (A.3).
This leads to the second expression in (iii)

$\mathbb{E}\left[W\log\left(1+\tfrac{X^{\prime}X}{\nu}\right){\rm
vec}\left(XX^{\prime}\right)\right]=\left[\psi\left(\tfrac{\nu+n}{2}+1\right)-\psi\left(\tfrac{\nu}{2}\right)\right]{\rm
vec}\left(I_{n}\right).$

If we set $r=1,s=\tfrac{n}{2},\mu=\tfrac{\nu}{2}$ in (A.3) then we obtain the
third expression in (iii)

$\mathbb{E}\left[\log\left(1+\tfrac{Q}{\nu}\right)\right]=-\mathbb{E}\left[\log\left(S\right)\right]=\frac{-1}{\mathrm{B}(\tfrac{\nu}{2},\tfrac{n}{2})}\int_{0}^{1}\log\left(x\right)x^{\frac{\nu}{2}-1}(1-x)^{\frac{n}{2}-1}dx=\psi\left(\tfrac{\nu+n}{2}\right)-\psi\left(\tfrac{\nu}{2}\right).$

Finally, for the fourth expression in (iii), we use

$\int_{0}^{1}(\log)^{2}x^{p-1}\left(1-x^{r}\right)^{q-1}dx=\tfrac{\mathrm{B}\left(\tfrac{p}{r},q\right)}{r^{3}}\left\\{\psi^{\prime}\left(\tfrac{p}{r}\right)-\psi^{\prime}\left(\tfrac{p}{r}+q\right)+\left[\psi\left(\tfrac{p}{r}\right)-\psi\left(\tfrac{p}{r}+q\right)\right]^{2}\right\\},$

see Gradshteyn and Ryzhik, (2007, p. 543), and set
$p=\frac{\nu}{2},r=1,q=\tfrac{n}{2}$, whereby we obtain

$\displaystyle\mathbb{E}\left[\log^{2}\left(1+\tfrac{Q}{\nu}\right)\right]$
$\displaystyle=\mathbb{E}\left[\log^{2}\left(S\right)\right]$
$\displaystyle=\frac{1}{\mathrm{B}(\tfrac{\nu}{2},\tfrac{n}{2})}\int_{0}^{1}\log^{2}\left(x\right)x^{\frac{\nu}{2}-1}(1-x)^{\frac{n}{2}-1}dx$
$\displaystyle=\frac{1}{\mathrm{B}(\tfrac{\nu}{2},\tfrac{n}{2})}\mathrm{B}\left(\tfrac{\nu}{2},\tfrac{n}{2}\right)\left\\{\psi^{\prime}\left(\tfrac{\nu}{2}\right)-\psi^{\prime}\left(\tfrac{\nu+n}{2}\right)+\left[\psi\left(\tfrac{\nu}{2}\right)-\psi\left(\tfrac{\nu+n}{2}\right)\right]^{2}\right\\}$
$\displaystyle=\psi^{\prime}\left(\tfrac{\nu}{2}\right)-\psi^{\prime}\left(\tfrac{\nu+n}{2}\right)+\left[\psi\left(\tfrac{\nu+n}{2}\right)-\psi\left(\tfrac{\nu}{2}\right)\right]^{2}.$

This completes the proof.

$\square$

Proof of Theorem 6 (Expected Scores). We have
$\mathbb{E}\left(\nabla_{\mu}\right)=\sum_{k=1}^{K}A^{\prime}e_{k}\mathbb{E}\left(W_{k}X_{k}\right)=0$,

$\displaystyle\mathbb{E}\left(\nabla_{\nu_{k}}\right)$
$\displaystyle=\tfrac{1}{2}\mathbb{E}\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)+1-W_{k}-\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)\right]$
$\displaystyle=\tfrac{1}{2}\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)+1-1-\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)\right]\right]=0.$

Next, observe that and
$\mathbb{E}\left(W_{k}X_{k}X_{k}^{\prime}\right)=I_{n_{k}}$ and for $k\neq l$
we have $\mathbb{E}\left(W_{k}X_{k}X_{l}^{\prime}\right)=0$, which we used
$\nu_{l}>1$. We therefore have
$\mathbb{E}\left(W_{k}X_{k}X^{\prime}\right)=e_{k}^{\prime}$ and it follows
that

$\displaystyle\mathbb{E}\left(\nabla_{\tilde{\Xi}}\right)$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{K}{\rm
vec}\left(A^{\prime}e_{k}\mathbb{E}\left(W_{k}X_{k}X^{\prime}\right)\right)-{\rm
vec}\left(A^{\prime}\right)$ $\displaystyle=$ $\displaystyle\sum_{k=1}^{K}{\rm
vec}\left(A^{\prime}e_{k}e_{k}^{\prime}\right)-{\rm
vec}\left(A^{\prime}\right)=0,$

since $\sum_{k=1}^{K}e_{k}e_{k}^{\prime}=I_{n}$.

$\square$

Proof of Theorem 6 (Information matrix). We have

$\displaystyle\mathbb{E}\left(\nabla_{\mu}\nabla_{\mu}^{\prime}\right)$
$\displaystyle=\sum_{k=1}^{K}\sum_{l=1}^{K}A^{\prime}e_{l}\mathbb{E}\left(W_{l}W_{k}X_{l}X_{k}^{\prime}\right)e_{k}^{\prime}A$
$\displaystyle=\sum_{k=1}^{K}A^{\prime}e_{k}\mathbb{E}\left(W_{k}^{2}X_{k}X_{k}^{\prime}\right)e_{k}^{\prime}A+\sum_{k=1}^{K}\sum_{l\neq
k}^{K}A^{\prime}e_{l}\mathbb{E}\left(W_{l}X_{l}\right)\mathbb{E}\left(W_{k}X_{k}^{\prime}\right)e_{k}^{\prime}A$
$\displaystyle=\sum_{k=1}^{K}A^{\prime}e_{k}\mathbb{E}\left(W_{k}^{2}X_{k}X_{k}^{\prime}\right)e_{k}^{\prime}A$
$\displaystyle=\sum_{k=1}^{K}\phi_{k}A^{\prime}e_{k}e_{k}^{\prime}A=\sum_{k=1}^{K}\phi_{k}A^{\prime}J_{k}A,$

where we used Lemma A.3
$\mathbb{E}\left(W_{k}^{2}X_{k}X_{k}^{\prime}\right)=\zeta_{4,2}I_{n_{k}}$,
where $\zeta_{4,2}=\frac{\nu_{k}+n_{k}}{\nu_{k}+n_{k}+2}=\phi_{k}$. Next we
have,

$\mathbb{E}\left(\nabla_{\mu}\nabla_{\tilde{\Xi}}^{\prime}\right)=\mathbb{E}\left(\nabla_{\mu}\nabla_{\nu}^{\prime}\right)=0,$

because $\nabla_{\tilde{\Xi}}^{\prime}$ and $\nabla_{\nu}^{\prime}$ are both
even functions of $X$, whereas $\nabla_{\mu}$ is an odd function.

##### The term:
$\mathbb{E}\left(\nabla_{\tilde{\Xi}}\nabla_{\tilde{\Xi}}^{\prime}\right)$

It is convenient to define $M=\left(I\otimes A^{\prime}\right)$ and
$\nabla_{\tilde{\Xi}}^{s}=\sum_{k=1}^{K}W_{k}{\rm
vec}\left(e_{k}X_{k}X^{\prime}\right)-{\rm vec}\left(I_{n}\right)$ (whose
distribution does not depend on $A$), such that

$\nabla_{\tilde{\Xi}}=\sum_{k=1}^{K}W_{k}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)-{\rm
vec}\left(A^{\prime}\right)=M\nabla_{\tilde{\Xi}}^{s}.$

It follows that
$\mathcal{I}_{\tilde{\Xi}}=M\mathbb{E}\left(\nabla_{\tilde{\Xi}}^{s}\nabla_{\tilde{\Xi}}^{s\prime}\right)M^{\prime}$,
where

$\displaystyle\mathbb{E}\left(\nabla_{\tilde{\Xi}}^{s}\nabla_{\tilde{\Xi}}^{s\prime}\right)$
$\displaystyle=\mathbb{E}\left[\sum_{k=1}^{K}\sum_{l=1}^{K}W_{k}W_{l}{\rm
vec}\left(e_{k}X_{k}X^{\prime}\right){\rm
vec}\left(e_{l}X_{l}X^{\prime}\right)^{\prime}-{\rm vec}\left(I_{n}\right){\rm
vec}\left(I_{n}\right)^{\prime}\right].$

Next we derive the expected value of

$\varphi\left(k,l\right)\equiv W_{k}W_{l}{\rm
vec}\left(e_{k}X_{k}X^{\prime}\right){\rm
vec}\left(e_{l}X_{l}X^{\prime}\right)^{\prime}.$

For $k=l$ we have

$\displaystyle\mathbb{E}\left[\varphi\left(k,k\right)\right]=$
$\displaystyle\mathbb{E}\left[W_{k}^{2}\sum_{p=1}^{K}{\rm
vec}\left(e_{k}X_{k}X_{p}^{\prime}e_{p}^{\prime}\right)\sum_{q=1}^{K}{\rm
vec}\left(e_{k}X_{k}X_{q}^{\prime}e_{q}^{\prime}\right)^{\prime}\right]$
$\displaystyle=$ $\displaystyle\mathbb{E}\left[W_{k}^{2}\sum_{p\neq k}{\rm
vec}\left(e_{k}X_{k}X_{p}^{\prime}e_{p}^{\prime}\right){\rm
vec}\left(e_{k}X_{k}X_{p}^{\prime}e_{p}^{\prime}\right)^{\prime}+W_{k}^{2}{\rm
vec}\left(e_{k}X_{k}X_{k}^{\prime}e_{k}^{\prime}\right){\rm
vec}\left(e_{k}X_{k}X_{k}^{\prime}e_{k}^{\prime}\right)^{\prime}\right].$

From Lemma A.3 we have the expression for $\phi_{k}=\zeta_{42}=\zeta_{44}$ and
$\zeta_{0,2}=\frac{\nu}{\nu-2}$, we have

$\displaystyle\sum_{p\neq k}\mathbb{E}\left[W_{k}^{2}{\rm
vec}\left(e_{k}X_{k}X_{p}^{\prime}e_{p}^{\prime}\right){\rm
vec}\left(e_{k}X_{k}X_{p}^{\prime}e_{p}^{\prime}\right)^{\prime}\right]$
$\displaystyle=$ $\displaystyle\sum_{p\neq k}\left(e_{p}\otimes
e_{k}\right)\mathbb{E}\left[W_{k}^{2}{\rm
vec}\left(X_{k}X_{p}^{\prime}\right){\rm
vec}\left(X_{k}X_{p}^{\prime}\right)^{\prime}\right]\left(e_{p}^{\prime}\otimes
e_{k}^{\prime}\right)$ $\displaystyle=$ $\displaystyle\phi_{k}\sum_{p\neq
k}\tfrac{\nu_{p}}{\nu_{p}-2}\left(e_{p}\otimes
e_{k}\right)\left(e_{p}^{\prime}\otimes e_{k}^{\prime}\right)$
$\displaystyle=$ $\displaystyle\phi_{k}\sum_{p\neq
k}\tfrac{\nu_{p}}{\nu_{p}-2}\left(J_{p}\otimes J_{k}\right),$

where we applied Lemma A.3 (ii) to the conditional expectation,
$\mathbb{E}[W_{k}^{2}g(X_{k})|X_{p}]=\zeta_{4,2}\mathbb{E}[g(Z)|X_{p}]$ with
$g(X_{k})={\rm vec}\left(X_{k}X_{p}^{\prime}\right){\rm
vec}\left(X_{k}X_{p}^{\prime}\right)^{\prime}$, using that $X_{k}$ and $X_{p}$
are independent, followed by taking unconditional expectation

$\mathbb{E}[g(Z)]=\mathbb{E}[{\rm vec}\left(ZX_{p}^{\prime}\right){\rm
vec}\left(ZX_{p}^{\prime}\right)^{\prime}]=\tfrac{\nu_{p}}{\nu_{p}-2}I_{n_{k}n_{p}}.$

Similarly, using Lemma A.3 and $\zeta_{4,4}=\phi_{k}$ we have

$\displaystyle\mathbb{E}\left[W_{k}^{2}{\rm
vec}\left(e_{k}X_{k}X_{k}^{\prime}e_{k}^{\prime}\right){\rm
vec}\left(e_{k}X_{k}X_{k}^{\prime}e_{k}^{\prime}\right)^{\prime}\right]$
$\displaystyle=$ $\displaystyle\left(e_{k}\otimes
e_{k}\right)\mathbb{E}\left[\omega_{k}^{2}{\rm
vec}\left(X_{k}X_{k}^{\prime}\right){\rm
vec}\left(X_{k}X_{k}^{\prime}\right)^{\prime}\right]\left(e_{k}^{\prime}\otimes
e_{k}^{\prime}\right)$ $\displaystyle=$
$\displaystyle\phi_{k}\left(e_{k}\otimes
e_{k}\right)\left[I_{n_{k}^{2}}+K_{n_{k}}+{\rm vec}(I_{n_{k}}){\rm
vec}(I_{n_{k}})^{\prime}\right]\left(e_{k}^{\prime}\otimes
e_{k}^{\prime}\right)$ $\displaystyle=$
$\displaystyle\phi_{k}\left[J_{k\otimes}\left(K_{n}+I_{n^{2}}\right)+{\rm
vec}(J_{k}){\rm vec}(J_{k})^{\prime}\right].$

Combining the results, we have

$\displaystyle\mathbb{E}\left[\varphi\left(k,k\right)\right]$
$\displaystyle=\phi_{k}\sum_{p\neq k}\tfrac{\nu_{p}}{\nu_{p}-2}J_{p}\otimes
J_{k}+\phi_{k}\left[J_{k\otimes}\left(K_{n}+I_{n^{2}}\right)+{\rm
vec}(J_{k}){\rm vec}(J_{k})^{\prime}\right]$
$\displaystyle=\phi_{k}J_{k}^{\bullet}\otimes
J_{k}-\phi_{k}J_{k\otimes}+\phi_{k}\left[J_{k\otimes}\left(K_{n}+I_{n^{2}}\right)+{\rm
vec}(J_{k}){\rm vec}(J_{k})^{\prime}\right]$
$\displaystyle=\phi_{k}J_{k}^{\bullet}\otimes
J_{k}+\phi_{k}\left[J_{k\otimes}K_{n}+{\rm vec}(J_{k}){\rm
vec}(J_{k})^{\prime}\right],$ (A.4)

where $J_{k}^{\bullet}=\sum_{p\neq k}\frac{\nu_{p}}{\nu_{p}-2}J_{p}+J_{k}$.

Next, for $k\neq l$, we have

$\displaystyle\mathbb{E}\left[\varphi\left(k,l\right)\right]=$
$\displaystyle\mathbb{E}\left[W_{k}W_{l}\sum_{p=1}^{K}{\rm
vec}\left(e_{k}X_{k}X_{p}^{\prime}e_{p}^{\prime}\right)\sum_{q=1}^{K}{\rm
vec}\left(e_{l}X_{l}X_{q}^{\prime}e_{q}^{\prime}\right)^{\prime}\right]$
$\displaystyle=$ $\displaystyle\mathbb{E}\left[W_{k}W_{l}{\rm
vec}\left(e_{k}X_{k}X_{k}^{\prime}e_{k}^{\prime}\right){\rm
vec}\left(e_{l}X_{l}X_{l}^{\prime}e_{l}^{\prime}\right)^{\prime}+W_{k}W_{l}{\rm
vec}\left(e_{k}X_{k}X_{l}^{\prime}e_{l}^{\prime}\right){\rm
vec}\left(e_{l}X_{l}X_{k}^{\prime}e_{k}^{\prime}\right)^{\prime}\right].$

From Lemma A.3 with $p=q=2$ we find that

$\displaystyle\mathbb{E}\left[W_{k}W_{l}{\rm
vec}\left(e_{k}X_{k}X_{k}^{\prime}e_{k}^{\prime}\right){\rm
vec}\left(e_{l}X_{l}X_{q}^{\prime}e_{q}^{\prime}\right)^{\prime}\right]$
$\displaystyle=$ $\displaystyle{\rm
vec}\left(e_{k}\mathbb{E}\left[W_{k}X_{k}X_{k}^{\prime}\right]e_{k}^{\prime}\right){\rm
vec}\left(e_{l}\mathbb{E}\left[W_{l}X_{l}X_{l}^{\prime}\right]e_{l}^{\prime}\right)^{\prime}$
$\displaystyle=$ $\displaystyle{\rm vec}\left(J_{k}\right){\rm
vec}\left(J_{l}\right)^{\prime},$

and we also have

$\displaystyle\mathbb{E}\left[W_{k}W_{l}{\rm
vec}\left(e_{k}X_{k}X_{l}^{\prime}e_{l}^{\prime}\right){\rm
vec}\left(e_{l}X_{l}X_{k}^{\prime}e_{k}^{\prime}\right)^{\prime}\right]$
$\displaystyle=$ $\displaystyle\mathbb{E}\left[W_{k}W_{l}{\rm
vec}\left(e_{k}X_{k}X_{l}^{\prime}e_{l}^{\prime}\right){\rm
vec}\left(e_{k}X_{k}X_{l}^{\prime}e_{l}^{\prime}\right)^{\prime}K_{n}\right]$
$\displaystyle=$ $\displaystyle\left(e_{l}\otimes
e_{k}\right)\mathbb{E}\left[{\rm vec}\left(Z_{k}Z_{l}^{\prime}\right){\rm
vec}\left(Z_{k}Z_{l}^{\prime}\right)^{\prime}\right]\left(e_{l}^{\prime}\otimes
e_{k}^{\prime}\right)K_{n}$ $\displaystyle=$ $\displaystyle\left(e_{l}\otimes
e_{k}\right)\left(e_{l}^{\prime}\otimes e_{k}^{\prime}\right)K_{n}$
$\displaystyle=$ $\displaystyle\left(J_{l}\otimes J_{k}\right)K_{n},$

where $Z_{k}\sim N(0,I_{n_{k}})$. Finally, we have

$\mathbb{E}\left[\varphi\left(k,l\right)\right]={\rm
vec}\left(J_{k}\right){\rm vec}\left(J_{l}\right)^{\prime}+\left(J_{l}\otimes
J_{k}\right)K_{n}.$

We are finally ready to derive the expression for
$\mathbb{E}\left(\nabla_{\tilde{\Xi}}^{s}\nabla_{\tilde{\Xi}}^{s\prime}\right)$,
which takes the form

$\displaystyle\mathbb{E}\left(\nabla_{\tilde{\Xi}}^{s}\nabla_{\tilde{\Xi}}^{s\prime}\right)$
$\displaystyle=\mathbb{E}\left[\sum_{k=1}^{K}\sum_{l=1}^{K}W_{k}W_{l}{\rm
vec}\left(e_{k}X_{k}X^{\prime}\right){\rm
vec}\left(e_{l}V_{l}X^{\prime}\right)^{\prime}\right]-{\rm
vec}\left(I_{n}\right){\rm vec}\left(I_{n}\right)^{\prime}$
$\displaystyle=\sum_{k=1}^{K}\sum_{l=1}^{K}\left[{\rm
vec}\left(J_{k}\right){\rm vec}\left(J_{l}\right)^{\prime}+\left(J_{l}\otimes
J_{k}\right)K_{n}\right]-{\rm vec}\left(I_{n}\right){\rm
vec}\left(I_{n}\right)^{\prime}$
$\displaystyle\quad+\sum_{k=1}^{K}\left[\mathbb{E}\left[\varphi\left(k,k\right)\right]-{\rm
vec}\left(J_{k}\right){\rm
vec}\left(J_{k}\right)^{\prime}-J_{k\otimes}K_{n}\right]$
$\displaystyle=K_{n}+\tilde{\Upsilon}_{K},$

where $\tilde{\Upsilon}_{K}=\sum_{k}\tilde{\Psi}_{k}$ with

$\displaystyle\tilde{\Psi}_{k}$ $\displaystyle=\phi_{k}J_{k}^{\bullet}\otimes
J_{k}+\phi_{k}\left[J_{k\otimes}K_{n}+{\rm vec}(J_{k}){\rm
vec}(J_{k})^{\prime}\right]-{\rm vec}\left(J_{k}\right){\rm
vec}\left(J_{k}\right)^{\prime}-J_{k\otimes}K_{n}$
$\displaystyle=\phi_{k}J_{k}^{\bullet}\otimes
J_{k}+\left(\phi_{k}-1\right)\left(J_{k\otimes}K_{n}+{\rm vec}(J_{k}){\rm
vec}(J_{k})^{\prime}\right).$

So, the final formula for the information matrix is given by

$\displaystyle\mathcal{I}_{\tilde{\Xi}}$
$\displaystyle=M\mathbb{E}\left(\nabla_{\tilde{\Xi}}^{s}\nabla_{\tilde{\Xi}}^{s\prime}\right)M^{\prime}=M\left(K_{n}+\tilde{\Upsilon}_{K}\right)M^{\prime}$
$\displaystyle=\left(A\otimes A^{\prime}\right)K_{n}+\Upsilon_{K},$

where $\Upsilon_{K}=\sum_{k=1}^{K}\Psi_{k}$ with

$\displaystyle\Psi_{k}=M\tilde{\Psi}_{k}M^{\prime}$
$\displaystyle=\phi_{k}J_{k}^{\bullet}\otimes
A^{\prime}J_{k}A+\left(\phi_{k}-1\right)\left[\left(J_{k}A\otimes
A^{\prime}J_{k}\right)K_{n}+{\rm vec}(A^{\prime}J_{k}){\rm
vec}(A^{\prime}J_{k})^{\prime}\right].$

##### Cross-Covariance:
$\mathbb{E}\left(\nabla_{\nu_{k}}\nabla_{\nu_{l}}\right)$

Because $X_{k}$ and $X_{l}$ are independent for $k\neq l$, we have
$\mathbb{E}\left(\nabla_{\nu_{k}}\nabla_{\nu_{l}}\right)=0$ for $k\neq l$. For
$k=l$, we first note that

$\displaystyle\mathbb{E}\left[W_{k}^{2}\right]$
$\displaystyle=\tfrac{\left(\nu_{k}+n_{k}\right)\left(\nu_{k}+2\right)}{\nu_{k}\left(\nu_{k}+n_{k}+2\right)},$
$\displaystyle\mathbb{E}\left[W_{k}\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)\right]$
$\displaystyle=\psi\left(\tfrac{\nu+n}{2}+1\right)-\psi\left(\tfrac{\nu}{2}+1\right),$
$\displaystyle\mathbb{E}\left[\log^{2}\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)\right]$
$\displaystyle=\psi^{\prime}\left(\tfrac{\nu_{k}}{2}\right)-\psi^{\prime}\left(\tfrac{\nu_{k}+n_{k}}{2}\right)+\left[\psi\left(\tfrac{\nu_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{\nu_{k}}{2}\right)\right]^{2},$

using Lemma A.3 (ii) and (iii) and, after some algebra, it follows that

$\displaystyle\mathbb{E}\left(\nabla_{\nu_{k}}^{2}\right)$ $\displaystyle=$
$\displaystyle\tfrac{1}{4}\mathbb{E}\left[W_{k}^{2}+\log^{2}\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)+2W_{k}\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)\right]-\tfrac{1}{4}\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)+1\right]^{2}.$
$\displaystyle=$
$\displaystyle\tfrac{1}{4}\left[\psi^{\prime}\left(\tfrac{\nu_{k}}{2}\right)-\psi^{\prime}\left(\tfrac{\nu_{k}+n_{k}}{2}\right)\right]-\tfrac{n_{k}\left(\nu_{k}+n_{k}+4\right)}{2\nu_{k}\left(\nu_{k}+n_{k}+2\right)\left(\nu_{k}+n_{k}\right)}.$

##### Cross-Covariance:
$\mathbb{E}\left(\nabla_{\nu}\nabla_{\tilde{\Xi}}^{\prime}\right)$

We first compute

$\displaystyle\mathbb{E}\left[\nabla_{\nu_{k}}\nabla_{\tilde{\Xi}}^{s}\right]$
$\displaystyle=\tfrac{1}{2}\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)+1-W_{k}-\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)\right]\left[\sum_{l=1}^{K}W_{l}{\rm
vec}\left(e_{l}X_{l}X^{\prime}\right)-{\rm vec}\left(I_{n}\right)\right]$
$\displaystyle=-\tfrac{1}{2}\mathbb{E}\left[\left[W_{k}+\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)\right]\sum_{l=1}^{K}W_{l}{\rm
vec}\left(e_{l}X_{l}X^{\prime}\right)\right]+\tfrac{1}{2}\left(\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)+1\right){\rm
vec}\left(I_{n}\right),$ (A.5)

For $k=l$, we have

$\displaystyle\mathbb{E}\left[W_{k}^{2}{\rm
vec}\left(e_{k}X_{k}X\right)\right]=$
$\displaystyle\mathbb{E}\left[W_{k}^{2}\sum_{p=1}^{K}{\rm
vec}\left(e_{k}X_{k}X_{p}^{\prime}e_{p}^{\prime}\right)\right]$
$\displaystyle=$ $\displaystyle\mathbb{E}\left[W_{k}^{2}{\rm
vec}\left(e_{k}X_{k}X_{k}^{\prime}e_{k}^{\prime}\right)\right]$
$\displaystyle=$ $\displaystyle e_{k\otimes}\mathbb{E}\left[W_{k}^{2}{\rm
vec}\left(X_{k}X_{k}^{\prime}\right)\right]$ $\displaystyle=$ $\displaystyle
e_{k\otimes}\phi_{k}{\rm vec}\left(I_{n}\right)=\phi_{k}{\rm
vec}\left(J_{k}\right),$

and for $k\neq l$ we have

$\displaystyle\mathbb{E}\left[W_{k}W_{l}{\rm
vec}\left(e_{l}X_{l}X^{\prime}\right)\right]=$
$\displaystyle\mathbb{E}\left[W_{k}W_{l}\sum_{p=1}^{K}{\rm
vec}\left(e_{l}X_{l}X_{p}^{\prime}e_{p}^{\prime}\right)\right]$
$\displaystyle=$ $\displaystyle\mathbb{E}\left[W_{k}W_{l}{\rm
vec}\left(e_{l}X_{l}X_{k}^{\prime}e_{k}^{\prime}\right)\right]+\mathbb{E}\left[W_{k}W_{l}{\rm
vec}\left(e_{l}X_{g}X_{l}^{\prime}e_{l}^{\prime}\right)\right]$
$\displaystyle=$ $\displaystyle\mathbb{E}\left[{\rm
vec}\left(e_{l}\left(W_{l}X_{l}\right)\left(W_{k}X_{k}^{\prime}\right)e_{k}^{\prime}\right)\right]+\mathbb{E}\left[W_{l}{\rm
vec}\left(e_{l}X_{l}X_{l}^{\prime}e_{l}^{\prime}\right)\right]$
$\displaystyle=$ $\displaystyle\mathbb{E}\left[W_{l}{\rm
vec}\left(e_{l}X_{l}X_{l}^{\prime}e_{l}^{\prime}\right)\right]$
$\displaystyle=$ $\displaystyle{\rm vec}(J_{l}),$

where we used $\zeta_{2,2}=1$. Thus, we have

$\displaystyle\mathbb{E}\left[W_{k}\sum_{l=1}^{K}W_{l}{\rm
vec}\left(e_{l}X_{l}X^{\prime}\right)\right]$
$\displaystyle=\mathbb{E}\left[W_{k}^{2}{\rm
vec}\left(e_{k}X_{k}X^{\prime}\right)\right]+\mathbb{E}\left[W_{k}\sum_{l\neq
k}^{K}W_{l}{\rm vec}\left(e_{l}X_{l}X^{\prime}\right)\right]$
$\displaystyle=\phi_{k}{\rm vec}\left(J_{k}\right)+\sum_{l\neq k}^{K}{\rm
vec}\left(J_{l}\right)$ $\displaystyle=\left(\phi_{k}-1\right){\rm
vec}\left(J_{k}\right)+{\rm vec}\left(I_{n}\right).$

For the remaining term in (A.5), we have, for $k=l$,

$\displaystyle\mathbb{E}\left[\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)W_{k}{\rm
vec}\left(e_{k}X_{k}X^{\prime}\right)\right]=$
$\displaystyle\mathbb{E}\left[\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)W_{k}{\rm
vec}\left(e_{k}X_{k}X_{k}^{\prime}e_{k}^{\prime}\right)\right]$
$\displaystyle=$ $\displaystyle
e_{k\otimes}\mathbb{E}\left[\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)W_{k}{\rm
vec}\left(X_{k}X_{k}^{\prime}\right)\right]$ $\displaystyle=$
$\displaystyle\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}+1\right)-\psi\left(\tfrac{v_{k}}{2}\right)\right]{\rm
vec}\left(J_{k}\right),$

where we used Lemma A.3 (iii). For $k\neq l$, we have

$\displaystyle\mathbb{E}\left[\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)W_{l}{\rm
vec}\left(e_{l}X_{g}X^{\prime}\right)\right]$ $\displaystyle=$
$\displaystyle\mathbb{E}\left[\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)W_{l}\sum_{p=1}^{K}{\rm
vec}\left(e_{l}X_{l}X_{p}^{\prime}e_{p}^{\prime}\right)\right]$
$\displaystyle=$
$\displaystyle\mathbb{E}\left[\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)W_{l}{\rm
vec}\left(e_{l}X_{l}X_{l}^{\prime}e_{l}^{\prime}\right)\right]+\mathbb{E}\left[\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)W_{l}{\rm
vec}\left(e_{l}X_{l}X_{k}^{\prime}e_{k}^{\prime}\right)\right]$
$\displaystyle=$
$\displaystyle\mathbb{E}\left[\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)\right]\times\mathbb{E}\left[W_{l}{\rm
vec}\left(e_{l}X_{l}X_{l}^{\prime}e_{l}^{\prime}\right)\right]$
$\displaystyle=$
$\displaystyle\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)\right]{\rm
vec}\left(J_{l}\right),$

and combined we have

$\displaystyle\mathbb{E}\left[\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)\sum_{l=1}^{K}W_{l}{\rm
vec}\left(e_{l}X_{l}X^{\prime}\right)\right]$ $\displaystyle=$
$\displaystyle\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}+1\right)-\psi\left(\tfrac{v_{k}}{2}\right)\right]{\rm
vec}\left(J_{k}\right)+\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)\right]\sum_{l\neq
k}{\rm vec}\left(J_{l}\right)$ $\displaystyle=$
$\displaystyle\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}+1\right)-\psi\left(\tfrac{v_{k}}{2}\right)\right]{\rm
vec}\left(J_{k}\right)+\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)\right]\left[{\rm
vec}\left(I_{n}\right)-{\rm vec}\left(J_{k}\right)\right]$ $\displaystyle=$
$\displaystyle\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)\right]{\rm
vec}\left(I_{n}\right)+{\rm
vec}\left(J_{k}\right)\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}+1\right)-\psi\left(\tfrac{v_{k}}{2}\right)-\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)+\psi\left(\tfrac{v_{k}}{2}\right)\right]$
$\displaystyle=$
$\displaystyle\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)\right]{\rm
vec}\left(I_{n}\right)+\tfrac{2}{v_{k}+n_{k}}{\rm vec}\left(J_{k}\right).$

where we used $\psi(z+1)=\psi(z)+\frac{1}{z}$ in the last identity. Thus, we
have

$\displaystyle\mathbb{E}\left[\left[W_{k}+\log\left(1+\tfrac{X_{k}^{\prime}X_{k}}{\nu_{k}}\right)\right]\sum_{l=1}^{K}W_{l}{\rm
vec}\left(e_{l}X_{l}X^{\prime}\right)\right]$ $\displaystyle=$
$\displaystyle\left(\phi_{k}-1\right){\rm vec}\left(J_{k}\right)+{\rm
vec}\left(I_{n}\right)+\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)\right]{\rm
vec}\left(I_{n}\right)+\tfrac{2}{v_{k}+n_{k}}{\rm vec}\left(J_{k}\right)$
$\displaystyle=$
$\displaystyle\left(\phi_{k}-1+\tfrac{2}{v_{k}+n_{k}}\right){\rm
vec}\left(J_{k}\right)+\left[\psi\left(\tfrac{v_{k}+n_{k}}{2}\right)-\psi\left(\tfrac{v_{k}}{2}\right)+1\right]{\rm
vec}\left(I_{n}\right),$

where the last term (scaled by $-\frac{1}{2}$) is cancelled by the last term
in (A.5), and we have

$\displaystyle\mathbb{E}\left[\nabla_{\nu_{k}}\nabla_{\tilde{\Xi}}^{s}\right]$
$\displaystyle=-\tfrac{1}{2}\left(\phi_{k}-1+\tfrac{2}{v_{k}+n_{k}}\right){\rm
vec}\left(J_{k}\right)$
$\displaystyle=-\tfrac{1}{2}\left(\tfrac{4}{\left(v_{k}+n_{k}\right)\left(v_{k}+n_{k}+2\right)}\right){\rm
vec}\left(J_{k}\right)$
$\displaystyle=-\tfrac{2}{\left(v_{k}+n_{k}\right)\left(v_{k}+n_{k}+2\right)}{\rm
vec}\left(J_{k}\right).$

It follows that

$\displaystyle\mathbb{E}\left[\nabla_{\nu_{k}}\nabla_{\tilde{\Xi}}\right]$
$\displaystyle=-\tfrac{2}{\left(v_{k}+n_{k}\right)\left(v_{k}+n_{k}+2\right)}M{\rm
vec}\left(J_{k}\right)=-\tfrac{2}{\left(v_{k}+n_{k}\right)\left(v_{k}+n_{k}+2\right)}{\rm
vec}\left(A^{\prime}J_{k}\right).$

$\square$

Proof of Theorem 6 (Expected Hessian matrix).

##### The terms: $\mathbb{E}(\nabla_{\mu\mu^{\prime}})$,
$\mathbb{E}(\nabla_{\mu\tilde{\Xi}^{\prime}})$,
$\mathbb{E}(\nabla_{\mu\nu_{k}})$

$\displaystyle\mathbb{E}[\nabla_{\mu\mu^{\prime}}]$
$\displaystyle=\sum_{k=1}^{K}\tfrac{2}{\nu_{k}+n_{k}}A^{\prime}e_{k}\mathbb{E}[W_{k}^{2}X_{k}X_{k}^{\prime}]e_{k}^{\prime}A-\mathbb{E}[W_{k}]A^{\prime}J_{k}A$
$\displaystyle=\sum_{k=1}^{K}\left(\tfrac{2}{\nu_{k}+n_{k}}\phi_{k}-1\right)A^{\prime}e_{k}e_{k}^{\prime}A-A^{\prime}J_{k}A$
$\displaystyle=-\sum_{k=1}^{K}\phi_{k}A^{\prime}J_{k}A.$

The expectations of $\nabla_{\mu\tilde{\Xi}^{\prime}}$ and
$\nabla_{\mu\nu_{k}}$ are both zeros, because they are odd functions of $X$.

##### The term: $\mathbb{E}(\nabla_{\Xi\Xi^{\prime}})$

Let $\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}=\left(I_{n}\otimes
A^{\prime}\right)\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}^{s}\left(I_{n}\otimes
A\right)$, so we have

$\displaystyle\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}^{s}$
$\displaystyle=-K_{n}+\sum_{k=1}^{K}\tfrac{2W_{k}^{2}}{\nu_{k}+n_{k}}{\rm
vec}\left(e_{k}X_{k}X^{\prime}\right){\rm
vec}\left(e_{k}X_{k}X^{\prime}\right)^{\prime}$
$\displaystyle\quad+\sum_{k=1}^{K}W_{k}\left[\left(XX_{k}^{\prime}e_{k}^{\prime}\otimes
I_{n}\right)K_{n}-\left(XX^{\prime}\otimes J_{k}\right)-\left(I_{n}\otimes
e_{k}X_{k}X^{\prime}\right)K_{n}\right].$

From (A.4) we have

$\mathbb{E}\left[{\rm vec}\left(e_{k}X_{k}X^{\prime}\right){\rm
vec}\left(e_{k}X_{k}X^{\prime}\right)^{\prime}\right]=\phi_{k}\left[J_{k}^{\bullet}\otimes
J_{k}+J_{k\otimes}K_{n}+{\rm vec}(J_{k}){\rm vec}(J_{k})^{\prime}\right],$

where $J_{k}^{\bullet}=\sum_{p\neq k}\xi_{p}J_{p}+J_{k}$. Next,

$\displaystyle\mathbb{E}\left[W_{k}\left(XX_{k}^{\prime}e_{k}^{\prime}\otimes
I_{n}\right)K_{n}\right]$
$\displaystyle=\mathbb{E}[\sum_{p}W_{k}\left(e_{p}X_{p}X_{k}^{\prime}e_{k}^{\prime}\otimes
I_{n}\right)K_{n}]$
$\displaystyle=\mathbb{E}[W_{k}\left(e_{k}X_{k}X_{k}^{\prime}e_{k}^{\prime}\otimes
I_{n}\right)K_{n}]$ $\displaystyle=\left(e_{k}e_{k}^{\prime}\otimes
I_{n}\right)K_{n},\qquad(\text{using
}\mathbb{E}[W_{k}X_{k}X_{k}^{\prime}]=I_{n_{k}})$
$\displaystyle=\left(J_{k}\otimes I_{n}\right)K_{n},$

similarly

$\mathbb{E}\left[W_{k}\left(I_{n}\otimes
e_{k}X_{k}X^{\prime}\right)K_{n}\right]K_{n}=\left(I_{n}\otimes
J_{k}\right)K_{n},$

and

$\displaystyle\mathbb{E}\left[W_{k}XX^{\prime}\right]$
$\displaystyle=\mathbb{E}\left[\sum_{l}\sum_{p}W_{k}e_{l}X_{l}X_{p}^{\prime}e_{p}^{\prime}\right]$
$\displaystyle=\mathbb{E}\left[\sum_{l\neq
k}W_{k}e_{l}X_{l}X_{l}^{\prime}e_{l}^{\prime}+W_{k}e_{k}X_{k}X_{k}^{\prime}e_{k}^{\prime}\right]$
$\displaystyle=\sum_{l\neq
k}e_{l}\mathbb{E}\left[X_{l}X_{l}^{\prime}\right]e_{l}^{\prime}+e_{k}\mathbb{E}\left[W_{k}X_{k}X_{k}^{\prime}\right]e_{k}^{\prime}$
$\displaystyle=J_{k}+\sum_{l\neq
k}\tfrac{\nu_{l}}{\nu_{l}-2}J_{l}=J_{k}^{\bullet},$

leads to $\mathbb{E}\left[W_{k}\left(XX^{\prime}\otimes
J_{k}\right)\right]=\left(J_{k}^{\bullet}\otimes J_{k}\right)$. Combining the
terms we have

$\displaystyle\mathbb{E}\left(\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}^{s}\right)$
$\displaystyle=-K_{n}+\sum_{k=1}^{K}\varphi_{k},$

where

$\displaystyle\varphi_{k}=$
$\displaystyle\tfrac{2}{\nu_{k}+n_{k}}\phi_{k}\left[J_{k}^{\bullet}\otimes
J_{k}+J_{k\otimes}K_{n}+{\rm vec}(J_{k}){\rm vec}(J_{k})^{\prime}\right]$
$\displaystyle+\left(J_{k}\otimes
I_{n}\right)K_{n}-\left(J_{k}^{\bullet}\otimes J_{k}\right)-\left(I_{n}\otimes
J_{k}\right)K_{n}$ $\displaystyle=$
$\displaystyle\left(\tfrac{2}{\nu_{k}+n_{k}+2}-1\right)J_{k}^{\bullet}\otimes
J_{k}+\tfrac{2}{\nu_{k}+n_{k}+2}\left[J_{k\otimes}K_{n}+{\rm vec}(J_{k}){\rm
vec}(J_{k})^{\prime}\right]+\left[\left(J_{k}\otimes
I_{n}\right)-\left(I_{n}\otimes J_{k}\right)\right]K_{n}$ $\displaystyle=$
$\displaystyle-\phi_{k}J_{k}^{\bullet}\otimes
J_{k}+\left(1-\phi_{k}\right)\left[J_{k\otimes}K_{n}+{\rm vec}(J_{k}){\rm
vec}(J_{k})^{\prime}\right]+\left(J_{k}\otimes
I_{n}\right)K_{n}-\left(I_{n}\otimes J_{k}\right)K_{n},$

such that

$\displaystyle\mathbb{E}\left(\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}^{s}\right)$
$\displaystyle=-K_{n}+\sum_{k=1}^{K}\varphi_{k}=-K_{n}-\sum_{k=1}^{K}\phi_{k}J_{k}^{\bullet}\otimes
J_{k}+\left(\phi_{k}-1\right)\left[J_{k\otimes}K_{n}+{\rm vec}(J_{k}){\rm
vec}(J_{k})^{\prime}\right],$

where we used $\sum_{k=1}^{K}\left(J_{k}\otimes
I_{n}\right)K_{n}-\left(I_{n}\otimes J_{k}\right)K_{n}=0$. Finally, we have
shown that

$\displaystyle\mathbb{E}\left(\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}\right)$
$\displaystyle=\left(I_{n}\otimes
A^{\prime}\right)\mathbb{E}\left(\nabla_{\Xi\Xi^{\prime}}^{s}\right)\left(I_{n}\otimes
A\right)$ $\displaystyle=-K_{n}-\sum_{k=1}^{K}\phi_{k}J_{k}^{\bullet}\otimes
A^{\prime}J_{k}A-\left(\phi_{k}-1\right)\left[\left(J_{k}A\otimes
A^{\prime}J_{k}\right)K_{n}+{\rm vec}\left(A^{\prime}J_{k}\right){\rm
vec}\left(A^{\prime}J_{k}\right)^{\prime}\right],$

which proves that
$\mathbb{E}\left(\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}\right)=-\mathcal{I}_{\tilde{\Xi}}$.

##### The term: $\mathbb{E}(\nabla_{\tilde{\Xi}\nu^{\prime}})$

We have

$\mathbb{E}[\nabla_{\tilde{\Xi}\nu_{k}}]=\tfrac{1}{\nu_{k}+n_{k}}\left[\mathbb{E}\left[W_{k}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)\right]-\mathbb{E}\left[W_{k}^{2}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)\right]\right],$

where

$\displaystyle\mathbb{E}\left[W_{k}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)\right]$
$\displaystyle=\sum_{l=1}^{K}\mathbb{E}\left[W_{k}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X_{l}^{\prime}e_{l}^{\prime}\right)\right]$
$\displaystyle=\mathbb{E}\left[W_{k}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X_{k}^{\prime}e_{k}^{\prime}\right)\right]$
$\displaystyle={\rm vec}\left(A^{\prime}J_{k}\right),$

and

$\displaystyle\mathbb{E}\left[W_{k}^{2}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X^{\prime}\right)\right]$
$\displaystyle=\sum_{l=1}^{K}\mathbb{E}\left[W_{k}^{2}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X_{l}^{\prime}e_{l}^{\prime}\right)\right]$
$\displaystyle=\mathbb{E}\left[W_{k}^{2}{\rm
vec}\left(A^{\prime}e_{k}X_{k}X_{k}^{\prime}e_{k}^{\prime}\right)\right]$
$\displaystyle=\phi_{k}{\rm vec}\left(A^{\prime}J_{k}\right).$

So, we have

$\displaystyle\mathbb{E}[\nabla_{\tilde{\Xi}\nu_{k}}]$
$\displaystyle=\tfrac{1}{\nu_{k}+n_{k}}\left[{\rm
vec}\left(A^{\prime}J_{k}\right)-\phi_{k}{\rm
vec}\left(A^{\prime}J_{k}\right)\right]$
$\displaystyle=\tfrac{1}{\left(\nu_{k}+n_{k}+2\right)\left(\nu_{k}+n_{k}\right)}{\rm
vec}\left(A^{\prime}J_{k}\right),$

which equals $-\mathcal{I}_{\tilde{\Xi}\nu_{k}}$.

##### The term: $\mathbb{E}(\nabla_{\nu\nu^{\prime}})$

Using
$\mathbb{E}\left[W_{k}^{2}\right]=\tfrac{\left(\nu_{k}+2\right)(\nu_{k}+n_{k})}{\nu_{k}(\nu_{k}+n_{k}+2)}$
and $\mathbb{E}\left[W_{k}\right]=1$, we have

$\displaystyle\nabla_{\nu_{k}\nu_{k}}$ $\displaystyle=$
$\displaystyle\tfrac{1}{4}\psi^{\prime}\left(\tfrac{v_{k}+n_{k}}{2}\right)-\tfrac{1}{4}\psi^{\prime}\left(\tfrac{v_{k}}{2}\right)+\tfrac{1}{2\nu_{k}}+\tfrac{1}{2}\tfrac{1}{\nu_{k}+n_{k}}W_{k}^{2}-\tfrac{1}{\left(\nu_{k}+n_{k}\right)}W_{k}$
$\displaystyle=$
$\displaystyle\tfrac{1}{2}\left[\tfrac{1}{2}\psi^{\prime}\left(\tfrac{v_{k}+n_{k}}{2}\right)-\tfrac{1}{2}\psi^{\prime}\left(\tfrac{v_{k}}{2}\right)+\tfrac{1}{\nu_{k}+n_{k}}\left(W_{k}^{2}-W_{k}\right)+\tfrac{1}{\nu_{k}}\left[1-\tfrac{\nu_{k}}{\nu_{k}+X_{k}^{\prime}X_{k}}\right]\right],$
$\displaystyle=$
$\displaystyle\tfrac{1}{4}\psi^{\prime}\left(\tfrac{v_{k}+n_{k}}{2}\right)-\tfrac{1}{4}\psi^{\prime}\left(\tfrac{v_{k}}{2}\right)+\tfrac{1}{2\nu_{k}}+\tfrac{1}{2}\tfrac{1}{\nu_{k}+n_{k}}\tfrac{\left(\nu_{k}+2\right)(\nu_{k}+n_{k})}{\nu_{k}(\nu_{k}+n_{k}+2)}-\tfrac{1}{\left(\nu_{k}+n_{k}\right)}$
$\displaystyle=$
$\displaystyle\tfrac{1}{4}\psi^{\prime}\left(\tfrac{v_{k}+n_{k}}{2}\right)-\tfrac{1}{4}\psi^{\prime}\left(\tfrac{v_{k}}{2}\right)+\tfrac{1}{2}\left[\tfrac{1}{\nu_{k}}+\tfrac{\nu_{k}+2}{\nu_{k}(\nu_{k}+n_{k}+2)}-\tfrac{2}{\nu_{k}+n_{k}}\right]$
$\displaystyle=$
$\displaystyle\tfrac{1}{4}\psi^{\prime}\left(\tfrac{v_{k}+n_{k}}{2}\right)-\tfrac{1}{4}\psi^{\prime}\left(\tfrac{v_{k}}{2}\right)+\tfrac{1}{2}\left[\tfrac{n_{k}\left(\nu_{k}+n_{k}+4\right)}{\nu_{k}(\nu_{k}+n_{k}+2)\left(\nu_{k}+n_{k}\right)}\right],$

which equals $-\mathcal{I}_{\nu_{k}}$.

$\square$

Proof of Lemma 1. First, we can express the $\Xi$ by using the cluster/group
structure
$\tilde{\Xi}=\left[\tilde{\Xi}_{1},\tilde{\Xi}_{2},\ldots,\tilde{\Xi}_{K}\right]$
with $\tilde{\Xi}_{k}\in\mathbb{R}^{n\times n_{k}}$, and
$\tilde{\Xi}_{k}=\left[\tilde{\Xi}_{1k}^{\prime},\tilde{\Xi}_{2k}^{\prime},\ldots,\tilde{\Xi}_{Kk}^{\prime}\right]^{\prime}$
with $\tilde{\Xi}_{lk}\in\mathbb{R}^{n_{l}\times n_{k}}$. Then, we have

$\frac{\partial{\rm vec}\left(\Xi\right)}{\partial{\rm
vec}\left(\tilde{\Xi}\right)^{\prime}}=\left[\begin{array}[]{cccc}\Gamma_{11}&\Gamma_{12}&\ldots&\Gamma_{1K}\\\
\Gamma_{21}&\Gamma_{22}&&\Gamma_{2K}\\\ \vdots&\vdots&\ddots&\vdots\\\
\Gamma_{K1}&\Gamma_{K2}&&\Gamma_{KK}\end{array}\right],\quad{\rm
where}\quad\Gamma_{kl}=\frac{\partial{\rm
vec}\left(\Xi_{k}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{l}\right)^{\prime}}\in\mathbb{R}^{nn_{k}\times
nn_{k}}.$

Because $\Xi_{k}=\tilde{\Xi}_{k}P_{kk}$ and $P_{kk}$ only depends on
$\tilde{\Xi}_{kk}$, we have $\Gamma_{kl}=0$ if $k\neq l$. Then, we have

$\Gamma_{kk}=\frac{\partial{\rm vec}\left(\Xi_{k}\right)}{\partial{\rm
vec}\left(\Xi_{k}^{\prime}\right)^{\prime}}\frac{\partial{\rm
vec}\left(\Xi_{k}^{\prime}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{k}^{\prime}\right)^{\prime}}\frac{\partial{\rm
vec}\left(\tilde{\Xi}_{k}^{\prime}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{k}\right)^{\prime}}=K_{n_{k},n}\frac{\partial{\rm
vec}\left(\Xi_{k}^{\prime}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{k}^{\prime}\right)^{\prime}}K_{n,n_{k}}$

and we have

$\frac{\partial{\rm vec}\left(\Xi_{k}^{\prime}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{k}^{\prime}\right)^{\prime}}=\left[\begin{array}[]{cccc}\Pi_{11}&\Pi_{12}&\ldots&\Pi_{1K}\\\
\Pi_{21}&\Pi_{22}&&\Pi_{2K}\\\ \vdots&\vdots&\ddots&\vdots\\\
\Pi_{K1}&\Pi_{K2}&\cdots&\Pi_{KK}\end{array}\right],$

and we have

$\Pi_{ij}=\frac{\partial{\rm vec}\left(\Xi_{ik}^{\prime}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{jk}^{\prime}\right)^{\prime}}=\frac{\partial{\rm
vec}\left(P_{kk}^{\prime}\tilde{\Xi}_{ik}^{\prime}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{jk}^{\prime}\right)^{\prime}}=\begin{cases}\bm{0}&i\neq
j,j\neq k\\\ I_{n_{i}}\otimes P_{kk}^{\prime}&i=j\neq k\\\
\left(\tilde{\Xi}_{ik}\otimes I_{n_{k}}\right)\frac{\partial{\rm
vec}\left(P_{kk}^{\prime}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}^{\prime}\right)^{\prime}}&i\neq j,j=k\\\
\left(I_{n_{i}}\otimes P_{kk}^{\prime}\right)+\left(\tilde{\Xi}_{ik}\otimes
I_{n_{k}}\right)\frac{\partial{\rm
vec}\left(P_{kk}^{\prime}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}^{\prime}\right)^{\prime}}&i=j=k.\end{cases}$

Because we have

$\displaystyle\mathrm{d}{\rm
vec}\left(P_{kk}^{\prime}\tilde{\Xi}_{ik}^{\prime}\right)$
$\displaystyle=\left(\tilde{\Xi}_{ik}\otimes I_{n_{k}}\right)d{\rm
vec}\left(P_{kk}^{\prime}\right)+\left(I_{n_{i}}\otimes
P_{kk}^{\prime}\right)d{\rm vec}\left(\tilde{\Xi}_{ik}^{\prime}\right),$

and

$\displaystyle\frac{\partial{\rm vec}\left(P_{kk}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}\right)^{\prime}}=\frac{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}^{\prime}(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}\right)^{\prime}}$ $\displaystyle=$
$\displaystyle\left((\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\otimes
I_{n_{k}}\right)\frac{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}^{\prime}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}\right)^{\prime}}+\left(I_{n_{k}}\otimes\tilde{\Xi}_{kk}^{\prime}\right)\frac{\partial{\rm
vec}\left((\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}\right)^{\prime}}$ $\displaystyle=$
$\displaystyle\left((\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\otimes
I_{n_{k}}\right)K_{n_{k}}+\left(I_{n_{k}}\otimes\tilde{\Xi}_{kk}^{\prime}\right)\frac{\partial{\rm
vec}\left((\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}\right)^{\prime}}$ $\displaystyle=$
$\displaystyle\left(I_{n_{k}}\otimes(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\right)K_{n_{k}}+\left(I_{n_{k}}\otimes\tilde{\Xi}_{kk}^{\prime}\right)\frac{\partial{\rm
vec}\left((\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\right)}{\partial{\rm
vec}\left(\left(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\right)^{-1}\right)^{\prime}}\frac{\partial{\rm
vec}\left(\left(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\right)^{-1}\right)^{\prime}}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\right)^{\prime}}\frac{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\right)^{\prime}}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}\right)^{\prime}}$ $\displaystyle=$
$\displaystyle\left((\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\otimes
I_{n_{k}}\right)K_{n_{k}}-\left\\{\left(I_{n_{k}}\otimes\tilde{\Xi}_{kk}^{\prime}\right)\left((\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\oplus(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\right)^{-1}\right.$
$\displaystyle\quad\left.\times\left[\left(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\right)^{-1}\otimes\left(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\right)^{-1}\right]\left(I_{n_{k}^{2}}+K_{n_{k}}\right)\left(\tilde{\Xi}_{kk}\otimes
I_{n_{k}}\right)\right\\}.$

Next, from

$\displaystyle\left((\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\oplus(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\right)^{-1}\left[\left(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\right)^{-1}\otimes\left(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\right)^{-1}\right]$
$\displaystyle=$
$\displaystyle\left[\left(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\otimes\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\right)\left[(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\oplus(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\right]\right]^{-1}$
$\displaystyle=$
$\displaystyle\left[\left(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\otimes\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\right)\left[(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\otimes
I+I\otimes(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\right]\right]^{-1}$
$\displaystyle=$
$\displaystyle\left[(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{\frac{1}{2}}\otimes\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}+\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\otimes(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{\frac{1}{2}}\right]^{-1}$
$\displaystyle=$
$\displaystyle\left[(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{\frac{1}{2}}\oplus\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\right]^{-1},$

we find

$\displaystyle\left(I_{n_{k}}\otimes\tilde{\Xi}_{kk}^{\prime}\right)\left[(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{\frac{1}{2}}\oplus\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\right]^{-1}$
$\displaystyle=$
$\displaystyle\left[\left((\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{\frac{1}{2}}\otimes\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}+\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\otimes(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{\frac{1}{2}}\right)\left(I_{n_{k}}\otimes\tilde{\Xi}_{kk}^{\prime-1}\right)\right]^{-1}$
$\displaystyle=$
$\displaystyle\left[\left((\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{\frac{1}{2}}\otimes\tilde{\Xi}_{kk}+\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\otimes(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{\frac{1}{2}}\tilde{\Xi}_{kk}^{\prime-1}\right)\right]^{-1}$
$\displaystyle=$
$\displaystyle\left[(\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{\frac{1}{2}}\otimes\tilde{\Xi}_{kk}+\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\otimes
P_{kk}^{-1}\right]^{-1}$ $\displaystyle=$
$\displaystyle\left[P_{kk}^{-1}\tilde{\Xi}_{kk}^{\prime}\otimes\tilde{\Xi}_{kk}+\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\otimes
P_{kk}^{-1}\right]^{-1}$ $\displaystyle=$
$\displaystyle\left[I_{n_{k}}\otimes\tilde{\Xi}_{kk}+\tilde{\Xi}_{kk}P_{kk}\otimes
P_{kk}^{-1}\left(P_{kk}^{-1}\tilde{\Xi}_{kk}^{\prime}\otimes
I_{n_{k}}\right)\right]^{-1}$ $\displaystyle=$
$\displaystyle\left(\tilde{\Xi}_{kk}^{\prime-1}P_{kk}\otimes
I_{n_{k}}\right)\left[I_{n_{k}}\otimes\tilde{\Xi}_{kk}+\tilde{\Xi}_{kk}P_{kk}\otimes
P_{kk}^{-1}\right]^{-1}.$

Thus, we have

$\displaystyle\frac{\partial{\rm vec}\left(P_{kk}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}\right)^{\prime}}$
$\displaystyle=\left((\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime})^{-\frac{1}{2}}\otimes
I_{n_{k}}\right)K_{n_{k}}-\left[P_{kk}^{-1}\tilde{\Xi}_{kk}^{\prime}\otimes\tilde{\Xi}_{kk}+\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\otimes
P_{kk}^{-1}\right]^{-1}\left(I_{n_{k}^{2}}+K_{n_{k}}\right)\left(\tilde{\Xi}_{kk}\otimes
I_{n_{k}}\right)$
$\displaystyle=\left(\tilde{\Xi}_{kk}^{\prime-1}P_{kk}\otimes
I_{n_{k}}\right)K_{n_{k}}-\left[P_{kk}^{-1}\tilde{\Xi}_{kk}^{\prime}\otimes\tilde{\Xi}_{kk}+\tilde{\Xi}_{kk}\tilde{\Xi}_{kk}^{\prime}\otimes
P_{kk}^{-1}\right]^{-1}\left(I_{n_{k}^{2}}+K_{n_{k}}\right)\left(\tilde{\Xi}_{kk}\otimes
I_{n_{k}}\right)$
$\displaystyle=\left(\tilde{\Xi}_{kk}^{\prime-1}P_{kk}\otimes
I_{n_{k}}\right)\left[K_{n_{k}}-\left[I_{n_{k}}\otimes\tilde{\Xi}_{kk}+\tilde{\Xi}_{kk}P_{kk}\otimes
P_{kk}^{-1}\right]^{-1}\left(I_{n_{k}^{2}}+K_{n_{k}}\right)\left(\tilde{\Xi}_{kk}\otimes
I_{n_{k}}\right)\right].$

Finally, we have

$\displaystyle\frac{\partial{\rm
vec}\left(P_{kk}^{\prime}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}^{\prime}\right)^{\prime}}$
$\displaystyle=\frac{\partial{\rm
vec}\left(P_{kk}^{\prime}\right)}{\partial{\rm
vec}\left(P_{kk}\right)^{\prime}}\frac{\partial{\rm
vec}\left(P_{kk}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}\right)^{\prime}}\frac{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}^{\prime}\right)^{\prime}}=K_{n_{k}}\frac{\partial{\rm
vec}\left(P_{kk}\right)}{\partial{\rm
vec}\left(\tilde{\Xi}_{kk}\right)^{\prime}}K_{n_{k}}$
$\displaystyle=K_{n_{k}}\left(\tilde{\Xi}_{kk}^{\prime-1}P_{kk}\otimes
I_{n_{k}}\right)$
$\displaystyle\quad\times\left[I_{n_{k}^{2}}-\left[I_{n_{k}}\otimes\tilde{\Xi}_{kk}+\tilde{\Xi}_{kk}P_{kk}\otimes
P_{kk}^{-1}\right]^{-1}\left(I_{n_{k}^{2}}+K_{n_{k}}\right)\left(I_{n_{k}}\otimes\tilde{\Xi}_{kk}\right)\right].$

$\square$

Proof of Theorem 7. Part (i). The consistency of the MLE, $\hat{\theta}_{T}$,
follows by verifying the four conditions of Newey and McFadden, (1994, theorem
2.5). (i) Our identification assumptions ensures that $\theta\neq\theta_{0}$
implies $f\left(y_{i}|\theta\right)\neq f\left(y_{i}|\theta_{0}\right)$. (ii)
The compactness condition is Assumption 3. (iii) the log-likelihood function

$\log
f(Y|\theta)=-\tfrac{1}{2}\log|\Xi\Xi^{\prime}|+\sum_{k=1}^{K}c_{k}-\tfrac{\nu_{k}+n_{k}}{2}\log\left[1+\tfrac{1}{\nu_{k}}(Y-\mu)^{\prime}A^{\prime}e_{k}e_{k}^{\prime}A(Y-\mu)\right],$
(A.6)

where $A=\Xi^{-1}$ is continuous for $\theta\in\Theta$ with probability one.
What remains to verified is the dominance condition: (iv)
$\mathbb{E}_{\theta_{0}}\left[\sup_{\theta\in\Theta}|\log
f(y|\theta)|\right]<\infty$.

First note that

$\left|\log
f(Y|\theta)\right|\leq\left|\tfrac{1}{2}\log|\Xi\Xi^{\prime}|\right|+\sum_{k=1}^{K}|c_{k}|+R_{k}(\theta)\leq
C_{0}+\sum_{k=1}^{K}R_{k}(\theta),$

where
$C_{0}=\sup_{\theta\in\Theta}\left|\tfrac{1}{2}\log|\Xi\Xi^{\prime}|\right|+\sum_{k=1}^{K}|c_{k}|<\infty$
(since $\log|\Xi\Xi^{\prime}|$ is continuous and $\Theta$ is compact) and

$\displaystyle R_{k}(\theta)$ $\displaystyle=$
$\displaystyle\tfrac{\nu_{k}+n_{k}}{2}\log\left[1+\tfrac{1}{\nu_{k}}(Y-\mu)^{\prime}A^{\prime}e_{k}e_{k}^{\prime}A(Y-\mu)\right].$
$\displaystyle=$
$\displaystyle\tfrac{\nu_{k}+n_{k}}{2}\log\left[1+\tfrac{1}{\nu_{k}}(X-\xi)^{\prime}\Psi(X-\xi)\right]$
$\displaystyle\leq$
$\displaystyle\tfrac{\nu_{k}+n_{k}}{2}\log\left[1+\tfrac{\rho(\Psi)}{\nu_{k}}(X-\xi)^{\prime}(X-\xi)\right],$

where we substituted, $Y=\mu_{0}+\Xi_{0}X$, such that

$(Y-\mu)=(\mu_{0}+\Xi_{0}X-\mu)=\Xi_{0}(X-A_{0}(\mu-\mu_{0}))=\Xi_{0}(X-\xi),$

with $\xi=A_{0}(\mu-\mu_{0})$, $A_{0}=\Xi_{0}^{-1}$,
$\Psi_{k}=\Xi_{0}^{\prime}A^{\prime}e_{k}e_{k}^{\prime}A\Xi_{0}$, and
$\rho(\Psi_{k})$ denoting the largest eigenvalue of $\Psi_{k}$.

Next we add the positive term
$\tfrac{\rho(\Psi_{k})}{\nu_{k}}(X+\xi)^{\prime}(X+\xi)$

$\displaystyle R_{k}(\theta)$ $\displaystyle\leq$
$\displaystyle\tfrac{\nu_{k}+n_{k}}{2}\log\left[1+\tfrac{\rho(\Psi_{k})}{\nu_{k}}(X-\xi)^{\prime}(X-\xi)+\tfrac{\rho(\Psi_{k})}{\nu_{k}}(X+\xi)^{\prime}(X+\xi)\right]$
$\displaystyle=$
$\displaystyle\tfrac{\nu_{k}+n_{k}}{2}\log\left[1+2\tfrac{\rho(\Psi_{k})}{\nu_{k}}X^{\prime}X+2\tfrac{\rho(\Psi_{k})}{\nu_{k}}\xi^{\prime}\xi\right]$
$\displaystyle=$
$\displaystyle\tfrac{\nu_{k}+n_{k}}{2}\log\left[1+2\tfrac{\rho(\Psi_{k})}{\nu_{k}}\xi^{\prime}\xi+\sum_{l=1}^{K}2\tfrac{\rho(\Psi_{k})}{\nu_{k}}X_{l}^{\prime}X_{l}\right]$
$\displaystyle\leq$
$\displaystyle\tfrac{\nu_{k}+n_{k}}{2}\log\left[1+2\tfrac{\rho(\Psi_{k})}{\nu_{k}}\xi^{\prime}\xi\right]+\tfrac{\nu_{k}+n_{k}}{2}\sum_{l=1}^{K}\log\left[1+2\tfrac{\rho(\Psi_{k})}{\nu_{k}}X_{l}^{\prime}X_{l}\right],$

where we used $\log(1+a_{1}+\cdots+a_{M})\leq\sum_{j=1}^{M}\log(1+a_{j})$ for
$a_{1},\ldots,a_{M}\geq 0$. The first term is bounded by

$C_{1,k}=\sup_{\theta\in\Theta}\tfrac{\nu_{k}+n_{k}}{2}\log\left[1+2\tfrac{\rho(\Psi_{k})}{\nu_{k}}\xi^{\prime}\xi\right]<\infty,$

since $\lim_{\nu\rightarrow\infty}\nu\log(1+a/\nu)\rightarrow a<\infty$. Next,
using $\log(1+ab)\leq\log(1+a)(1+b)]$ for $a,b>0$ we find

$\displaystyle\log\left[1+2\tfrac{\rho(\Psi_{k})}{\nu_{k}}X_{l}^{\prime}X_{l}\right]$
$\displaystyle=$
$\displaystyle\log\left[1+2\tfrac{\rho(\Psi_{k})\nu_{l,0}}{\nu_{k}}\tfrac{X_{l}^{\prime}X_{l}}{\nu_{l,0}}\right]$
$\displaystyle\leq$
$\displaystyle\log\left(1+2\tfrac{\rho(\Psi_{k})\nu_{l,0}}{\nu_{k}}\right)+\log\left(1+\tfrac{X_{l}^{\prime}X_{l}}{\nu_{l,0}}\right),$

where

$C_{2,k}\equiv\sup_{\theta\in\Theta}\tfrac{\nu_{k}+n_{k}}{2}\log\left(1+2\tfrac{\rho(\Psi_{k})\nu_{l,0}}{\nu_{k}}\right)<\infty,$

and since

$\mathbb{E}_{\theta_{0}}\left[\log\left(1+\tfrac{X_{l}^{\prime}X_{l}}{\nu_{l,0}}\right)\right]=\psi\left(\tfrac{\nu_{l,0}+n_{l}}{2}\right)-\psi\left(\tfrac{\nu_{l,0}}{2}\right),$

it follows that

$\mathbb{E}_{\theta_{0}}\left[\sup_{\theta\in\Theta}\left|\log
f(Y|\theta)\right|\right]\leq
C_{0}+\sum_{k=1}^{K}C_{1,k}+C_{2,k}+\tfrac{\nu_{k}+n_{k}}{2}\sum_{l=1}^{K}\left[\psi\left(\tfrac{\nu_{l,0}+n_{l}}{2}\right)-\psi\left(\tfrac{\nu_{l,0}}{2}\right)\right],$

which proves that
$\mathbb{E}_{\theta_{0}}\left[\sup_{\theta\in\Theta}R_{k}(\theta)\right]<\infty$.

Next we prove part (ii) concerning the limit distribution. Given the four
conditions of Newey and McFadden, (1994, theorem 2.5) in the proof of
consistency, we proceed to establish the asymptotic normality by verifying the
conditions of Amemiya, (1985, theorem 2.4). These are as follows: (i)
$\frac{\partial^{2}f(y\mid\theta)}{\partial\theta\partial\theta^{\prime}}$
exists and is continuous in an open, convex neighborhood $\mathscr{N}$ of
$\theta_{0}$ (ii) $\int\frac{\partial f(y\mid\theta)}{\partial\theta}dy=0$,
(iii)
$\int\frac{\partial^{2}f\left(y|\theta\right)}{\partial\theta\partial\theta^{\prime}}dy=0$,
(iv) ${\rm plim}\frac{1}{T}\sum_{t=1}^{T}\frac{\partial^{2}\log
f(y_{t}\mid\theta)}{\partial\theta\partial\theta^{\prime}}=\mathbb{E}_{\theta_{0}}\left[\frac{\partial^{2}\log
f\left(y|\theta\right)}{\partial\theta\partial\theta^{\prime}}\right]$
uniformly in $\theta$ in an open neighborhood of $\theta_{0}$. Then, we have
$\sqrt{T}\left(\hat{\theta}_{T}-\theta_{0}\right)\rightarrow
N\left\\{0,\mathcal{J}^{-1}\right\\}$, where
$\mathcal{J}=-\mathbb{E}_{\theta_{0}}\left[\left.\frac{\partial^{2}\log
f\left(y|\theta\right)}{\partial\theta\partial\theta^{\prime}}\right|_{\theta=\theta_{0}}\right]$.

Note that because $\mathscr{N}\subset\Theta$ where $\Theta$ is the parameter
sets, it is sufficient to verify the Conditions (i), (ii), (iii), and (iv)
hold for $\forall\theta\in\Theta$ and $\theta_{0}$ is interior to $\Theta$,
where $\Theta$ is compact by our assumption in the proof of consistency.
Before giving a formal proof, we first define the random vector
$Z=Z(Y|\theta)$, $Z_{k}=Z_{k}(Y|\theta)$, and random variable
$S_{k}=S_{k}\left(Y|\theta\right)$, given by

$Z=A\left(Y-\mu\right),\quad Z_{k}=e_{k}^{\prime}Z,\quad
S_{k}=\tfrac{\nu_{k}+n_{k}}{\nu_{k}+Z_{k}^{\prime}Z_{k}}$

where $A=\Xi^{-1}$ and $Y=\mu_{0}+\Xi_{0}X$. When $\mu=\mu_{0}$ and
$\Xi=\Xi_{0}$, we have $Z=X$, $Z_{k}=X_{k}$, and
$S_{k}=W_{k}=\tfrac{\nu_{k}+n_{k}}{\nu_{k}+X_{k}^{\prime}X_{k}}$.

Condition (i): That $\theta_{0}$ is interior to $\Theta$ is assumed. We prove
that
$\frac{\partial^{2}f(y\mid\theta)}{\partial\theta\partial\theta^{\prime}}$
exists and is continuous for $\forall\theta\in\Theta$ directly by deriving the
expression for $\frac{\partial^{2}\log
f\left(Y|\theta\right)}{\partial\theta\partial\theta^{\prime}}$. The form of
such hessian matrix can be easily adapted from e Theorem 5 by replacing $X$,
$X_{k}$, and $W_{k}$ with $Z$, $Z_{k}$, and $S_{k}$.

Condition (ii): It is sufficient to show that for $\forall\theta\in\Theta$, we
have

$\displaystyle\int\frac{\partial f(y|\theta)}{\partial\theta}\mathrm{d}y$
$\displaystyle=\int\nabla_{\theta}f(Y|\theta)\mathrm{d}Y=\mathbb{E}_{\theta}\left[\nabla_{\theta}\right]=0,\quad\nabla_{\theta}=\frac{\partial\log
f\left(Y|\theta\right)}{\partial\theta}$

Adapted from Theorem 5, we can easily obtain that
$\mathbb{E}_{\theta}\left[\nabla_{\theta}\right]=0$ when $\min_{k}\nu_{k}>1$.
Note that
$\mathbb{E}\left[\nabla_{\Xi}\right]=\mathbb{E}\left[M_{\Xi}^{+\prime}\nabla_{\Xi}\right]=M_{\Xi}^{+\prime}\mathbb{E}\left[\nabla_{\Xi}\right]=0$.

Condition (iii): It is sufficient to show that for $\forall\ \theta\in\Theta$,
we have

$\displaystyle\int\frac{\partial^{2}f\left(y|\theta\right)}{\partial\theta\partial\theta^{\prime}}dy$
$\displaystyle=\int\left[\nabla_{\theta}\nabla_{\theta}^{\prime}+\nabla_{\theta\theta^{\prime}}\right]f(y\mid\theta)\mathrm{d}y=\underbrace{\mathbb{E}_{\theta}\left[\nabla_{\theta}\nabla_{\theta}^{\prime}\right]}_{\mathcal{I}^{\theta}}+\underbrace{\mathbb{E}_{\theta}\left[\nabla_{\theta\theta^{\prime}}\right]}_{-\mathcal{J^{\theta}}}=0.$

This requires to verify the existences both of the Fisher information matrix
$\mathcal{I}^{\theta}=\mathbb{E}_{\theta}\left[\nabla_{\theta}\nabla_{\theta}^{\prime}\right]$
and the expectation of Hessian matrix
$\mathcal{J^{\theta}}=-\mathbb{E}_{\theta}\left[\nabla_{\theta\theta^{\prime}}\right]$,
and show that $\mathcal{I^{\theta}}=\mathcal{J^{\theta}}$. Such results can be
easily verified by adapting from Theorem 5, which requires
$\min_{k}\nu_{k}>2$. Note that for $\Xi$, instead of $\tilde{\Xi}$, we have

$\displaystyle\mathcal{I}_{\Xi}$
$\displaystyle=\mathbb{E}\left[\nabla_{\Xi}\nabla_{\Xi}^{\prime}\right]=M_{\Xi}^{+\prime}\mathbb{E}\left[\nabla_{\tilde{\Xi}}\nabla_{\tilde{\Xi}}^{\prime}\right]M_{\Xi}^{+}=M_{\Xi}^{+\prime}\mathcal{I}_{\tilde{\Xi}}M_{\Xi}^{+}$
$\displaystyle\mathcal{J}_{\Xi}$
$\displaystyle=\mathbb{E}\left[\nabla_{\Xi\Xi^{\prime}}\right]=M_{\Xi}^{+\prime}\mathbb{E}\left[\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}\right]M_{\Xi}^{+\prime}-\left(\mathbb{E}(\nabla_{\tilde{\Xi}}^{\prime})\otimes
I_{n}\right)\frac{\partial{\rm vec}(M_{\Xi}^{+\prime})}{\partial{\rm
vec}(\Xi)^{\prime}}=M_{\Xi}^{+\prime}\mathcal{J}_{\tilde{\Xi}}M_{\Xi}^{+},$

and from $\mathcal{I}_{\tilde{\Xi}}=\mathcal{J}_{\tilde{\Xi}}$, we have
$\mathcal{I}_{\Xi}=\mathcal{J}_{\Xi}$.

Condition (iv): If there exist a $g(y)$ function with
$\|\frac{\partial^{2}\log
f\left(y|\theta\right)}{\partial\theta\partial\theta^{\prime}}\|\leqslant
g(y)$ for all $\theta\in\Theta$ and
$\mathbb{E}_{\theta_{0}}\left[g(z)\right]<\infty$, then we have ${\rm plim}\
\frac{1}{T}\sum_{t=1}^{T}\frac{\partial^{2}\log
f(y_{t}\mid\theta)}{\partial\theta\partial\theta^{\prime}}=\mathbb{E}_{\theta_{0}}\left[\frac{\partial^{2}\log
f\left(y|\theta\right)}{\partial\theta\partial\theta^{\prime}}\right]$
uniformly in $\theta\in\Theta$, see e.g. Newey and McFadden, (1994, lemma
2.4). This section, we determine the dominating function, $g(y)$.

First, adapted from Theorem 5, by replacing $X$, $X_{k}$, and $W_{k}$ with
$Z$, $Z_{k}$, and $S_{k}$, the expression of Hessian matrix of $\log
f\left(Y|\theta\right)$ with respective to $\theta$ are given by

$\displaystyle\nabla_{\mu\mu^{\prime}}^{\theta}$
$\displaystyle=\sum_{k=1}^{K}\tfrac{2S_{k}^{2}}{\nu_{k}+n_{k}}A^{\prime}e_{k}Z_{k}Z_{k}^{\prime}e_{k}^{\prime}A-S_{k}A^{\prime}J_{k}A$
$\displaystyle\nabla_{\mu\tilde{\Xi}^{\prime}}^{\theta}$
$\displaystyle=\sum_{k=1}^{K}S_{k}\left(AZ_{k}^{\prime}e_{k}^{\prime}\otimes
A^{\prime}\right)K_{n}+\tfrac{2S_{k}^{2}}{\nu_{k}+n_{k}}A^{\prime}e_{k}Z_{k}{\rm
vec}\left(A^{\prime}e_{k}Z_{k}Z^{\prime}\right)^{\prime}-S_{k}A^{\prime}\left(Z^{\prime}\otimes
J_{k}A\right)$ $\displaystyle\nabla_{\mu\nu_{k}}^{\theta}$
$\displaystyle=\sum_{k=1}^{K}\tfrac{1}{\nu_{k}+n_{k}}A^{\prime}e_{k}Z_{k}\left(S_{k}-S_{k}^{2}\right)$
$\displaystyle\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}^{\theta}$
$\displaystyle=-\left(A\otimes
A^{\prime}\right)K_{n}+\sum_{k=1}^{K}\tfrac{2S_{k}^{2}}{\nu_{k}+n_{k}}{\rm
vec}\left(A^{\prime}e_{k}Z_{k}Z^{\prime}\right){\rm
vec}\left(A^{\prime}e_{k}Z_{k}Z^{\prime}\right)^{\prime}$
$\displaystyle\quad+\sum_{k=1}^{K}S_{k}\left[\left(ZZ_{k}^{\prime}e_{k}^{\prime}A\otimes
A^{\prime}\right)K_{n}-\left(ZZ^{\prime}\otimes
A^{\prime}J_{k}A\right)-\left(A\otimes
A^{\prime}e_{k}Z_{k}Z^{\prime}\right)K_{n}\right]$
$\displaystyle\nabla_{\tilde{\Xi}\nu_{k}}^{\theta}$
$\displaystyle=\tfrac{1}{\nu_{k}+n_{k}}\left(S_{k}-S_{k}^{2}\right){\rm
vec}\left(A^{\prime}e_{k}Z_{k}Z^{\prime}\right)$
$\displaystyle\nabla_{\nu_{k}\nu_{k}}^{\theta}$
$\displaystyle=\tfrac{1}{4}\psi^{\prime}\left(\tfrac{v_{k}+n_{k}}{2}\right)-\tfrac{1}{4}\psi^{\prime}\left(\tfrac{v_{k}}{2}\right)+\tfrac{1}{2\nu_{k}}+\tfrac{1}{2}\tfrac{1}{\nu_{k}+n_{k}}S_{k}^{2}-\tfrac{1}{\left(\nu_{k}+n_{k}\right)}S_{k}$
$\displaystyle\nabla_{\nu_{k}\nu_{l}}^{\theta}$ $\displaystyle=0,\quad{\rm
for}\quad k\neq l,$

where the $\theta$ in the superscript means that the Score and Hessian matrix
is evaluated at $\theta$.

Note that we have
$\nabla_{\mu\Xi^{\prime}}^{\theta}=\nabla_{\mu\tilde{\Xi}^{\prime}}^{\theta}M_{\Xi}^{+}$,
and
$\nabla_{\Xi\Xi^{\prime}}^{\theta}=M_{\Xi}^{+\prime}\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}M_{\Xi}^{+\prime}+\left(\nabla_{\tilde{\Xi}}^{\prime}\otimes
I_{n}\right)\frac{\partial{\rm vec}(M_{\Xi}^{+\prime})}{\partial{\rm
vec}(\Xi)^{\prime}}$, and
$\nabla_{\Xi\nu_{k}}^{\theta}=M_{\Xi}^{+\prime}\nabla_{\tilde{\Xi}\nu_{k}}^{\theta}$,
and from the following inequalities

$\displaystyle\left\|\nabla_{\mu\Xi^{\prime}}^{\theta}\right\|$
$\displaystyle\leq\left\|\nabla_{\mu\tilde{\Xi}^{\prime}}^{\theta}\right\|\left\|M_{\Xi}^{+}\right\|,\quad\left\|\nabla_{\Xi\nu_{k}}^{\theta}\right\|=\left\|M_{\Xi}^{+\prime}\right\|\left\|\nabla_{\tilde{\Xi}\nu_{k}}^{\theta}\right\|$
$\displaystyle\left\|\nabla_{\Xi\Xi^{\prime}}^{\theta}\right\|$
$\displaystyle\leq\left\|M_{\Xi}^{+}\right\|^{2}\left\|\nabla_{\tilde{\Xi}\tilde{\Xi}^{\prime}}\right\|+n\left\|\nabla_{\tilde{\Xi}}\right\|\left\|\tfrac{\partial{\rm
vec}(M_{\Xi}^{+\prime})}{\partial{\rm vec}(\Xi)^{\prime}}\right\|,$

and because
$\left\|\nabla_{\tilde{\Xi}}\right\|\leq\sum_{k=1}^{K}\left\|A^{\prime}e_{k}\right\|\left\|S_{k}Z_{k}Z^{\prime}\right\|+\left\|A^{\prime}\right\|$,
and from Equation (A.8), we have

$\mathbb{E}_{\theta_{0}}\sup_{\theta\in\Theta}\left|S_{k}Z_{k,i}Z_{l,j}\right|<\infty,\quad{\rm
if\ all}\ \nu_{k}>1$

It is enough to determine the dominating function, $g(y)$ under
$\theta=\\{\mu,\tilde{\Xi},\bm{\nu}\\}$, instead of
$\theta=\\{\mu,\Xi,\bm{\nu}\\}$.

Next we establish bounds for each of the terms. First, we have

$\displaystyle\left\|\nabla_{\mu\mu^{\prime}}^{\theta}\right\|$
$\displaystyle=\left\|\sum_{k=1}^{K}\tfrac{2S_{k}^{2}}{\nu_{k}+n_{k}}A^{\prime}e_{k}Z_{k}Z_{k}^{\prime}e_{k}^{\prime}A-S_{k}A^{\prime}J_{k}A\right\|$
$\displaystyle\leq\sum_{k=1}^{K}\tfrac{2}{\nu_{k}+n_{k}}\left\|A^{\prime}e_{k}\right\|^{2}\left\|S_{k}^{2}Z_{k}Z_{k}^{\prime}\right\|+S_{k}\left\|A^{\prime}J_{k}A\right\|$
$\displaystyle\leq\sum_{k=1}^{K}\tfrac{2}{\nu_{k}+n_{k}}\left\|A^{\prime}e_{k}\right\|^{2}\left\|S_{k}^{2}Z_{k}Z_{k}^{\prime}\right\|+\tfrac{\nu_{k}+n_{k}}{\nu_{k}}\left\|A^{\prime}J_{k}A\right\|$

which we use the fact that
$S_{k}=\frac{\nu_{k}+n_{k}}{\nu_{k}+Z_{k}^{\prime}Z_{k}}\leq\frac{\nu_{k}+n_{k}}{\nu_{k}}$.
As for the $S_{k}^{2}Z_{k}Z_{k}^{\prime}$, it involves the following terms

$\displaystyle\left|S_{k}^{2}Z_{k,i}Z_{k,j}\right|$
$\displaystyle=\tfrac{\left(\nu_{k}+n_{k}\right)^{2}}{\left(\nu_{k}+\sum_{p}Z_{k,p}^{2}\right)^{2}}\left|Z_{k,i}Z_{k,j}\right|$
$\displaystyle\leq\tfrac{\left(\nu_{k}+n_{k}\right)^{2}}{\nu_{k}\left(\nu_{k}+\sum_{p}Z_{k,p}^{2}\right)}\left|Z_{k,i}Z_{k,j}\right|,\quad{\rm
use}\ \tfrac{1}{\nu_{k}+\sum_{p}Z_{k,p}^{2}}<\tfrac{1}{\nu_{k}}$
$\displaystyle\leq\tfrac{\left(\nu_{k}+n_{k}\right)^{2}}{\nu_{k}\left(\nu_{k}+\sum_{p}Z_{k,p}^{2}\right)}\left|Z_{k}\right|_{\max}^{2},\quad\left|Z_{k}\right|_{\max}=\max_{i=1,\ldots,n_{k}}\
\left|Z_{k,i}\right|$
$\displaystyle\leq\tfrac{\left(\nu_{k}+n_{k}\right)^{2}}{\nu_{k}\left(\nu_{k}+\left|Z_{k}\right|_{\max}^{2}\right)}\left|Z_{k}\right|_{\max}^{2}$
$\displaystyle\leq\tfrac{1}{\nu_{k}}\left(\nu_{k}+n_{k}\right)^{2}$

So the term
$\left\|A^{\prime}e_{k}\left(S_{k}^{2}Z_{k}Z_{k}^{\prime}\right)e_{k}^{\prime}A\right\|$
is bounded. This leads to $\left\|\nabla_{\mu\mu^{\prime}}^{\theta}\right\|$
is bounded.

Next, we have

$\displaystyle\left\|\nabla_{\mu\tilde{\Xi^{\prime}}}^{\theta}\right\|$
$\displaystyle=\left\|\sum_{k=1}^{K}S_{k}\left(AZ_{k}^{\prime}e_{k}^{\prime}\otimes
A^{\prime}\right)K_{n}+\tfrac{2S_{k}^{2}}{\nu_{k}+n_{k}}A^{\prime}e_{k}Z_{k}{\rm
vec}\left(A^{\prime}e_{k}Z_{k}Z^{\prime}\right)^{\prime}-S_{k}A^{\prime}\left(Z^{\prime}\otimes
J_{k}A\right)\right\|$
$\displaystyle\leq\sum_{k=1}^{K}\left\|S_{k}\left(AZ_{k}^{\prime}e_{k}^{\prime}\otimes
A^{\prime}\right)K_{n}\right\|+\tfrac{2S_{k}^{2}}{\nu_{k}+n_{k}}\left\|A^{\prime}e_{k}Z_{k}{\rm
vec}\left(A^{\prime}e_{k}Z_{k}Z^{\prime}\right)^{\prime}\right\|+\left\|S_{k}A^{\prime}\left(Z^{\prime}\otimes
J_{k}A\right)\right\|,$

where the first term involves

$\displaystyle\left|S_{k}Z_{k,i}\right|$
$\displaystyle=\tfrac{\nu_{k}+n_{k}}{\nu_{k}+\sum_{p}Z_{k,p}^{2}}\left|Z_{k,i}\right|$
$\displaystyle\leq\tfrac{\nu_{k}+n_{k}}{\nu_{k}+Z_{k,i}^{2}}\left|Z_{k,i}\right|$
$\displaystyle\leq\tfrac{\nu_{k}+n_{k}}{2\sqrt{\nu_{k}}\left|Z_{k,i}\right|}\left|Z_{k,i}\right|$
$\displaystyle=\tfrac{\nu_{k}+n_{k}}{2\sqrt{\nu_{k}}},$ (A.7)

the second term involves

$\displaystyle\left|S_{k}^{2}Z_{k,p}Z_{k,i}Z_{l,j}\right|$
$\displaystyle=S_{k}^{2}\left|Z_{k,p}Z_{k,i}\right|\left|Z_{l,j}\right|$
$\displaystyle\leq\left(\nu_{k}+n_{k}\right)^{2}\left|Z_{l,j}\right|$
$\displaystyle=\left(\nu_{k}+n_{k}\right)^{2}\left|\tilde{e}_{l,i}Z\right|,\quad\tilde{e}_{l,i}Z=Z_{l,j}$
$\displaystyle=\left(\nu_{k}+n_{k}\right)^{2}\left|\tilde{e}_{l,i}A\left(Y-\mu\right)\right|$
$\displaystyle=\left(\nu_{k}+n_{k}\right)^{2}\left|\tilde{e}_{l,i}A\Xi_{0}\left(X-\xi\right)\right|,\quad\xi=A_{0}(\mu-\mu_{0})$
$\displaystyle\leq\left(\nu_{k}+n_{k}\right)^{2}\left[\left|\tilde{e}_{l,i}A\Xi_{0}X\right|+\left|\tilde{e}_{l,i}A\Xi_{0}\xi\right|\right]$
$\displaystyle\leq
C_{1}\left(\theta\right)+\sum_{k=1}^{K}C_{2}\left(\theta\right)\left|X_{k}\right|,\quad
C_{1}\left(\theta\right),C_{2}\left(\theta\right)>0$ $\displaystyle\leq
C_{1}^{*}+\sum_{k=1}^{K}C_{2}^{*}\left|X_{k}\right|,\quad
C_{i}^{*}=\sup_{\theta\in\Theta}C_{i}\left(\theta\right)$
$\displaystyle\sup_{\theta\in\Theta}\left|S_{k}^{2}Z_{k,p}Z_{k,i}Z_{l,j}\right|$
$\displaystyle\leq C_{1}^{*}+\sum_{k=1}^{K}C_{2}^{*}\left|X_{k}\right|$
$\displaystyle\mathbb{E}_{\theta_{0}}\sup_{\theta\in\Theta}\left|S_{k}^{2}Z_{k,p}Z_{k,i}Z_{l,j}\right|$
$\displaystyle\leq
C_{1}^{*}+\sum_{k=1}^{K}C_{2}^{*}\mathbb{E}_{\theta_{0}}\left[\left|X_{k}\right|\right]<\infty,\quad{\rm
if\ all}\ \nu_{k}>1,$

and the third term and final term involves

$\displaystyle\left|S_{k}Z_{l,i}\right|$
$\displaystyle\leq\tfrac{\nu_{k}+n_{k}}{\nu_{k}}\left|Z_{l,i}\right|\leq\tilde{C}_{1}+\sum_{k=1}^{K}\tilde{C}_{2}\left|X_{k}\right|,$
$\displaystyle\mathbb{E}_{\theta_{0}}\sup_{\theta\in\Theta}\left|S_{k}^{2}Z_{k,p}Z_{k,i}Z_{l,j}\right|$
$\displaystyle<\tilde{C}_{1}+\sum_{k=1}^{K}\tilde{C}_{2}\mathbb{E}_{\theta_{0}}\left[\left|X_{k}\right|\right]<\infty,\quad{\rm
if\ all}\ \nu_{k}>1.$

So, we have shown that
$\mathbb{E}_{\theta_{0}}\left\|\sup_{\theta\in\Theta}\nabla_{\mu\Xi^{\prime}}^{\theta}\right\|<\infty$
for all $\nu_{k}>1$.

Next, we have

$\displaystyle\left\|\nabla_{\mu\nu_{k}}^{\theta}\right\|$
$\displaystyle=\left\|\sum_{k=1}^{K}\tfrac{1}{\nu_{k}+n_{k}}A^{\prime}e_{k}Z_{k}\left(S_{k}-S_{k}^{2}\right)\right\|$
$\displaystyle\leq\sum_{k=1}^{K}\tfrac{1}{\nu_{k}+n_{k}}\left\|A^{\prime}e_{k}S_{k}Z_{k}\right\|+\tfrac{1}{\nu_{k}+n_{k}}\left\|A^{\prime}e_{k}S_{k}^{2}Z_{k}\right\|.$

Again, the first term involves

$\left|S_{k}Z_{k}\right|=\tfrac{\nu_{k}+n_{k}}{\nu_{k}+Z_{k}^{\prime}Z_{k}}\left|Z_{k}\right|\leq\tfrac{\nu_{k}+n_{k}}{2\sqrt{\nu_{k}}\left|Z_{k}\right|}\left|Z_{k}\right|=\tfrac{\nu_{k}+n_{k}}{2\sqrt{\nu_{k}}}$

The second term involves

$\left|S_{k}^{2}Z_{k}\right|=\tfrac{\left(\nu_{k}+n_{k}\right)^{2}}{\left(\nu_{k}+Z_{k}^{\prime}Z_{k}\right)^{2}}\left|Z_{k}\right|\leq\tfrac{1}{\nu_{k}}\tfrac{\left(\nu_{k}+n_{k}\right)^{2}}{\left(\nu_{k}+Z_{k}^{\prime}Z_{k}\right)}\left|Z_{k}\right|\leq\tfrac{1}{\nu_{k}}\tfrac{\nu_{k}+n_{k}}{2\sqrt{\nu_{k}}\left|Z_{k}\right|}\left|Z_{k}\right|=\tfrac{1}{\nu_{k}}\tfrac{\nu_{k}+n_{k}}{2\sqrt{\nu_{k}}},$

which shows that $\left\|\nabla_{\mu\nu_{k}}^{\theta}\right\|$ is bounded for
all $\nu_{k}>0$.

Next, we have

$\displaystyle\nabla_{\tilde{\Xi}\tilde{\Xi^{\prime}}}^{\theta}$
$\displaystyle=-\left(A\otimes
A^{\prime}\right)K_{n}+\sum_{k=1}^{K}\tfrac{2S_{k}^{2}}{\nu_{k}+n_{k}}{\rm
vec}\left(A^{\prime}e_{k}Z_{k}Z^{\prime}\right){\rm
vec}\left(A^{\prime}e_{k}Z_{k}Z^{\prime}\right)^{\prime}$
$\displaystyle\quad+\sum_{k=1}^{K}S_{k}\left[\left(ZZ_{k}^{\prime}e_{k}^{\prime}A\otimes
A^{\prime}\right)K_{n}-\left(ZZ^{\prime}\otimes
A^{\prime}J_{k}A\right)-\left(A\otimes
A^{\prime}e_{k}Z_{k}Z^{\prime}\right)K_{n}\right].$

The first term involves

$\displaystyle\left|S_{k}^{2}Z_{k,i}Z_{k,j}Z_{l,p}Z_{m,q}\right|$
$\displaystyle=\tfrac{\left(\nu_{k}+n_{k}\right)^{2}}{\left(\nu_{k}+Z_{k}^{\prime}Z_{k}\right)^{2}}\left|Z_{k,i}Z_{k,j}Z_{l,p}Z_{m,q}\right|$
$\displaystyle\leq\tfrac{\left(\nu_{k}+n_{k}\right)^{2}}{\nu_{k}}\left|Z_{l,p}Z_{m,q}\right|$
$\displaystyle=\tfrac{\left(\nu_{k}+n_{k}\right)^{2}}{\nu_{k}}\left|\tilde{e}_{l,p}ZZ^{\prime}\tilde{e}_{m,q}^{\prime}\right|$
$\displaystyle=\tfrac{\left(\nu_{k}+n_{k}\right)^{2}}{\nu_{k}}\left|\tilde{e}_{l,p}A\Xi_{0}\left(X-\xi\right)\left(X-\xi\right)^{\prime}\Xi_{0}^{\prime}A^{\prime}\tilde{e}_{m,q}^{\prime}\right|$
$\displaystyle=\tfrac{\left(\nu_{k}+n_{k}\right)^{2}}{\nu_{k}}\left|\tilde{e}_{l,p}A\Xi_{0}\left(XX^{\prime}-\xi
X^{\prime}-X\xi^{\prime}+\xi\xi^{\prime}\right)\Xi_{0}^{\prime}A^{\prime}\tilde{e}_{m,q}^{\prime}\right|,\quad\Psi_{l,p}=\tilde{e}_{l,p}A\Xi_{0}$
$\displaystyle\leq
C_{0}+C_{1}\left|\Psi_{l,p}XX^{\prime}\Psi_{l,p}^{\prime}\right|+C_{2}\left|\Psi_{l,p}\xi
X^{\prime}\Psi_{l,p}^{\prime}\right|+C_{3}\left|\Psi_{l,p}X\xi^{\prime}\Psi_{l,p}^{\prime}\right|$
$\displaystyle\leq
C_{0}^{*}+C_{1}^{*}\sum_{i}\sum_{j}\left|X_{i}X_{j}\right|+C_{1}^{2}\sum_{i}\left|X_{i}\right|.$

Thus, if $\nu_{k}>2$ for all $k$, we have

$\mathbb{E}_{\theta_{0}}\sup_{\theta\in\Theta}\left|S_{k}^{2}Z_{k,i}Z_{k,j}Z_{l,p}Z_{m,q}\right|\leq
C_{0}^{*}+C_{1}^{*}\sum_{i}\sum_{j}\mathbb{E}_{\theta_{0}}\left|X_{i}X_{j}\right|+C_{1}^{2}\sum_{i}\mathbb{E}_{\theta_{0}}\left|X_{i}\right|<\infty.$

The second term involves

$\displaystyle\left|S_{k}Z_{k,i}Z_{l,j}\right|$
$\displaystyle=\left|S_{k}Z_{k,i}\right|\left|Z_{l,j}\right|\leq\tfrac{\nu_{k}+n_{k}}{2\sqrt{\nu_{k}}}\left|Z_{l,j}\right|$
$\displaystyle\leq C_{1}^{*}+\sum_{k=1}^{K}C_{2}^{*}\left|X_{k}\right|,$
$\displaystyle\mathbb{E}_{\theta_{0}}\sup_{\theta\in\Theta}\left|S_{k}Z_{k,i}Z_{l,j}\right|$
$\displaystyle\leq
C_{1}^{*}+\sum_{k=1}^{K}C_{2}^{*}\mathbb{E}_{\theta_{0}}\left|X_{k}\right|<\infty,\quad\quad{\rm
if}\ \min_{k}\nu_{k}>1.$ (A.8)

The third term, $S_{k}\left(ZZ^{\prime}\otimes A^{\prime}J_{k}A\right)$,
involves

$\displaystyle\left|S_{k}Z_{p,i}Z_{q,j}\right|$
$\displaystyle\leq\tfrac{\nu_{k}+n_{k}}{\nu_{k}}\left|Z_{p,i}Z_{q,j}\right|$
$\displaystyle\leq
C_{0}^{*}+C_{1}^{*}\sum_{i}\sum_{j}\left|X_{i}X_{j}\right|+C_{1}^{2}\sum_{i}\left|X_{i}\right|,$
$\displaystyle\mathbb{E}_{\theta_{0}}\sup_{\theta\in\Theta}\left|S_{k}Z_{p,i}Z_{q,j}\right|$
$\displaystyle\leq
C_{0}^{*}+C_{1}^{*}\sum_{i}\sum_{j}\mathbb{E}_{\theta_{0}}\left|X_{i}X_{j}\right|+C_{1}^{2}\sum_{i}\mathbb{E}_{\theta_{0}}\left|X_{i}\right|$
$\displaystyle<\infty,\quad{\rm if}\ \min_{k}\nu_{k}>2$

Finally, the fourth term, $S_{k}\left(A\otimes
A^{\prime}e_{k}Z_{k}Z^{\prime}\right)K_{n}$, involves
$\left|S_{k}Z_{k,i}Z_{l,j}\right|$, which satisfies
$\mathbb{E}_{\theta_{0}}\sup_{\theta\in\Theta}\left|S_{k}Z_{k,i}Z_{l,j}\right|<\infty$,
if $\min_{k}\nu_{k}>1$.

Next, we have

$\displaystyle\left\|\nabla_{\tilde{\Xi}\nu_{k}}^{\theta}\right\|$
$\displaystyle=\tfrac{1}{\nu_{k}+n_{k}}\left\|\left(S_{k}-S_{k}^{2}\right){\rm
vec}\left(A^{\prime}e_{k}Z_{k}Z^{\prime}\right)\right\|$
$\displaystyle\leq\tfrac{1}{\nu_{k}+n_{k}}\left\|S_{k}{\rm
vec}\left(A^{\prime}e_{k}Z_{k}Z^{\prime}\right)\right\|+\tfrac{1}{\nu_{k}+n_{k}}\left\|S_{k}^{2}{\rm
vec}\left(A^{\prime}e_{k}Z_{k}Z^{\prime}\right)\right\|,$

which involves $\left|S_{k}Z_{k,i}Z_{l,k}\right|$ and
$\left|S_{k}^{2}Z_{k,i}Z_{l,k}\right|$. We need $\nu_{k}>1$ to ensure
$\mathbb{E}_{\theta_{0}}\sup_{\theta\in\Theta}\left\|\nabla_{\tilde{\Xi}\nu_{k}}^{\theta}\right\|<\infty$
as shown earlier.

Nest, we have

$\displaystyle\left|\nabla_{\nu_{k}\nu_{k}}^{\theta}\right|$
$\displaystyle=\left|\tfrac{1}{2}\left[\psi^{\prime}\left(\tfrac{v_{k}+n_{k}}{2}\right)-\tfrac{1}{2}\psi^{\prime}\left(\tfrac{v_{k}}{2}\right)\right]+\tfrac{1}{\nu_{k}+n_{k}}\left(S_{k}^{2}-S_{k}\right)+\tfrac{1}{\nu_{k}}-\tfrac{1}{\nu_{k}+Z_{k}^{\prime}Z_{k}}\right|$
$\displaystyle\leq\tfrac{1}{2}\psi^{\prime}\left(\tfrac{v_{k}+n_{k}}{2}\right)+\tfrac{1}{2}\psi^{\prime}\left(\tfrac{v_{k}}{2}\right)+\tfrac{1}{\nu_{k}+n_{k}}S_{k}^{2}+\tfrac{1}{\nu_{k}+n_{k}}S_{k}+\tfrac{1}{\nu_{k}}+\tfrac{1}{\nu_{k}+Z_{k}^{\prime}Z_{k}}$
$\displaystyle\leq\tfrac{1}{2}\psi^{\prime}\left(\tfrac{v_{k}+n_{k}}{2}\right)+\tfrac{1}{2}\psi^{\prime}\left(\tfrac{v_{k}}{2}\right)+\tfrac{1}{\nu_{k}+n_{k}}\tfrac{\left(\nu_{k}+n_{k}\right)^{2}}{\nu_{k}^{2}}+\tfrac{1}{\nu_{k}+n_{k}}\tfrac{\nu_{k}+n_{k}}{\nu_{k}}+\tfrac{1}{\nu_{k}}+\tfrac{1}{\nu_{k}}$
$\displaystyle<\infty.$

This verifies condition (iv) and completes the proof.

$\square$

## Supplemental Material for ‘Convolution-$t$ Distributions’

### General Moments of Convolution-$t$ Distributions

The $r$-th moment of $Y_{1}=\mu+\beta^{\prime}X$ exists if and only if
$r<\nu_{\min}\equiv\min_{k}(\nu_{k})$, and all odd moments, less than
$\nu_{\min}$, are zero because a convolution of symmetric distributions is
symmetric. I.e. $\mathbb{E}\left(Y_{1}-\mu\right)^{r}=0$ for $r=2m-1$ for some
$m\in\mathbb{N}$, provided $r<\nu_{\min}$. We derived the 2nd and 4th
centralized moments, $\mathbb{E}(Y_{1}-\mu)^{2}$ and
$\mathbb{E}(Y_{1}-\mu)^{4}$, in Theorem 2.

The $r$-th moment of $Y_{1}=\mu+\beta^{\prime}X$ exists if and only if
$r<\nu_{\min}\equiv\min_{k}(\nu_{k})$, and all odd moments, less than
$\nu_{\min}$, are zero because a convolution of symmetric distributions is
symmetric. General moments, including fractional moments, $r>0$, can be
obtained with the method in Kawata, (1972, Theorem 11.4.4), which states,

$\mathbb{E}|Y_{1}|^{r}=C_{p,r}\int_{0}^{\infty}s^{-(1+r)}\left[-{\rm
Re}[\varphi_{Y_{1}}(s)]+\sum_{k=0}^{p}\frac{s^{2k}}{(2k)!}\varphi_{Y_{1}}^{(2k)}(0)\right]\mathrm{d}s,$

where $p=\lfloor r/2\rfloor$ and $C_{p,r}$ is the positive constant given by

$C_{p,r}=\int_{0}^{\infty}\left[L_{p}(x)/x^{1+r}\right]\mathrm{d}x,\quad
L_{p}(x)=-\cos x+\sum_{k=0}^{p}(-1)^{k}\frac{x^{2k}}{(2k)!}.$

For small moments, $0<r<2$, this expression simplifies to

$\mathbb{E}|Y_{1}|^{r}=\frac{r(1-r)}{\Gamma(2-r)}\frac{1}{\sin\left[(1-r)\pi/2\right]}\int_{0}^{\infty}\frac{1-{\rm
Re}[\varphi_{Y_{1}}(s)]}{s^{1+r}}\mathrm{d}s.$

The coefficient of the integral on the right-hand side is interpreted to be
$2/\pi$ when $r=1$.

And because
$\varphi_{Y_{1}}(s)=e^{is\mu}\prod_{k=1}^{K}\phi_{\nu_{k}}(\omega_{k}s)$ with

$\phi_{\nu}(s)=\frac{K_{\frac{\nu}{2}}(\sqrt{\nu}|s|)(\sqrt{\nu}|s|)^{\frac{1}{2}\nu}}{\Gamma\left(\frac{\nu}{2}\right)2^{\frac{\nu}{2}-1}},\qquad\text{for}\quad
s\in\mathbb{R},$

The derivatives of $\varphi_{Y_{1}}\left(s\right)$ with respect to $s$ uses
the following formula

$\frac{\partial K_{\nu}(z)}{\partial z}=\frac{\nu}{z}K_{\nu}(z)-K_{\nu+1}(z).$

### Simulation Results for Non-Standard Situations

#### Simulations with Small Samples

In Table S.1, we conduct a simulation study with $T=50,100,200,400$, based on
50,000 Monte-Carlo simulations. Distributional specification is the same as in
Section 3.4.

Table S.1: Simulations with Small Samples

|  True |  |  Mean |  Std |  aStd |  $\alpha_{L}^{0.025}$ |  $\alpha_{R}^{0.025}$ |  |  Mean |  Std |  aStd |  $\alpha_{L}^{0.025}$ |  $\alpha_{R}^{0.025}$
---|---|---|---|---|---|---|---|---|---|---|---|---|---
|  |  | $T=50$ |  | $T=100$
$\mu_{1}$ |  0.1000 |  |  0.1006 |  0.1170 |  0.1117 |  0.0309 |  0.0315 |  |  0.1006 |  0.0811 |  0.0790 |  0.0284 |  0.0293
$\mu_{2}$ |  0.2000 |  |  0.2011 |  0.1457 |  0.1404 |  0.0290 |  0.0302 |  |  0.2006 |  0.1013 |  0.0993 |  0.0270 |  0.0284
$\mu_{3}$ |  0.3000 |  |  0.3007 |  0.1486 |  0.1442 |  0.0283 |  0.0287 |  |  0.3004 |  0.1035 |  0.1020 |  0.0263 |  0.0263
$\rm{vec}(\Xi)$ |  0.6000 |  |  0.5119 |  0.1434 |  0.1203 |  0.1352 |  0.0109 |  |  0.5562 |  0.1057 |  0.0851 |  0.0953 |  0.0185
0.5000 |  |  0.4265 |  0.2029 |  0.1581 |  0.1072 |  0.0295 |  |  0.4630 |  0.1562 |  0.1118 |  0.0908 |  0.0429
0.4000 |  |  0.3398 |  0.2261 |  0.1645 |  0.1031 |  0.0424 |  |  0.3708 |  0.1735 |  0.1163 |  0.0897 |  0.0572
0.3000 |  |  0.3126 |  0.1770 |  0.1426 |  0.0530 |  0.0660 |  |  0.3074 |  0.1368 |  0.1009 |  0.0608 |  0.0750
0.7000 |  |  0.6720 |  0.1650 |  0.1429 |  0.0687 |  0.0216 |  |  0.6880 |  0.1256 |  0.1010 |  0.0652 |  0.0358
|  0.2000 |  |  0.2069 |  0.1180 |  0.0993 |  0.0509 |  0.0469 |  |  0.2040 |  0.0900 |  0.0702 |  0.0550 |  0.0585
|  0.1000 |  |  0.1188 |  0.1564 |  0.1237 |  0.0462 |  0.0760 |  |  0.1093 |  0.1217 |  0.0874 |  0.0579 |  0.0794
|  0.2000 |  |  0.2069 |  0.1180 |  0.0993 |  0.0509 |  0.0469 |  |  0.2040 |  0.0900 |  0.0702 |  0.0550 |  0.0585
|  0.8000 |  |  0.7532 |  0.1581 |  0.1345 |  0.0829 |  0.0153 |  |  0.7774 |  0.1176 |  0.0951 |  0.0718 |  0.0243
$1/\nu_{1}$ |  0.2500 |  |  0.2702 |  0.1699 |  0.1464 |  0.0000 |  0.0595 |  |  0.2531 |  0.1147 |  0.1035 |  0.0543 |  0.0368
$1/\nu_{2}$ |  0.1250 |  |  0.1232 |  0.0998 |  0.0884 |  0.0000 |  0.0508 |  |  0.1195 |  0.0698 |  0.0625 |  0.0752 |  0.0350
|  |  |  |  |  |  |  |  |  |  |  |  |
|  |  | $T=200$ |  | $T=400$
$\mu_{1}$ |  0.1000 |  |  0.1000 |  0.0567 |  0.0559 |  0.0279 |  0.0268 |  |  0.1000 |  0.0397 |  0.0395 |  0.0237 |  0.0245
$\mu_{2}$ |  0.2000 |  |  0.1997 |  0.0711 |  0.0702 |  0.0270 |  0.0256 |  |  0.2001 |  0.0498 |  0.0496 |  0.0248 |  0.0261
$\mu_{3}$ |  0.3000 |  |  0.3001 |  0.0725 |  0.0721 |  0.0262 |  0.0259 |  |  0.3000 |  0.0512 |  0.0510 |  0.0252 |  0.0249
$\rm{vec}(\Xi)$ |  0.6000 |  |  0.5824 |  0.0698 |  0.0602 |  0.0577 |  0.0246 |  |  0.5937 |  0.0449 |  0.0425 |  0.0351 |  0.0271
0.5000 |  |  0.4855 |  0.1034 |  0.0790 |  0.0621 |  0.0445 |  |  0.4949 |  0.0636 |  0.0559 |  0.0411 |  0.0379
0.4000 |  |  0.3882 |  0.1119 |  0.0822 |  0.0616 |  0.0506 |  |  0.3960 |  0.0672 |  0.0581 |  0.0415 |  0.0407
0.3000 |  |  0.3017 |  0.0923 |  0.0713 |  0.0527 |  0.0555 |  |  0.3002 |  0.0574 |  0.0504 |  0.0395 |  0.0398
0.7000 |  |  0.6945 |  0.0861 |  0.0714 |  0.0530 |  0.0364 |  |  0.6976 |  0.0554 |  0.0505 |  0.0393 |  0.0305
|  0.2000 |  |  0.2013 |  0.0601 |  0.0496 |  0.0464 |  0.0458 |  |  0.2003 |  0.0386 |  0.0351 |  0.0366 |  0.0336
|  0.1000 |  |  0.1036 |  0.0803 |  0.0618 |  0.0492 |  0.0580 |  |  0.1010 |  0.0499 |  0.0437 |  0.0388 |  0.0413
|  0.2000 |  |  0.2013 |  0.0601 |  0.0496 |  0.0464 |  0.0458 |  |  0.2003 |  0.0386 |  0.0351 |  0.0366 |  0.0336
|  0.8000 |  |  0.7915 |  0.0792 |  0.0672 |  0.0513 |  0.0293 |  |  0.7965 |  0.0509 |  0.0475 |  0.0385 |  0.0263
$1/\nu_{1}$ |  0.2500 |  |  0.2504 |  0.0781 |  0.0732 |  0.0356 |  0.0307 |  |  0.2498 |  0.0533 |  0.0518 |  0.0306 |  0.0264
$1/\nu_{2}$ |  0.1250 |  |  0.1210 |  0.0476 |  0.0442 |  0.0462 |  0.0261 |  |  0.1230 |  0.0324 |  0.0313 |  0.0345 |  0.0251

Note: The mean, standard deviation of the MLE estimated parameters for
convolution-$t$ distribution. The asymptotic standard deviations are also
included for comparison. we conduct a simulation study with sample sizes
$T=50,100,200,400$, based on 50,000 Monte-Carlo simulations.

#### Simulations when $\min_{k}\nu_{k}\leq 2$

We conduct a simulation study to investigate the asymptotic properties of the
maximum likelihood estimators when $\min_{k}\nu_{k}\leq 2$. To provide a
straightforward investigation for the effects of different values of degree of
freedom, we focus on the case when $\Xi$ is an identity matrix, $\mu$ are
zeros, and $\nu_{1}=\nu_{2}=\nu^{*}$. The trivariate system is given by

$\mu=\left[\begin{array}[]{c}\pagecolor{black!10}0\\\ \pagecolor{black!5}0\\\
\pagecolor{black!5}0\end{array}\right],\quad\Xi=\left[\begin{array}[]{ccc}\pagecolor{black!10}1&\pagecolor{black!5}0&\pagecolor{black!5}0\\\
\pagecolor{black!10}0&\pagecolor{black!5}1&\pagecolor{black!5}0\\\
\pagecolor{black!10}0&\pagecolor{black!5}0&\pagecolor{black!5}1\end{array}\right]\quad\nu=\left[\begin{array}[]{c}\pagecolor{black!10}\nu^{*}\\\
\pagecolor{black!5}\nu^{*}\end{array}\right]$

where the first two elements belong to one group, i.e. $n_{1}=1$ and
$n_{2}=2$. We conduct a simulation study for several values of
$\nu^{*}=2,3/2,1,1/2$. To examine the converge rate of different parameters,
for each value of degree of freedom, we adopt $T=500,1000,2000,4000$, based on
50,000 Monte-Carlo simulations. We report the mean, standard deviation of the
estimated parameters ${\rm Std}(\hat{\theta})$, and two tails parameters,
$\alpha_{L}^{0.025}$ and $\alpha_{R}^{0.025}$, defined below

$\alpha_{L}^{0.025}=\frac{1}{T}\sum_{t=1}^{T}\left[\frac{\hat{\theta}_{t}-\theta}{{\rm
Std}(\hat{\theta})}<-1.96\right],\quad\alpha_{R}^{0.025}=\frac{1}{T}\sum_{t=1}^{T}\left[\frac{\hat{\theta}_{t}-\theta}{{\rm
Std}(\hat{\theta})}>1.96\right]$

and we will have $\alpha_{L}^{0.025}\rightarrow 0.025$ and
$\alpha_{R}^{0.025}\rightarrow 0.025$ if $\hat{\theta}_{t}$ is approximately
normally distributed. We also include the ratio of standard deviations
$R_{\sigma}$, defined as

$R_{\sigma}\left(T\right)=\frac{{\rm Std}(\hat{\theta}|\frac{T}{2})}{{\rm
Std}(\hat{\theta}|T)},\quad{\rm for}\quad T=500,1000,2000,4000.$

If the converge rate is $\sqrt{T}$, we should have
$R_{\sigma}\left(T\right)\rightarrow\sqrt{2}\approx 1.4142$ when
$T\rightarrow\infty$. Note that we also conduct a simulation with $T=250$ to
provide the ${\rm Std}(\hat{\theta})$ for $R_{\sigma}\left(500\right)$.

Tables S.2-S.5 present the simulation results for $\nu^{*}=2,3/2,1,1/2$,
accordingly. Interestingly, the estimates of the inverse degree of freedom
parameter, appear to be well-approximated by a Gaussian distribution in all
these designs. Figures S.1-S.4 present histograms of the elements of
$\hat{\Xi}$ for the sample size $T=4,000$ for $\nu^{*}=2,3/2,1,1/2$,
accordingly. These designs have degrees of freedom that are smaller than the
range covered by Theorem 7, and the histograms reveal a non-Gaussian limit
distribution for all $\hat{\Xi}$-elements, which is most pronounced for the
elements in the off-diagonal blocks.

Table S.2: Simulations with $\nu=2$

|  True |  |  Mean |  Std |  $\alpha_{L}^{0.025}$ |  $\alpha_{R}^{0.025}$ |  $R_{\sigma}$ |  |  Mean |  Std |  $\alpha_{L}^{0.025}$ |  $\alpha_{R}^{0.025}$ |  $R_{\sigma}$
---|---|---|---|---|---|---|---|---|---|---|---|---|---
|  |  | $T=500$ |  | $T=1,000$
$\mu_{1}$ |  0 |  |  0.0002 |  0.0579 |  0.0250 |  0.0256 |  1.4332 |  |  0.0001 |  0.0407 |  0.0246 |  0.0246 |  1.4220
$\mu_{2}$ |  0 |  |  0.0003 |  0.0548 |  0.0250 |  0.0242 |  1.4184 |  |  0.0002 |  0.0387 |  0.0253 |  0.0255 |  1.4152
$\mu_{3}$ |  0 |  |  0.0006 |  0.0550 |  0.0231 |  0.0261 |  1.4125 |  |  0.0001 |  0.0390 |  0.0253 |  0.0250 |  1.4089
$\rm{vec}(\Xi)$ |  1 |  |  0.9971 |  0.0635 |  0.0236 |  0.0272 |  1.4172 |  |  0.9986 |  0.0443 |  0.0232 |  0.0270 |  1.4324
0 |  |  0.0000 |  0.0233 |  0.0296 |  0.0291 |  1.5733 |  |  0.0000 |  0.0154 |  0.0290 |  0.0290 |  1.5108
0 |  |  0.0002 |  0.0235 |  0.0281 |  0.0290 |  1.5541 |  |  0.0001 |  0.0154 |  0.0280 |  0.0293 |  1.5256
0 |  |  0.0001 |  0.0255 |  0.0296 |  0.0284 |  1.5660 |  |  0.0000 |  0.0166 |  0.0280 |  0.0279 |  1.5324
1 |  |  0.9979 |  0.0526 |  0.0230 |  0.0273 |  1.4247 |  |  0.9990 |  0.0371 |  0.0237 |  0.0255 |  1.4179
|  0 |  |  0.0002 |  0.0274 |  0.0252 |  0.0250 |  1.4121 |  |  0.0000 |  0.0194 |  0.0247 |  0.0249 |  1.4113
|  0 |  |  0.0000 |  0.0253 |  0.0279 |  0.0288 |  1.5845 |  |  0.0000 |  0.0166 |  0.0281 |  0.0276 |  1.5259
|  0 |  |  0.0002 |  0.0274 |  0.0252 |  0.0250 |  1.4121 |  |  0.0000 |  0.0194 |  0.0247 |  0.0249 |  1.4113
|  1 |  |  0.9974 |  0.0527 |  0.0244 |  0.0264 |  1.4205 |  |  0.9989 |  0.0373 |  0.0235 |  0.0270 |  1.4121
$1/\nu_{1}$ |  0.5 |  |  0.4997 |  0.0574 |  0.0256 |  0.0253 |  1.4402 |  |  0.4998 |  0.0404 |  0.0254 |  0.0239 |  1.4202
$1/\nu_{2}$ |  0.5 |  |  0.5001 |  0.0452 |  0.0239 |  0.0265 |  1.4277 |  |  0.4995 |  0.0318 |  0.0247 |  0.0243 |  1.4205
|  |  |  |  |  |  |  |  |  |  |  |  |
|  |  | $T=2,000$ |  | $T=4,000$
$\mu_{1}$ |  0 |  |  0.0000 |  0.0290 |  0.0249 |  0.0251 |  1.4019 |  |  0.0001 |  0.0203 |  0.0246 |  0.0251 |  1.4308
$\mu_{2}$ |  0 |  |  0.0001 |  0.0275 |  0.0241 |  0.0251 |  1.4098 |  |  0.0000 |  0.0194 |  0.0252 |  0.0243 |  1.4196
$\mu_{3}$ |  0 |  |  0.0002 |  0.0275 |  0.0244 |  0.0259 |  1.4190 |  |  0.0001 |  0.0194 |  0.0240 |  0.0251 |  1.4188
$\rm{vec}(\Xi)$ |  1 |  |  0.9992 |  0.0316 |  0.0255 |  0.0249 |  1.4036 |  |  0.9997 |  0.0222 |  0.0246 |  0.0264 |  1.4228
0 |  |  0.0000 |  0.0101 |  0.0282 |  0.0279 |  1.5207 |  |  0.0000 |  0.0068 |  0.0287 |  0.0290 |  1.4903
0 |  |  0.0001 |  0.0102 |  0.0278 |  0.0288 |  1.5194 |  |  0.0001 |  0.0068 |  0.0292 |  0.0283 |  1.4906
0 |  |  0.0000 |  0.0109 |  0.0283 |  0.0276 |  1.5258 |  |  0.0000 |  0.0074 |  0.0276 |  0.0275 |  1.4805
1 |  |  0.9994 |  0.0262 |  0.0254 |  0.0247 |  1.4175 |  |  0.9997 |  0.0185 |  0.0246 |  0.0258 |  1.4158
|  0 |  |  0.0001 |  0.0137 |  0.0254 |  0.0257 |  1.4156 |  |  0.0000 |  0.0097 |  0.0250 |  0.0254 |  1.4177
|  0 |  |  0.0001 |  0.0109 |  0.0278 |  0.0268 |  1.5125 |  |  0.0000 |  0.0073 |  0.0274 |  0.0279 |  1.4951
|  0 |  |  0.0001 |  0.0137 |  0.0254 |  0.0257 |  1.4156 |  |  0.0000 |  0.0097 |  0.0250 |  0.0254 |  1.4177
|  1 |  |  0.9995 |  0.0262 |  0.0241 |  0.0263 |  1.4239 |  |  0.9998 |  0.0186 |  0.0239 |  0.0251 |  1.4087
$1/\nu_{1}$ |  0.5 |  |  0.4999 |  0.0285 |  0.0253 |  0.0253 |  1.4172 |  |  0.5000 |  0.0201 |  0.0245 |  0.0247 |  1.4176
$1/\nu_{2}$ |  0.5 |  |  0.4999 |  0.0223 |  0.0255 |  0.0246 |  1.4241 |  |  0.4999 |  0.0158 |  0.0255 |  0.0251 |  1.4127

Note: The mean, standard deviation of the MLE estimated parameters for
convolution-$t$ distribution when $\nu_{1}=\nu_{2}=2$. We conduct a simulation
study with sample sizes $T=500,1000,2000,4000$, based on 50,000 Monte-Carlo
simulations. This design has $\nu_{1}=\nu_{2}=2$ that is outside the range
covered by the results in Theorem 7. The shaded regions highlights the
coefficients in the off-diagonal blocks of $\Xi$, which appear to converge at
a faster rate.

Table S.3: Simulations with $\nu=3/2$

|  True |  |  Mean |  Std |  $\alpha_{L}^{0.025}$ |  $\alpha_{R}^{0.025}$ |  $R_{\sigma}$ |  |  Mean |  Std |  $\alpha_{L}^{0.025}$ |  $\alpha_{R}^{0.025}$ |  $R_{\sigma}$
---|---|---|---|---|---|---|---|---|---|---|---|---|---
|  |  | $T=500$ |  | $T=1,000$
$\mu_{1}$ |  0 |  |  0.0002 |  0.0604 |  0.0246 |  0.0256 |  1.4275 |  |  0.0002 |  0.0423 |  0.0246 |  0.0255 |  1.4279
$\mu_{2}$ |  0 |  |  0.0003 |  0.0564 |  0.0246 |  0.0248 |  1.4171 |  |  0.0003 |  0.0398 |  0.0259 |  0.0247 |  1.4167
$\mu_{3}$ |  0 |  |  0.0005 |  0.0565 |  0.0243 |  0.0243 |  1.4226 |  |  0.0001 |  0.0397 |  0.0254 |  0.0256 |  1.4230
$\rm{vec}(\Xi)$ |  1 |  |  0.9964 |  0.0679 |  0.0233 |  0.0260 |  1.4210 |  |  0.9982 |  0.0474 |  0.0242 |  0.0260 |  1.4336
0 |  |  0.0000 |  0.0142 |  0.0321 |  0.0320 |  1.6632 |  |  0.0001 |  0.0086 |  0.0324 |  0.0310 |  1.6536
0 |  |  0.0000 |  0.0142 |  0.0320 |  0.0317 |  1.6913 |  |  0.0000 |  0.0086 |  0.0314 |  0.0307 |  1.6453
0 |  |  0.0000 |  0.0160 |  0.0298 |  0.0306 |  1.6909 |  |  0.0001 |  0.0099 |  0.0292 |  0.0314 |  1.6173
1 |  |  0.9971 |  0.0559 |  0.0240 |  0.0255 |  1.4228 |  |  0.9986 |  0.0397 |  0.0238 |  0.0256 |  1.4066
|  0 |  |  0.0000 |  0.0280 |  0.0246 |  0.0258 |  1.4208 |  |  0.0001 |  0.0199 |  0.0242 |  0.0252 |  1.4077
|  0 |  |  0.0001 |  0.0162 |  0.0293 |  0.0315 |  1.6665 |  |  0.0000 |  0.0098 |  0.0299 |  0.0299 |  1.6539
|  0 |  |  0.0000 |  0.0280 |  0.0246 |  0.0258 |  1.4208 |  |  0.0001 |  0.0199 |  0.0242 |  0.0252 |  1.4077
|  1 |  |  0.9970 |  0.0557 |  0.0234 |  0.0259 |  1.4266 |  |  0.9987 |  0.0396 |  0.0236 |  0.0261 |  1.4069
$1/\nu_{1}$ |  0.6667 |  |  0.6669 |  0.0650 |  0.0245 |  0.0262 |  1.4305 |  |  0.6669 |  0.0457 |  0.0242 |  0.0266 |  1.4219
$1/\nu_{2}$ |  0.6667 |  |  0.6674 |  0.0527 |  0.0228 |  0.0268 |  1.4173 |  |  0.6667 |  0.0372 |  0.0243 |  0.0260 |  1.4167
|  |  |  |  |  |  |  |  |  |  |  |  |
|  |  | $T=2,000$ |  | $T=4,000$
$\mu_{1}$ |  0 |  |  0.0000 |  0.0300 |  0.0250 |  0.0255 |  1.4115 |  |  0.0001 |  0.0212 |  0.0250 |  0.0257 |  1.4120
$\mu_{2}$ |  0 |  |  0.0001 |  0.0280 |  0.0246 |  0.0259 |  1.4215 |  |  0.0001 |  0.0198 |  0.0245 |  0.0245 |  1.4152
$\mu_{3}$ |  0 |  |  0.0000 |  0.0279 |  0.0247 |  0.0240 |  1.4219 |  |  0.0000 |  0.0198 |  0.0244 |  0.0253 |  1.4073
$\rm{vec}(\Xi)$ |  1 |  |  0.9990 |  0.0335 |  0.0247 |  0.0252 |  1.4136 |  |  0.9994 |  0.0236 |  0.0258 |  0.0239 |  1.4205
0 |  |  0.0000 |  0.0053 |  0.0310 |  0.0312 |  1.6105 |  |  0.0000 |  0.0033 |  0.0319 |  0.0320 |  1.6321
0 |  |  0.0000 |  0.0053 |  0.0322 |  0.0307 |  1.6211 |  |  0.0000 |  0.0033 |  0.0325 |  0.0313 |  1.6264
0 |  |  0.0000 |  0.0060 |  0.0305 |  0.0291 |  1.6457 |  |  0.0000 |  0.0037 |  0.0310 |  0.0297 |  1.6102
1 |  |  0.9992 |  0.0279 |  0.0250 |  0.0238 |  1.4268 |  |  0.9997 |  0.0198 |  0.0241 |  0.0252 |  1.4080
|  0 |  |  0.0000 |  0.0140 |  0.0250 |  0.0241 |  1.4205 |  |  0.0000 |  0.0100 |  0.0254 |  0.0256 |  1.4084
|  0 |  |  0.0000 |  0.0060 |  0.0311 |  0.0289 |  1.6245 |  |  0.0000 |  0.0037 |  0.0306 |  0.0313 |  1.6164
|  0 |  |  0.0000 |  0.0140 |  0.0250 |  0.0241 |  1.4205 |  |  0.0000 |  0.0100 |  0.0254 |  0.0256 |  1.4084
|  1 |  |  0.9992 |  0.0280 |  0.0250 |  0.0249 |  1.4160 |  |  0.9997 |  0.0197 |  0.0239 |  0.0256 |  1.4166
$1/\nu_{1}$ |  0.6667 |  |  0.6669 |  0.0321 |  0.0235 |  0.0257 |  1.4226 |  |  0.6667 |  0.0227 |  0.0260 |  0.0242 |  1.4135
$1/\nu_{2}$ |  0.6667 |  |  0.6666 |  0.0262 |  0.0239 |  0.0250 |  1.4183 |  |  0.6668 |  0.0185 |  0.0241 |  0.0261 |  1.4209

Note: The mean, standard deviation of the MLE estimated parameters for
convolution-$t$ distribution $\nu_{1}=\nu_{2}=3/2$. We conduct a simulation
study with sample sizes $T=500,1000,2000,4000$, based on 50,000 Monte-Carlo
simulations. This design has $\nu_{1}=\nu_{2}=3/2$ that is outside the range
covered by the results in Theorem 7. The shaded regions highlights the
coefficients in the off-diagonal blocks of $\Xi$, which appear to converge at
a faster rate.

Table S.4: Simulations with $\nu=1$

|  True |  |  Mean |  Std |  $\alpha_{L}^{0.025}$ |  $\alpha_{R}^{0.025}$ |  $R_{\sigma}$ |  |  Mean |  Std |  $\alpha_{L}^{0.025}$ |  $\alpha_{R}^{0.025}$ |  $R_{\sigma}$
---|---|---|---|---|---|---|---|---|---|---|---|---|---
|  |  | $T=500$ |  | $T=1,000$
$\mu_{1}$ |  0 |  |  0.0012 |  0.0571 |  0.0251 |  0.0273 |  1.4230 |  |  0.0006 |  0.0406 |  0.0254 |  0.0281 |  1.4068
$\mu_{2}$ |  0 |  |  0.0009 |  0.0530 |  0.0253 |  0.0275 |  1.4164 |  |  0.0006 |  0.0377 |  0.0243 |  0.0279 |  1.4067
$\mu_{3}$ |  0 |  |  0.0015 |  0.0531 |  0.0240 |  0.0287 |  1.4141 |  |  0.0006 |  0.0375 |  0.0256 |  0.0266 |  1.4168
$\rm{vec}(\Xi)$ |  1 |  |  0.9970 |  0.0662 |  0.0236 |  0.0291 |  1.4115 |  |  0.9981 |  0.0467 |  0.0256 |  0.0275 |  1.4180
0 |  |  0.0000 |  0.0046 |  0.0311 |  0.0316 |  2.0418 |  |  0.0000 |  0.0023 |  0.0322 |  0.0310 |  2.0194
0 |  |  0.0000 |  0.0046 |  0.0312 |  0.0297 |  1.9968 |  |  0.0000 |  0.0023 |  0.0308 |  0.0308 |  2.0126
0 |  |  0.0000 |  0.0058 |  0.0308 |  0.0317 |  2.0385 |  |  0.0000 |  0.0030 |  0.0320 |  0.0319 |  1.9756
1 |  |  0.9982 |  0.0542 |  0.0243 |  0.0281 |  1.4208 |  |  0.9995 |  0.0388 |  0.0244 |  0.0280 |  1.3956
|  0 |  |  0.0004 |  0.0286 |  0.0254 |  0.0257 |  1.4143 |  |  0.0001 |  0.0201 |  0.0249 |  0.0258 |  1.4202
|  0 |  |  0.0000 |  0.0059 |  0.0313 |  0.0314 |  2.0388 |  |  0.0000 |  0.0029 |  0.0314 |  0.0312 |  1.9978
|  0 |  |  0.0004 |  0.0286 |  0.0254 |  0.0257 |  1.4143 |  |  0.0001 |  0.0201 |  0.0249 |  0.0258 |  1.4202
|  1 |  |  0.9975 |  0.0546 |  0.0244 |  0.0281 |  1.4134 |  |  0.9993 |  0.0388 |  0.0238 |  0.0282 |  1.4048
$1/\nu_{1}$ |  1 |  |  1.0016 |  0.0690 |  0.0245 |  0.0294 |  1.4058 |  |  1.0008 |  0.0492 |  0.0262 |  0.0275 |  1.4023
$1/\nu_{2}$ |  1 |  |  1.0017 |  0.0612 |  0.0232 |  0.0290 |  1.4062 |  |  1.0006 |  0.0431 |  0.0245 |  0.0293 |  1.4188
|  |  |  |  |  |  |  |  |  |  |  |  |
|  |  | $T=2,000$ |  | $T=4,000$
$\mu_{1}$ |  0 |  |  0.0005 |  0.0287 |  0.0246 |  0.0283 |  1.4114 |  |  0.0003 |  0.0204 |  0.0257 |  0.0279 |  1.4078
$\mu_{2}$ |  0 |  |  0.0003 |  0.0266 |  0.0260 |  0.0269 |  1.4149 |  |  0.0001 |  0.0189 |  0.0262 |  0.0259 |  1.4100
$\mu_{3}$ |  0 |  |  0.0003 |  0.0266 |  0.0259 |  0.0272 |  1.4115 |  |  0.0002 |  0.0188 |  0.0259 |  0.0267 |  1.4125
$\rm{vec}(\Xi)$ |  1 |  |  0.9990 |  0.0331 |  0.0253 |  0.0281 |  1.4108 |  |  0.9996 |  0.0234 |  0.0268 |  0.0274 |  1.4113
0 |  |  0.0000 |  0.0011 |  0.0308 |  0.0311 |  1.9808 |  |  0.0000 |  0.0006 |  0.0301 |  0.0319 |  2.0367
0 |  |  0.0000 |  0.0011 |  0.0304 |  0.0312 |  2.0205 |  |  0.0000 |  0.0006 |  0.0309 |  0.0313 |  2.0302
0 |  |  0.0000 |  0.0015 |  0.0320 |  0.0304 |  2.0041 |  |  0.0000 |  0.0007 |  0.0315 |  0.0311 |  2.0121
1 |  |  0.9999 |  0.0275 |  0.0237 |  0.0286 |  1.4132 |  |  0.9997 |  0.0196 |  0.0257 |  0.0274 |  1.3995
|  0 |  |  0.0001 |  0.0142 |  0.0250 |  0.0259 |  1.4159 |  |  0.0001 |  0.0100 |  0.0243 |  0.0268 |  1.4223
|  0 |  |  0.0000 |  0.0015 |  0.0306 |  0.0314 |  2.0211 |  |  0.0000 |  0.0007 |  0.0303 |  0.0308 |  1.9923
|  0 |  |  0.0001 |  0.0142 |  0.0250 |  0.0259 |  1.4159 |  |  0.0001 |  0.0100 |  0.0243 |  0.0268 |  1.4223
|  1 |  |  0.9997 |  0.0276 |  0.0262 |  0.0278 |  1.4079 |  |  0.9999 |  0.0197 |  0.0252 |  0.0278 |  1.3992
$1/\nu_{1}$ |  1 |  |  1.0003 |  0.0347 |  0.0263 |  0.0282 |  1.4183 |  |  1.0004 |  0.0246 |  0.0259 |  0.0286 |  1.4120
$1/\nu_{2}$ |  1 |  |  1.0004 |  0.0306 |  0.0254 |  0.0282 |  1.4095 |  |  1.0002 |  0.0217 |  0.0251 |  0.0278 |  1.4110

Note: The mean, standard deviation of the MLE estimated parameters for
convolution-$t$ distribution $\nu_{1}=\nu_{2}=1$. We conduct a simulation
study with sample sizes $T=500,1000,2000,4000$, based on 50,000 Monte-Carlo
simulations. This design has $\nu_{1}=\nu_{2}=1$ that is outside the range
covered by the results in Theorem 7. The shaded regions highlights the
coefficients in the off-diagonal blocks of $\Xi$, which appear to converge at
a faster rate.

Table S.5: Simulations with $\nu=1/2$

|  True |  |  Mean |  Std |  $\alpha_{L}^{0.025}$ |  $\alpha_{R}^{0.025}$ |  $R_{\sigma}$ |  |  Mean |  Std |  $\alpha_{L}^{0.025}$ |  $\alpha_{R}^{0.025}$ |  $R_{\sigma}$
---|---|---|---|---|---|---|---|---|---|---|---|---|---
|  |  | $T=500$ |  | $T=1,000$
$\mu_{1}$ |  0 |  |  0.0008 |  0.0519 |  0.0320 |  0.0287 |  1.4181 |  |  0.0006 |  0.0362 |  0.0311 |  0.0292 |  1.4329
$\mu_{2}$ |  0 |  |  0.0008 |  0.0469 |  0.0303 |  0.0303 |  1.4207 |  |  0.0001 |  0.0332 |  0.0315 |  0.0287 |  1.4112
$\mu_{3}$ |  0 |  |  0.0004 |  0.0473 |  0.0310 |  0.0286 |  1.4094 |  |  0.0002 |  0.0333 |  0.0325 |  0.0282 |  1.4213
$\rm{vec}(\Xi)$ |  1 |  |  0.9962 |  0.0646 |  0.0333 |  0.0286 |  1.4107 |  |  0.9977 |  0.0459 |  0.0336 |  0.0273 |  1.4083
0 |  |  0.0000 |  1.0261† |  0.0180 |  0.0188 |  3.9108 |  |  0.0000 |  0.2622† |  0.0183 |  0.0178 |  3.9136
0 |  |  0.0000 |  1.0403† |  0.0178 |  0.0190 |  3.7635 |  |  0.0000 |  0.2644† |  0.0184 |  0.0179 |  3.9347
0 |  |  0.0000 |  1.5685† |  0.0217 |  0.0217 |  4.1445 |  |  0.0000 |  0.4179† |  0.0208 |  0.0213 |  3.7537
1 |  |  0.9989 |  0.0506 |  0.0306 |  0.0297 |  1.4193 |  |  0.9995 |  0.0363 |  0.0311 |  0.0312 |  1.3929
|  0 |  |  0.0004 |  0.0271 |  0.0269 |  0.0277 |  1.4190 |  |  0.0002 |  0.0190 |  0.0279 |  0.0283 |  1.4270
|  0 |  |  0.0000 |  1.5737† |  0.0211 |  0.0215 |  4.0704 |  |  0.0000 |  0.4232† |  0.0207 |  0.0211 |  3.7182
|  0 |  |  0.0004 |  0.0271 |  0.0269 |  0.0277 |  1.4190 |  |  0.0002 |  0.0190 |  0.0279 |  0.0283 |  1.4270
|  1 |  |  0.9989 |  0.0509 |  0.0299 |  0.0312 |  1.4046 |  |  0.9993 |  0.0360 |  0.0315 |  0.0289 |  1.4131
$1/\nu_{1}$ |  2 |  |  2.0026 |  0.1201 |  0.0221 |  0.0283 |  1.4040 |  |  2.0012 |  0.0841 |  0.0226 |  0.0267 |  1.4293
$1/\nu_{2}$ |  2 |  |  2.0018 |  0.1089 |  0.0219 |  0.0286 |  1.4223 |  |  2.0010 |  0.0768 |  0.0236 |  0.0262 |  1.4173
|  |  |  |  |  |  |  |  |  |  |  |  |
|  |  | $T=2,000$ |  | $T=4,000$
$\mu_{1}$ |  0 |  |  0.0001 |  0.0263 |  0.0332 |  0.0288 |  1.3797 |  |  0.0001 |  0.0190 |  0.0318 |  0.0292 |  1.3828
$\mu_{2}$ |  0 |  |  0.0002 |  0.0239 |  0.0339 |  0.0271 |  1.3910 |  |  0.0000 |  0.0173 |  0.0314 |  0.0281 |  1.3815
$\mu_{3}$ |  0 |  |  0.0000 |  0.0237 |  0.0313 |  0.0280 |  1.4020 |  |  0.0001 |  0.0173 |  0.0308 |  0.0288 |  1.3673
$\rm{vec}(\Xi)$ |  1 |  |  0.9989 |  0.0325 |  0.0353 |  0.0290 |  1.4099 |  |  0.9993 |  0.0240 |  0.0350 |  0.0282 |  1.3547
0 |  |  0.0000 |  0.0658† |  0.0176 |  0.0185 |  3.9835 |  |  0.0000 |  0.0160† |  0.0173 |  0.0175 |  4.1040
0 |  |  0.0000 |  0.0675† |  0.0171 |  0.0179 |  3.9150 |  |  0.0000 |  0.0153† |  0.0188 |  0.0188 |  4.4151
0 |  |  0.0000 |  0.1058† |  0.0195 |  0.0215 |  3.9487 |  |  0.0000 |  0.0257† |  0.0213 |  0.0218 |  4.1253
1 |  |  0.9997 |  0.0259 |  0.0313 |  0.0296 |  1.4041 |  |  0.9998 |  0.0190 |  0.0314 |  0.0302 |  1.3608
|  0 |  |  0.0001 |  0.0136 |  0.0282 |  0.0277 |  1.3949 |  |  0.0001 |  0.0097 |  0.0280 |  0.0280 |  1.3985
|  0 |  |  0.0000 |  0.1071† |  0.0200 |  0.0216 |  3.9518 |  |  0.0000 |  0.0262† |  0.0207 |  0.0210 |  4.0892
|  0 |  |  0.0001 |  0.0136 |  0.0282 |  0.0277 |  1.3949 |  |  0.0001 |  0.0097 |  0.0280 |  0.0280 |  1.3985
|  1 |  |  0.9997 |  0.0259 |  0.0315 |  0.0303 |  1.3908 |  |  0.9997 |  0.0190 |  0.0320 |  0.0287 |  1.3608
$1/\nu_{1}$ |  2 |  |  2.0006 |  0.0596 |  0.0242 |  0.0257 |  1.4109 |  |  2.0004 |  0.0423 |  0.0240 |  0.0264 |  1.4086
$1/\nu_{2}$ |  2 |  |  2.0008 |  0.0545 |  0.0219 |  0.0271 |  1.4092 |  |  2.0005 |  0.0387 |  0.0236 |  0.0261 |  1.4106

Note: The mean, standard deviation of the MLE estimated parameters for
convolution-$t$ distribution $\nu_{1}=\nu_{2}=1/2$. We conduct a simulation
study with sample sizes $T=500,1000,2000,4000$, based on 50,000 Monte-Carlo
simulations. The unit of the values with $\dagger$ in the superscript is
$10^{-4}$. This design has $\nu_{1}=\nu_{2}=1/2$ that is outside the range
covered by the results in Theorem 7. The shaded regions highlights the
coefficients in the off-diagonal blocks of $\Xi$, which appear to converge at
a faster rate.

Figure S.1: Histograms for the elements of $\hat{\Xi}$ when $\nu=2$ and
$T=4,000$. We also report the standard deviations in each subtitles. Figure
S.2: Histograms for the elements of $\hat{\Xi}$ when $\nu=3/2$ and $T=4,000$.
We also report the standard deviations in each subtitles. Figure S.3:
Histograms for the elements of $\hat{\Xi}$ when $\nu=1$ and $T=4,000$. We also
report the standard deviations in each subtitles. Figure S.4: Histograms for
the elements of $\hat{\Xi}$ when $\nu=1/2$ and $T=4,000$. We also report the
standard deviations in each subtitles.