# The level crossings of random sums

Christopher Corley  Department of Mathematics, The University of Tennessee at
Chattanooga, 304 Lupton Hall (Dept. 6956), 615 McCallie Avenue, Chattanooga,
Tennessee 37403, United States of America<EMAIL_ADDRESS>and
Andrew Ledoan  Department of Mathematics, The University of Tennessee at
Chattanooga, 304 Lupton Hall (Dept. 6956), 615 McCallie Avenue, Chattanooga,
Tennessee 37403, United States of America<EMAIL_ADDRESS>
###### Abstract.

Let $\\{\eta_{j}\\}_{j=0}^{N}$ be a sequence of independent and identically
distributed complex normal random variables with mean zero and variances
$\\{\sigma_{j}^{2}\\}_{j=0}^{N}$. Let $\\{f_{j}(z)\\}_{j=0}^{N}$ be a sequence
of holomorphic functions that are real-valued on the real line. The purpose of
the present study is that of examining the number of times that the random sum
$\sum_{j=0}^{N}\eta_{j}f_{j}(z)$ crosses the complex level
$\boldsymbol{K}=K_{1}+iK_{2}$, where $K_{1}$ and $K_{2}$ are constants
independent of $z$. More specifically, we establish an exact formula for the
expected density function for the complex zeros. We then reformulate the
problem in terms of successive observations of a Brownian motion. We further
answer the basic question about the expected number of complex zeros for
coefficients of nonvanishing mean values.

###### Key words and phrases:

Density function; Level crossing; Multivariate normal distribution; Rice
formula; Random field; Random summation; Standard Brownian motion

###### 2010 Mathematics Subject Classification:

Primary 30C15; Secondary 30B20, 60B99, 60E05, 60G50, 62H10

##  1\. Introduction and statement of results

An exact formula for the expected number of real zeros of a random polynomial
was obtained by Kac [12] under independent and identically distributed
(i.i.d.) real standard normal coefficients. For complex coefficients, Dunnage
[5, 6] gave some estimates for the number of real zeros. For complex zeros,
the expected density of zeros was studied by Shepp and Vanderbei [17] for
i.i.d. real standard normal coefficients and generalized by Ibragimov and
Zeitouni [11] for a wider class of distributions of coefficients. Relevant to
these investigations is the work of Kostlan [13]. The expected density was
dealt with, also, by Hammersley [10], Edelman and Kostlan [7], and Farahmand
and Grigorash [9]. Vanderbei [18] generalized the work in [17] to random sums
with holomorphic functions that are real-valued on the real line as basis
functions. Motivated by the studies conducted by Vanderbei [18] and Farahmand
[8], the present authors [4] studied the number of times that these random
sums cross a complex level and obtained certain results on the level
crossings. The aim of the present paper is to extend these results in the
following direction.

Let $z$ be the complex variable $x+iy$. Let $\\{a_{j}\\}_{j=0}^{N}$ and
$\\{b_{j}\\}_{j=0}^{N}$ be sequences of mutually i.i.d. real normal random
variables defined on the complete probability space
$(\Omega,\mathscr{F},\operatorname{Prob})$ with mean zero and variances
$\\{\sigma_{a_{j}}^{2}\\}_{j=0}^{N}$ and $\\{\sigma_{b_{j}}^{2}\\}_{j=0}^{N}$,
so that $\mathscr{F}$ is a $\sigma$-field of subsets of $\Omega$ and
$\operatorname{Prob}$ is a probability measure on $(\Omega,\mathscr{F})$.
Assume all sub $\sigma$-fields contain all sets of measure zero. Then, let
$\\{\eta_{j}\\}_{j=0}^{N}$ be a sequence of i.i.d. complex normal random
variables with density $e^{-z\bar{z}}/\pi$ and given by
$\eta_{j}=a_{j}+ib_{j}$ for $0\leq j\leq N$. Let, further,
$\\{f_{j}(z)\\}_{j=0}^{N}$ be a sequence of holomorphic functions
$f_{j}(z)=u_{j}(x,y)+iv_{j}(x,y)$ for $0\leq j\leq N$ that are real-valued on
the real line. By the Schwarz reflection principle,
$\overline{f_{j}(z)}=f_{j}(\overline{z})$ for $0\leq j\leq N$. Then, define

$S_{N}(z)=\eta_{0}f_{0}(z)+\eta_{1}f_{1}(z)+\dots+\eta_{N}f_{N}(z)=\sum_{j=0}^{N}\eta_{j}f_{j}(z).$
(1)

If, for each compact subset $T$ of the complex plane,
$N_{\boldsymbol{K}}^{S_{N}}(T)$ denotes the random number of complex zeros,
counted with multiplicity, in $T$ of $S_{N}(z)$ that cross the complex level
$\boldsymbol{K}=K_{1}+iK_{2}$, where $K_{1}$ and $K_{2}$ are constants
independent of $z$, then from [4], with probability one, the expected density
$h_{N,\boldsymbol{K}}(z)$ of the complex zeros of

$S_{N}(z)=\boldsymbol{K}$ (2)

is given by

$E(N_{\boldsymbol{K}}^{S}(T))=\int_{T}h_{N,\boldsymbol{K}}(z)\,dz.$ (3)

The explicit derivation of $h_{N,\boldsymbol{K}}(z)$ constitutes the primary
reason for studying the complex zeros of (2). The main device for treating
$h_{N,\boldsymbol{K}}(z)$ throughout the complex plane is the Rice formula.
This remarkable result provides a representation for the expected number of
zeros of certain random fields. It is reproduced below from [3, Theorem 6.2,
pp. 163-164]. (See, also, [1, Theorem 11.2.3, Corollary 11.2.4, pp. 269-271]
and [2, Theorem 2.1, p. 256].)

######  Theorem 1.

Let $Z\colon U\rightarrow\mathds{R}^{N}$ be a random field, let $U$ be an open
subset of $\mathds{R}^{N}$, and let $u\in\mathds{R}^{N}$ be a fixed point in
the codomain. Assume the following conditions are satisfied with probability
one:

  1. (i)

$Z$ is normal.

  2. (ii)

Almost surely the function $t\rightsquigarrow Z(t)$ is of class $C^{1}$.

  3. (iii)

For each $t\in U$, $Z(t)$ has a nondegenerate distribution—i.e.,
$\operatorname{Var}(Z(t))\succ 0$.

  4. (iv)

For each $u\in\mathds{R}^{N}$, $\operatorname{Prob}(\exists\,t\in U\colon
Z(t)=u,\det(Z^{\prime}(t))=0)=0$.

If $N_{u}^{Z}(B)$ denotes the number of zeros of $Z(t)=u$ that belong to the
Borel subset $B\subset U$, then one has

$E(N_{u}^{Z}(B))=\int_{B}E(\lvert\det(Z^{\prime}(t))\rvert\mid
Z(t)=u)\,p_{Z(t)}(u)\,dt,$ (4)

where $p_{Z(t)}(u)$ is the probability density function of $Z(t)$ at $u$. If
$B$ is compact, then both sides of (4) are finite.

The function $Z$ in (4) is defined on $\mathds{R}^{N}$. In our application, we
need to find the real and complex zeros of (2)—i.e., the real zeros of
$\operatorname*{Re}(S_{N}(x+iy))=K_{1}$ and
$\operatorname*{Im}(S_{N}(x+iy))=K_{2}$ for $(x,y)\in\mathds{R}^{2}$. The
conditions $(i)$–$(iv)$ are easy to check. Formula (4) is interesting. It
shows that $h_{N,\boldsymbol{K}}(z)$, as defined by (3), can be expressed
through a conditioned mean function of a quadratic form of i.i.d. real normal
random variables conditioned on certain linear combinations.

######  Theorem 2.

Provided all the conditions imposed on $S_{N}(z)$ in (1) and $T$ are
satisfied, then for all $N>1$ one has

$\begin{split}&h_{N,\boldsymbol{K}}(z)=\frac{1}{2\pi
D_{0}(z)}\exp\left(-\frac{K_{1}^{2}Y_{3}(z)+K_{2}^{2}Y_{1}(z)-2K_{1}K_{2}Y_{2}(z)}{2D_{0}(z)^{2}}\right)\\\
&\hskip 5.0pt\times\left\\{D_{3}(z)-\frac{\lvert
D_{1}(z)\rvert^{2}}{D_{0}(z)}\left(\frac{Y_{2}(z)+Y_{3}(z)}{D_{0}(z)}-\frac{(K_{1}Y_{3}(z)-K_{2}Y_{2}(z))(K_{1}(Y_{2}(z)+Y_{3}(z))-K_{2}(Y_{1}(z)+Y_{2}(z)))}{D_{0}(z)^{3}}\right)\right.\\\
&\hskip 35.00005pt-\frac{\lvert
D_{2}(z)\rvert^{2}}{D_{0}(z)}\left(\frac{Y_{1}(z)+Y_{2}(z)}{D_{0}(z)}-\frac{(K_{1}Y_{2}(z)-K_{2}Y_{1}(z))(K_{1}(Y_{2}(z)+Y_{3}(z))-K_{2}(Y_{1}(z)+Y_{2}(z)))}{D_{0}(z)^{3}}\right)\\\
&\hskip 65.00009pt\left.+\frac{\lvert
D_{1}(z)+iD_{2}(z)\rvert^{2}}{D_{0}(z)}\left(\frac{Y_{2}(z)}{D_{0}(z)}-\frac{(K_{1}Y_{3}(z)-K_{2}Y_{2}(z))(K_{1}Y_{2}(z)-K_{2}Y_{1}(z))}{D_{0}(z)^{3}}\right)\right\\},\end{split}$

where

$\begin{array}[]{c@{\qquad}c}\displaystyle
Y_{1}(z)=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}u_{j}^{2}+\sigma_{b_{j}}^{2}v_{j}^{2}),\qquad\displaystyle
Y_{2}(z)=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}-\sigma_{b_{j}}^{2})u_{j}v_{j},\qquad\displaystyle
Y_{3}(z)=\sum_{j=0}^{N}(\sigma_{b_{j}}^{2}u_{j}^{2}+\sigma_{a_{j}}^{2}v_{j}^{2}),\end{array}$

and

$\begin{array}[]{c@{\qquad}c}\displaystyle
D_{0}(z)=\sqrt{Y_{1}(z)Y_{3}(z)-Y_{2}^{2}(z)},\qquad\displaystyle
D_{1}(z)=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}u_{j}-i\sigma_{b_{j}}^{2}v_{j})\left(\frac{\partial
u_{j}}{\partial x}+i\frac{\partial v_{j}}{\partial x}\right),\\\\[12.91663pt]
\displaystyle
D_{2}(z)=\sum_{j=0}^{N}(\sigma_{b_{j}}^{2}u_{j}-i\sigma_{a_{j}}^{2}v_{j})\left(\frac{\partial
u_{j}}{\partial x}+i\frac{\partial v_{j}}{\partial
x}\right),\qquad\displaystyle
D_{3}(z)=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}+\sigma_{b_{j}}^{2})\left(\left(\frac{\partial
u_{j}}{\partial x}\right)^{2}+\left(\frac{\partial v_{j}}{\partial
x}\right)^{2}\right).\end{array}$

In relation to the work in [4], observe that when
$\sigma_{a_{j}}^{2}=\sigma_{b_{j}}^{2}=\sigma^{2}$ for $0\leq j\leq N$

$\begin{split}\begin{array}[]{c@{\qquad}c}\displaystyle
Y_{1}(z)=Y_{3}(z)=\sigma^{2}B_{0}(z),\qquad\displaystyle
Y_{2}(z)=0,\end{array}\end{split}$

and

$\begin{split}\begin{array}[]{c@{\qquad}c}\displaystyle
D_{0}(z)=\sigma^{2}B_{0}(z),\qquad\displaystyle
D_{1}(z)=D_{2}(z)=\sigma^{2}B_{1}(z),\qquad\displaystyle
D_{3}(z)=2\sigma^{2}B_{2}(z).\end{array}\end{split}$

Then

$\lvert D_{1}(z)+iD_{2}(z)\rvert^{2}=\lvert D_{1}(z)\rvert^{2}+\lvert
D_{2}(z)\rvert^{2}=2\sigma^{4}\lvert B_{1}(z)\rvert^{2}.$

The following result is obtained by using these substitutions in Theorem 2.

######  Theorem 3.

If $\sigma_{a_{j}}^{2}=\sigma_{b_{j}}^{2}=\sigma^{2}$ for $0\leq j\leq N$,
then for all $N>1$ one has

$\begin{split}h_{N,\boldsymbol{K}}(z)=\frac{1}{\pi
B_{0}(z)}\exp\left(-\frac{K_{1}^{2}+K_{2}^{2}}{2\sigma^{2}B_{0}(z)}\right)\left\\{B_{2}(z)-\left(\frac{\lvert
B_{1}(z)\rvert}{B_{0}(z)}\right)^{2}\left(B_{0}(z)-\frac{K_{1}^{2}+K_{2}^{2}}{2\sigma^{2}}\right)\right\\},\end{split}$

where

$\begin{split}\begin{array}[]{c@{\qquad}c}\displaystyle
B_{0}(z)=\sum_{j=0}^{N}\lvert f_{j}(z)\rvert^{2},\qquad\displaystyle
B_{1}(z)=\sum_{j=0}^{N}\overline{f_{j}(z)}f_{j}^{\prime}(z),\qquad\displaystyle
B_{2}(z)=\sum_{j=0}^{N}\lvert
f_{j}^{\prime}(z)\rvert^{2}.\end{array}\end{split}$

Consequently, when $\sigma^{2}$ is set to be one, Theorem 1 in [4] is
recovered. Further, if $\boldsymbol{K}$ is the zero vector, Corollary 3 in [4]
is recovered. This latter result was proved independently by Yeager [19] and
one of the authors [14].

In addition, the following result follows from Theorem 2.

######  Corollary 1.

If the vector $\boldsymbol{K}$ is restricted to a circle of radius $K>0$, then
for all $N>1$ one has

$\begin{split}h_{N,\boldsymbol{K}}(z)&=\frac{1}{2\pi
D_{0}(z)}\exp\left(-\frac{K^{2}(Y_{1}(z)-Y_{2}(z)+Y_{3}(z))}{2D_{0}(z)^{2}}\right)\\\
&\hskip 30.00005pt\times\left\\{D_{3}(z)-\frac{\lvert
D_{1}(z)\rvert^{2}}{D_{0}(z)}\left(\frac{Y_{2}(z)+Y_{3}(z)}{D_{0}(z)}-\frac{K^{2}(Y_{2}(z)-Y_{3}(z))(Y_{1}(z)-Y_{2}(z)-1)}{D_{0}(z)^{3}}\right)\right.\\\
&\hskip 60.00009pt-\frac{\lvert
D_{2}(z)\rvert^{2}}{D_{0}(z)}\left(\frac{Y_{1}(z)+Y_{2}(z)}{D_{0}(z)}-\frac{K^{2}(Y_{1}(z)-Y_{2}(z))(Y_{2}(z)-Y_{3}(z)+1)}{D_{0}(z)^{3}}\right)\\\
&\hskip 90.00014pt\left.+\frac{\lvert
D_{1}(z)+iD_{2}(z)\rvert^{2}}{D_{0}(z)}\left(\frac{Y_{2}(z)}{D_{0}(z)}-\frac{K^{2}(Y_{1}(z)-Y_{2}(z))(Y_{2}(z)-Y_{3}(z))}{D_{0}(z)^{3}}\right)\right\\}.\end{split}$

A special case of Corollary 1 follows.

######  Corollary 2.

If $\boldsymbol{K}$ is the zero vector, then for all $N>1$ one has

$\begin{split}h_{N,\boldsymbol{0}}(z)&=\frac{D_{0}(z)^{2}D_{3}(z)-\lvert
D_{1}(z)\rvert^{2}(Y_{2}(z)+Y_{3}(z))-\lvert
D_{2}(z)\rvert^{2}(Y_{1}(z)+Y_{2}(z))+\lvert
D_{1}(z)+iD_{2}(z)\rvert^{2}Y_{2}(z)}{2\pi D_{0}(z)^{3}}.\end{split}$

The proof of Theorem 2, in the spirit of the method credited to Ibragimov and
Zeitouni [11], is presented in Section 2. In relation to the works of Rezakhah
and Shemehsavar [15] and Rezakhah and Soltani [16], an application of Theorem
2 entailing a sequence of successive observations of a Brownian motion is
given in Section 3. Theorem 2 can be extended further by letting the $a_{j}$
and $b_{j}$ to be mutually i.i.d. real random variables such that
$a_{j}\sim\mathcal{N}(\mu_{a_{j}},\sigma_{a_{j}}^{2})$ and
$b_{j}\sim\mathcal{N}(\mu_{b_{j}},\sigma_{b_{j}}^{2})$ for $0\leq j\leq N$.

Toward this end, let

$\begin{split}Y_{1}^{\ast}(z)&=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}u_{j}^{2}+\sigma_{b_{j}}^{2}v_{j}^{2})-\left(\sum_{j=0}^{N}(\mu_{a_{j}}u_{j}-\mu_{b_{j}}v_{j})\right)^{2},\\\
Y_{2}^{\ast}(z)&=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}-\sigma_{b_{j}}^{2})u_{j}v_{j}-\left(\sum_{j=0}^{N}(\mu_{a_{j}}u_{j}-\mu_{b_{j}}v_{j})\right)\left(\sum_{j=0}^{N}(\mu_{a_{j}}v_{j}+\mu_{b_{j}}u_{j})\right),\\\
Y_{3}^{\ast}(z)&=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}v_{j}^{2}+\sigma_{b_{j}}^{2}u_{j}^{2})-\left(\sum_{j=0}^{N}(\mu_{a_{j}}v_{j}+\mu_{b_{j}}u_{j})\right)^{2}.\end{split}$

Then define

$M(z)=\sum_{j=0}^{N}E(\eta_{j})f_{j}^{\prime}(z)$

and

$\begin{split}D_{0}^{\ast}(z)&=\sqrt{Y_{1}^{\ast}(z)Y_{3}^{\ast}(z)-Y_{2}^{\ast}(z)^{2}},\\\
D_{2}^{\ast}(z)&=\sum_{j=0}^{N}(B_{j,2}(z)-iA_{j,2}(z))f_{j}^{\prime}(z),\end{split}\qquad\begin{split}D_{1}^{\ast}(z)&=\sum_{j=0}^{N}(A_{j,1}(z)-iB_{j,1}(z))f_{j}^{\prime}(z),\\\
D_{3}^{\ast}(z)&=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}+\sigma_{b_{j}}^{2})\lvert
f_{j}^{\prime}(z)\rvert^{2},\end{split}$

where

$\begin{split}A_{j,1}(z)&=\sigma_{a_{j}}^{2}u_{j}-\mu_{a_{j}}E(X_{1}),\\\
B_{j,1}(z)&=\sigma_{b_{j}}^{2}v_{j}+\mu_{b_{j}}E(X_{1}),\end{split}\qquad\begin{split}A_{j,2}(z)&=\sigma_{a_{j}}^{2}v_{j}-\mu_{a_{j}}E(X_{2}),\\\
B_{j,2}(z)&=\sigma_{b_{j}}^{2}u_{j}-\mu_{b_{j}}E(X_{2}).\end{split}$

The following theorem is proved in Section 4.

######  Theorem 4.

Provided all the conditions imposed on $S_{N}(z)$ in (1) and $T$ are
satisfied, then for all $N>1$ one has

$\begin{split}&h_{N,\boldsymbol{K}}(z)=\frac{1}{2\pi D_{0}^{\ast}(z)}\\\
&\hskip
5.0pt\times\exp\left(-\frac{(K_{1}-E(X_{1}))^{2}Y_{3}^{\ast}(z)+(K_{2}-E(X_{2}))^{2}Y_{1}^{\ast}(z)-2(K_{1}-E(X_{1}))(K_{2}-E(X_{2}))Y_{2}^{\ast}(z)}{2D_{0}^{\ast}(z)^{2}}\right)\\\
&\hskip 5.0pt\times\left\\{D_{3}^{\ast}(z)-\frac{\lvert
D_{1}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\left(Y_{3}^{\ast}(z)-\frac{((K_{1}-E(X_{1}))Y_{3}^{\ast}(z)-(K_{2}-E(X_{2}))Y_{2}^{\ast}(z))^{2}}{D_{0}^{\ast}(z)^{2}}\right)\right.\\\
&\hskip 5.0pt\left.-\frac{\lvert
D_{2}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\left(Y_{1}^{\ast}(z)-\frac{((K_{1}-E(X_{1}))Y_{2}^{\ast}(z)-(K_{2}-E(X_{2}))Y_{1}^{\ast}(z))^{2}}{D_{0}^{\ast}(z)^{2}}\right)\right.\\\
&\hskip 5.0pt\left.+\left(\frac{\lvert
D_{1}^{\ast}(z)+iD_{2}^{\ast}(z)\rvert^{2}-\lvert
D_{1}^{\ast}(z)\rvert^{2}-\lvert
D_{2}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\right)\right.\\\ &\hskip
5.0pt\left.\times\left(Y_{2}^{\ast}(z)-\frac{((K_{1}-E(X_{1}))Y_{3}^{\ast}(z)-(K_{2}-E(X_{2}))Y_{2}^{\ast}(z))((K_{1}-E(X_{1}))Y_{2}^{\ast}(z)-(K_{2}-E(X_{2}))Y_{1}^{\ast}(z))}{D_{0}^{\ast}(z)^{2}}\right)\right.\\\
&\hskip 5.0pt\left.-\left(\frac{\lvert M(z)+D_{1}^{\ast}(z)\rvert^{2}-\lvert
M(z)\rvert^{2}-\lvert
D_{1}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\right)((K_{1}-E(X_{1}))Y_{3}^{\ast}(z)-(K_{2}-E(X_{2}))Y_{2}^{\ast}(z))\right.\\\
&\hskip 5.0pt\left.+\left(\frac{\lvert M(z)+iD_{2}^{\ast}(z)\rvert^{2}-\lvert
M(z)\rvert^{2}-\lvert
D_{2}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\right)((K_{1}-E(X_{1}))Y_{2}^{\ast}(z)-(K_{2}-E(X_{2}))Y_{1}^{\ast}(z))\right\\}.\end{split}$

Several consequences of Theorem 4 are of special interest. These are derived
in Section 5.

##  2\. Proof of Theorem 2

The proof of Theorem 2 starts with the decomposition

$S_{N}(z)=X_{1}+iX_{2},$

where

$X_{1}=\sum_{j=0}^{N}(a_{j}u_{j}-b_{j}v_{j}),\qquad
X_{2}=\sum_{j=0}^{N}(a_{j}v_{j}+b_{j}u_{j}).$

If the column vector

$\boldsymbol{X}=(X_{1},X_{2})^{\prime}$

genuinely represents a two-dimensional random field, then for $z=x+iy$ the
Jacobian matrix of the random transformation $(x,y)\rightarrow(X_{1},X_{2})$
is

$\nabla\boldsymbol{X}=\begin{pmatrix}[l]\displaystyle\sum_{j=0}^{N}\left(a_{j}\frac{\partial
u_{j}}{\partial x}-b_{j}\frac{\partial v_{j}}{\partial
x}\right)&\displaystyle\sum_{j=0}^{N}\left(a_{j}\frac{\partial v_{j}}{\partial
x}+b_{j}\frac{\partial u_{j}}{\partial x}\right)\\\
\displaystyle\sum_{j=0}^{N}\left(-a_{j}\frac{\partial v_{j}}{\partial
x}-b_{j}\frac{\partial u_{j}}{\partial
x}\right)&\displaystyle\sum_{j=0}^{N}\left(a_{j}\frac{\partial u_{j}}{\partial
x}-b_{j}\frac{\partial v_{j}}{\partial x}\right)\end{pmatrix}$

and

$\displaystyle\det(\nabla\boldsymbol{X})$
$\displaystyle=\sum_{j=0}^{N}\sum_{k=0}^{N}\left((a_{j}a_{k}+b_{j}b_{k})\left(\frac{\partial
u_{j}}{\partial x}\frac{\partial u_{k}}{\partial x}+\frac{\partial
v_{j}}{\partial x}\frac{\partial v_{k}}{\partial
x}\right)+(a_{j}b_{k}-b_{j}a_{k})\left(\frac{\partial v_{j}}{\partial
x}\frac{\partial u_{k}}{\partial x}-\frac{\partial u_{j}}{\partial
x}\frac{\partial v_{k}}{\partial x}\right)\right)$
$\displaystyle=\sum_{j=0}^{N}(a_{j}^{2}+b_{j}^{2})\left(\left(\frac{\partial
u_{j}}{\partial x}\right)^{2}+\left(\frac{\partial v_{j}}{\partial
x}\right)^{2}\right)+\sum_{j=0}^{N}\sum_{\begin{subarray}{c}k=0\\\ k\neq
j\end{subarray}}^{N}\left((a_{j}a_{k}+b_{j}b_{k})\left(\frac{\partial
u_{j}}{\partial x}\frac{\partial u_{k}}{\partial x}+\frac{\partial
v_{j}}{\partial x}\frac{\partial v_{k}}{\partial x}\right)\right.$
$\displaystyle\hskip
220.00034pt\left.+(a_{j}b_{k}-b_{j}a_{k})\left(\frac{\partial v_{j}}{\partial
x}\frac{\partial u_{k}}{\partial x}-\frac{\partial u_{j}}{\partial
x}\frac{\partial v_{k}}{\partial x}\right)\right).$ (5)

The evaluation of $h_{N,\boldsymbol{K}}(z)$ leads to the computation of the
expected value of a quadratic form $\det(\nabla\boldsymbol{X})$ of i.i.d. real
normal random variables conditioned on two linear combinations. Observe that
$\det(\nabla\boldsymbol{X})$ is always nonnegative. Since $N$ is fixed, $T$
contains not more than a finite number of zeros of

$\boldsymbol{X}=\boldsymbol{K},$ (6)

where

$\boldsymbol{K}=(K_{1},K_{2})^{\prime}.$ (7)

Since the set of zeros of (6) is of measure zero, assume the boundary
$\partial T$ does not contain any zeros of (6) and $T$ does not contain any
such zeros such that $\det(\nabla\boldsymbol{X})=0$. Theorem 1 applies. Thus,

$h_{N,\boldsymbol{K}}(z)=E(\det(\nabla\boldsymbol{X})\mid\boldsymbol{X}=\boldsymbol{K})\,p_{X_{1},X_{2}}(\boldsymbol{K}^{\prime}),$
(8)

where $p_{X_{1},X_{2}}(\boldsymbol{K}^{\prime})$ denotes the probability
density of the random vector $\boldsymbol{X}$ at $\boldsymbol{K}^{\prime}$. By
(6), and since $X_{1}$ and $X_{2}$ are linear forms with respect to $\eta_{j}$
for $0\leq j\leq N$, $h_{N,\boldsymbol{K}}(z)$ is the conditional mean of a
quadratic form with respect to $\eta_{j}$ for $0\leq j\leq N$. This form can
be calculated in terms of components by means of multivariate analysis.

Based on the assumption that the scalar random variables are independent and
normally distributed, the multivariate random vectors

$\boldsymbol{a}=(a_{0},a_{1},\ldots,a_{N})^{\prime},\qquad\boldsymbol{b}=(b_{0},b_{1},\ldots,b_{N})^{\prime}$

are such that

$\operatorname{Cov}(\boldsymbol{a},\boldsymbol{b}\mid\boldsymbol{X}=\boldsymbol{K})=\begin{pmatrix}[l]\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{a},\boldsymbol{X}}&\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{b},\boldsymbol{X}}\\\
\boldsymbol{\Sigma}_{\boldsymbol{b}\boldsymbol{a},\boldsymbol{X}}&\boldsymbol{\Sigma}_{\boldsymbol{b}\boldsymbol{b},\boldsymbol{X}}\end{pmatrix}.$
(9)

The elements of this covariance matrix are computed using

$\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{b},\boldsymbol{X}}=\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{b}}-\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{X}}\boldsymbol{\Sigma}_{\boldsymbol{X}\boldsymbol{X}}^{-1}\boldsymbol{\Sigma}_{\boldsymbol{X}\boldsymbol{b}}$
(10)

and the corresponding expression

$\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{b}}=E((\boldsymbol{a}-E(\boldsymbol{a}))(\boldsymbol{b}-E(\boldsymbol{b}))^{\prime}).$

Since the distribution of $a_{j}$ and $b_{j}$ is central for $0\leq j\leq N$,
$E(\boldsymbol{a})=0$ and $E(\boldsymbol{b})=0$. Clearly,

$\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{b}}=E(\boldsymbol{a}\boldsymbol{b}^{\prime}).$
(11)

Thusly, the conditional expected values are expressed in terms of
unconditional expected values and covariances.

Then, if $E(X_{1})=0$ and $E(X_{2})=0$, $E(\boldsymbol{X})=0$, whence, by
(11),

$\boldsymbol{\Sigma}_{\boldsymbol{X}\boldsymbol{X}}=\begin{pmatrix}[l]E(X_{1}X_{1})&E(X_{1}X_{2})\\\
E(X_{2}X_{1})&E(X_{2}X_{2})\end{pmatrix}=\begin{pmatrix}[l]Y_{1}&Y_{2}\\\
Y_{2}&Y_{3}\end{pmatrix}.$ (12)

It follows that, if $X_{1}$ and $X_{2}$ are not strictly correlated,

$\det(\boldsymbol{\Sigma}_{\boldsymbol{X}\boldsymbol{X}})=Y_{1}Y_{3}-Y_{2}^{2}.$

This quantity is strictly positive. Thus,

$\boldsymbol{\Sigma}_{\boldsymbol{X}\boldsymbol{X}}^{-1}=\frac{1}{Y_{1}Y_{3}-Y_{2}^{2}}\begin{pmatrix}[r]Y_{3}&-Y_{2}\\\
-Y_{2}&Y_{1}\end{pmatrix}.$ (13)

Expanding our definitions,
$\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{a},\boldsymbol{X}},\boldsymbol{\Sigma}_{\boldsymbol{b}\boldsymbol{b},\boldsymbol{X}},\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{b},\boldsymbol{X}}$,
and $\boldsymbol{\Sigma}_{\boldsymbol{b}\boldsymbol{a},\boldsymbol{X}}$ are
obtained as follows. Direct evaluation shows that

$\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{a}}=E(a_{j}a_{k})=\delta_{jk}\sigma_{a_{j}}^{2}$
(14)

and

$\boldsymbol{\Sigma}_{\boldsymbol{b}\boldsymbol{b}}=E(b_{j}b_{k})=\delta_{jk}\sigma_{b_{j}}^{2},$
(15)

where

$\delta_{jk}=\left\\{\begin{array}[]{ll}1&\mbox{if $j=k$,}\\\ 0&\mbox{if
$j\neq k$.}\end{array}\right.$

Further, notice that

$\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{b}}=E(a_{j}b_{k})=0$ (16)

and

$\boldsymbol{\Sigma}_{\boldsymbol{b}\boldsymbol{a}}=0.$ (17)

Next, since $E(a_{j}X_{1})=\sigma_{a_{j}}^{2}u_{j}$ and
$E(a_{j}X_{2})=\sigma_{a_{j}}^{2}v_{j}$ for $0\leq j\leq N$,

$\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{X}}=(\begin{matrix}[l]\sigma_{a_{j}}^{2}u_{j}&\sigma_{a_{j}}^{2}v_{j}\end{matrix}),$
(18)

whence

$\boldsymbol{\Sigma}_{\boldsymbol{X}\boldsymbol{a}}=\begin{pmatrix}[l]\sigma_{a_{k}}^{2}u_{k}\\\
\sigma_{a_{k}}^{2}v_{k}\end{pmatrix}.$ (19)

Analogously, since $E(b_{j}X_{1})=-\sigma_{b_{j}}^{2}v_{j}$ and
$E(b_{j}X_{2})=\sigma_{b_{j}}^{2}u_{j}$ for $0\leq j\leq N$,

$\boldsymbol{\Sigma}_{\boldsymbol{b}\boldsymbol{X}}=(\begin{matrix}[l]-\sigma_{b_{j}}^{2}v_{j}&\sigma_{b_{j}}^{2}u_{j}\end{matrix}),$
(20)

whence

$\boldsymbol{\Sigma}_{\boldsymbol{X}\boldsymbol{b}}=\begin{pmatrix}[r]-\sigma_{b_{k}}^{2}v_{k}\\\
\sigma_{b_{k}}^{2}u_{k}\end{pmatrix}$ (21)

for $0\leq k\leq N$.

Then, from (10), (13), (14), (18) and (19) for the $j$th row and $k$th column

$\begin{split}\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{a},\boldsymbol{X}}&=\delta_{jk}\sigma_{a_{j}}^{2}-\sigma_{a_{j}}^{2}\sigma_{a_{k}}^{2}\left(\frac{Y_{1}v_{j}v_{k}-Y_{2}(u_{j}v_{k}+v_{j}u_{k})+Y_{3}u_{j}u_{k}}{Y_{1}Y_{3}-Y_{2}^{2}}\right).\end{split}$
(22)

From (10), (13), (15), (20) and (21)

$\begin{split}\boldsymbol{\Sigma}_{\boldsymbol{b}\boldsymbol{b},\boldsymbol{X}}&=\delta_{jk}\sigma_{b_{j}}^{2}-\sigma_{b_{j}}^{2}\sigma_{b_{k}}^{2}\left(\frac{Y_{1}u_{j}u_{k}+Y_{2}(u_{j}v_{k}+v_{j}u_{k})+Y_{3}v_{j}v_{k}}{Y_{1}Y_{3}-Y_{2}^{2}}\right).\end{split}$
(23)

From (10), (13), (16), (18) and (21)

$\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{b},\boldsymbol{X}}=-\sigma_{a_{j}}^{2}\sigma_{b_{k}}^{2}\left(\frac{Y_{1}v_{j}u_{k}-Y_{2}(u_{j}u_{k}-v_{j}v_{k})-Y_{3}u_{j}v_{k}}{Y_{1}Y_{3}-Y_{2}^{2}}\right).$
(24)

From (10), (13), (17), (19) and (20)

$\boldsymbol{\Sigma}_{\boldsymbol{b}\boldsymbol{a},\boldsymbol{X}}=-\sigma_{b_{j}}^{2}\sigma_{a_{k}}^{2}\left(\frac{Y_{1}u_{j}v_{k}-Y_{2}(u_{j}u_{k}-v_{j}v_{k})-Y_{3}v_{j}u_{k}}{Y_{1}Y_{3}-Y_{2}^{2}}\right).$
(25)

The mean function in (8) is then found by applications of

$E(\boldsymbol{a}\mid\boldsymbol{X}=\boldsymbol{K})=E(\boldsymbol{a})+\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{X}}\boldsymbol{\Sigma}_{\boldsymbol{X}\boldsymbol{X}}^{-1}(\boldsymbol{K}-E(\boldsymbol{X})),$

which, for the aforesaid reasons, reduces to

$E(\boldsymbol{a}\mid\boldsymbol{X}=\boldsymbol{K})=\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{X}}\boldsymbol{\Sigma}_{\boldsymbol{X}\boldsymbol{X}}^{-1}\boldsymbol{K}.$
(26)

From (7), (13), (18) and (26)

$E(a_{j}\mid\boldsymbol{X}=\boldsymbol{K})=\sigma_{a_{j}}^{2}\left(\frac{(K_{1}Y_{3}-K_{2}Y_{2})u_{j}-(K_{1}Y_{2}-K_{2}Y_{1})v_{j}}{Y_{1}Y_{3}-Y_{2}^{2}}\right).$
(27)

The formula for $E(\boldsymbol{b}\mid\boldsymbol{X}=\boldsymbol{K})$ is
analogous. Then, from (7), (13), (20) and (26) with the obvious substitution

$E(b_{j}\mid\boldsymbol{X}=\boldsymbol{K})=-\sigma_{b_{j}}^{2}\left(\frac{(K_{1}Y_{3}-K_{2}Y_{2})v_{j}+(K_{1}Y_{2}-K_{2}Y_{1})u_{j}}{Y_{1}Y_{3}-Y_{2}^{2}}\right).$
(28)

Then, from (9), (22)–(25) and (27)

$\begin{split}&E(a_{j}a_{k}\mid\boldsymbol{X}=\boldsymbol{K})=E(a_{j}\mid\boldsymbol{X}=\boldsymbol{K})E(a_{k}\mid\boldsymbol{X}=\boldsymbol{K})+\operatorname{Cov}(a_{j},a_{k}\mid\boldsymbol{X}=\boldsymbol{K})=\sigma_{a_{j}}^{2}\sigma_{a_{k}}^{2}\\\
&\hskip
10.00002pt\times\left(\frac{(K_{1}Y_{3}-K_{2}Y_{2})^{2}u_{j}u_{k}+(K_{1}Y_{2}-K_{2}Y_{1})^{2}v_{j}v_{k}-(K_{1}Y_{3}-K_{2}Y_{2})(K_{1}Y_{2}-K_{2}Y_{1})(u_{j}v_{k}+v_{j}u_{k})}{(Y_{1}Y_{3}-Y_{2}^{2})^{2}}\right)\\\
&\hskip
20.00003pt+\delta_{jk}\sigma_{a_{j}}^{2}-\sigma_{a_{j}}^{2}\sigma_{a_{k}}^{2}\left(\frac{Y_{3}u_{j}u_{k}+Y_{1}v_{j}v_{k}-Y_{2}(u_{j}v_{k}+v_{j}u_{k})}{Y_{1}Y_{3}-Y_{2}^{2}}\right).\end{split}$
(29)

From (9), (22)–(25) and (28)

$\begin{split}&E(b_{j}b_{k}\mid\boldsymbol{X}=\boldsymbol{K})=E(b_{j}\mid\boldsymbol{X}=\boldsymbol{K})E(b_{k}\mid\boldsymbol{X}=\boldsymbol{K})+\operatorname{Cov}(b_{j},b_{k}\mid\boldsymbol{X}=\boldsymbol{K})=\sigma_{b_{j}}^{2}\sigma_{b_{k}}^{2}\\\
&\hskip
10.00002pt\times\left(\frac{(K_{1}Y_{2}-K_{2}Y_{1})^{2}u_{j}u_{k}+(K_{1}Y_{3}-K_{2}Y_{2})^{2}v_{j}v_{k}+(K_{1}Y_{2}-K_{2}Y_{1})(K_{1}Y_{3}-K_{2}Y_{2})(u_{j}v_{k}+v_{j}u_{k})}{(Y_{1}Y_{3}-Y_{2}^{2})^{2}}\right)\\\
&\hskip
20.00003pt+\delta_{jk}\sigma_{b_{j}}^{2}-\sigma_{b_{j}}^{2}\sigma_{b_{k}}^{2}\left(\frac{Y_{1}u_{j}u_{k}+Y_{3}v_{j}v_{k}+Y_{2}(u_{j}v_{k}+v_{j}u_{k})}{Y_{1}Y_{3}-Y_{2}^{2}}\right).\end{split}$
(30)

From (9), (22)–(25), (27) and (28)

$\begin{split}&E(a_{j}b_{k}\mid\boldsymbol{X}=\boldsymbol{K})=E(a_{j}\mid\boldsymbol{X}=\boldsymbol{K})E(b_{k}\mid\boldsymbol{X}=\boldsymbol{K})+\operatorname{Cov}(a_{j},b_{k}\mid\boldsymbol{X}=\boldsymbol{K})=\sigma_{a_{j}}^{2}\sigma_{b_{k}}^{2}\\\
&\hskip
10.00002pt\times\left(\frac{(K_{1}Y_{2}-K_{2}Y_{1})^{2}v_{j}u_{k}-(K_{1}Y_{3}-K_{2}Y_{2})^{2}u_{j}v_{k}-(K_{1}Y_{3}-K_{2}Y_{2})(K_{1}Y_{2}-K_{2}Y_{1})(u_{j}u_{k}-v_{j}v_{k})}{(Y_{1}Y_{3}-Y_{2}^{2})^{2}}\right)\\\
&\hskip
20.00003pt-\sigma_{a_{j}}^{2}\sigma_{b_{k}}^{2}\left(\frac{Y_{1}v_{j}u_{k}-Y_{2}(u_{j}u_{k}-v_{j}v_{k})-Y_{3}u_{j}v_{k}}{Y_{1}Y_{3}-Y_{2}^{2}}\right)\end{split}$
(31)

and, likewise,

$\begin{split}&E(b_{j}a_{k}\mid\boldsymbol{X}=\boldsymbol{K})=E(b_{j}\mid\boldsymbol{X}=\boldsymbol{K})E(a_{k}\mid\boldsymbol{X}=\boldsymbol{K})+\operatorname{Cov}(b_{j},a_{k}\mid\boldsymbol{X}=\boldsymbol{K})=\sigma_{b_{j}}^{2}\sigma_{a_{k}}^{2}\\\
&\hskip
10.00002pt\times\left(\frac{(K_{1}Y_{2}-K_{2}Y_{1})^{2}u_{j}v_{k}-(K_{1}Y_{3}-K_{2}Y_{2})^{2}v_{j}u_{k}-(K_{1}Y_{3}-K_{2}Y_{2})(K_{1}Y_{2}-K_{2}Y_{1})(u_{j}u_{k}-v_{j}v_{k})}{(Y_{1}Y_{3}-Y_{2}^{2})^{2}}\right)\\\
&\hskip
20.00003pt-\sigma_{b_{j}}^{2}\sigma_{a_{k}}^{2}\left(\frac{Y_{1}u_{j}v_{k}-Y_{2}(u_{j}u_{k}-v_{j}v_{k})-Y_{3}v_{j}u_{k}}{Y_{1}Y_{3}-Y_{2}^{2}}\right).\end{split}$
(32)

Then, from (29) and (30)

$\begin{split}&E(a_{j}a_{k}+b_{j}b_{k}\mid\boldsymbol{X}=\boldsymbol{K})=\frac{1}{(Y_{1}Y_{3}-Y_{2}^{2})^{2}}((K_{1}Y_{3}-K_{2}Y_{2})^{2}(\sigma_{a_{j}}^{2}\sigma_{a_{k}}^{2}u_{j}u_{k}+\sigma_{b_{j}}^{2}\sigma_{b_{k}}^{2}v_{j}v_{k})\\\
&\hskip
50.00008pt+(K_{1}Y_{2}-K_{2}Y_{1})^{2}(\sigma_{a_{j}}^{2}\sigma_{a_{k}}^{2}v_{j}v_{k}+\sigma_{b_{j}}^{2}\sigma_{b_{k}}^{2}u_{j}u_{k})\\\
&\hskip
50.00008pt-(K_{1}Y_{3}-K_{2}Y_{2})(K_{1}Y_{2}-K_{2}Y_{1})(u_{j}v_{k}+v_{j}u_{k})(\sigma_{a_{j}}^{2}\sigma_{a_{k}}^{2}-\sigma_{b_{j}}^{2}\sigma_{b_{k}}^{2}))\\\
&\hskip
50.00008pt-\frac{1}{Y_{1}Y_{3}-Y_{2}^{2}}(Y_{1}(\sigma_{a_{j}}^{2}\sigma_{a_{k}}^{2}v_{j}v_{k}+\sigma_{b_{j}}^{2}\sigma_{b_{k}}^{2}u_{j}u_{k})+Y_{2}(u_{j}v_{k}+v_{j}u_{k})(\sigma_{b_{j}}^{2}\sigma_{b_{k}}^{2}-\sigma_{a_{j}}^{2}\sigma_{a_{k}}^{2})\\\
&\hskip
50.00008pt+Y_{3}(\sigma_{a_{j}}^{2}\sigma_{a_{k}}^{2}u_{j}u_{k}+\sigma_{b_{j}}^{2}\sigma_{b_{k}}^{2}v_{j}v_{k}))+\delta_{jk}(\sigma_{a_{j}}^{2}+\sigma_{b_{j}}^{2}).\end{split}$
(33)

From (31) and (32)

$\begin{split}&E(a_{j}b_{k}-b_{j}a_{k}\mid\boldsymbol{X}=\boldsymbol{K})=\frac{1}{(Y_{1}Y_{3}-Y_{2}^{2})^{2}}((K_{1}Y_{2}-K_{2}Y_{1})^{2}(\sigma_{a_{j}}^{2}\sigma_{b_{k}}^{2}v_{j}u_{k}-\sigma_{b_{j}}^{2}\sigma_{a_{k}}^{2}u_{j}v_{k})\\\
&\hskip
50.00008pt-(K_{1}Y_{3}-K_{2}Y_{2})^{2}(\sigma_{a_{j}}^{2}\sigma_{b_{k}}^{2}u_{j}v_{k}-\sigma_{b_{j}}^{2}\sigma_{a_{k}}^{2}v_{j}u_{k})\\\
&\hskip
50.00008pt-(K_{1}Y_{3}-K_{2}Y_{2})(K_{1}Y_{2}-K_{2}Y_{1})(u_{j}u_{k}-v_{j}v_{k})(\sigma_{a_{j}}^{2}\sigma_{b_{k}}^{2}-\sigma_{b_{j}}^{2}\sigma_{a_{k}}^{2}))\\\
&\hskip
50.00008pt-\frac{1}{Y_{1}Y_{3}-Y_{2}^{2}}(Y_{1}(\sigma_{a_{j}}^{2}\sigma_{b_{k}}^{2}v_{j}u_{k}-\sigma_{b_{j}}^{2}\sigma_{a_{k}}^{2}u_{j}v_{k})+Y_{2}(v_{j}v_{k}-u_{j}u_{k})(\sigma_{a_{j}}^{2}\sigma_{b_{k}}^{2}-\sigma_{b_{j}}^{2}\sigma_{a_{k}}^{2})\\\
&\hskip 50.00008pt-
Y_{3}(\sigma_{a_{j}}^{2}\sigma_{b_{k}}^{2}u_{j}v_{k}-\sigma_{b_{j}}^{2}\sigma_{a_{k}}^{2}v_{j}u_{k})).\end{split}$
(34)

Altogether, in view of (2), (33) and (34), after all the necessary
simplifications,

$\begin{split}&E(\det(\nabla\boldsymbol{X})\mid\boldsymbol{X}=\boldsymbol{K})=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}+\sigma_{b_{j}}^{2})\left(\left(\frac{\partial
u_{j}}{\partial x}\right)^{2}+\left(\frac{\partial v_{j}}{\partial
x}\right)^{2}\right)\\\ &\hskip
40.00006pt-\left(\frac{Y_{1}+Y_{2}}{Y_{1}Y_{3}-Y_{2}^{2}}-\frac{(K_{1}Y_{2}-K_{2}Y_{1})^{2}+(K_{1}Y_{3}-K_{2}Y_{2})(K_{1}Y_{2}-K_{2}Y_{1})}{(Y_{1}Y_{3}-Y_{2}^{2})^{2}}\right)\\\
&\hskip
120.00018pt\times\left|\sum_{j=0}^{N}(\sigma_{b_{j}}^{2}u_{j}-i\sigma_{a_{j}}^{2}v_{j})\left(\frac{\partial
u_{j}}{\partial x}+i\frac{\partial v_{j}}{\partial x}\right)\right|^{2}\\\
&\hskip
40.00006pt-\left(\frac{Y_{2}+Y_{3}}{Y_{1}Y_{3}-Y_{2}^{2}}-\frac{(K_{1}Y_{3}-K_{2}Y_{2})^{2}+(K_{1}Y_{3}-K_{2}Y_{2})(K_{1}Y_{2}-K_{2}Y_{1})}{(Y_{1}Y_{3}-Y_{2}^{2})^{2}}\right)\\\
&\hskip
120.00018pt\times\left|\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}u_{j}-i\sigma_{b_{j}}^{2}v_{j})\left(\frac{\partial
u_{j}}{\partial x}+i\frac{\partial v_{j}}{\partial x}\right)\right|^{2}\\\
&\hskip
40.00006pt+\left(\frac{Y_{2}}{Y_{1}Y_{3}-Y_{2}^{2}}-\frac{(K_{1}Y_{3}-K_{2}Y_{2})(K_{1}Y_{2}-K_{2}Y_{1})}{(Y_{1}Y_{3}-Y_{2}^{2})^{2}}\right)\\\
&\hskip
120.00018pt\times\left|\sum_{j=0}^{N}((\sigma_{a_{j}}^{2}u_{j}-i\sigma_{b_{j}}^{2}v_{j})+i(\sigma_{b_{j}}^{2}u_{j}-i\sigma_{a_{j}}^{2}v_{j}))\left(\frac{\partial
u_{j}}{\partial x}+i\frac{\partial v_{j}}{\partial
x}\right)\right|^{2}.\end{split}$ (35)

Since $X_{1}$ and $X_{2}$ are random variables distributed according to the
normal law, their joint density is

$\begin{split}p_{X_{1},X_{2}}(\boldsymbol{K}^{\prime})&=\frac{1}{2\pi\sqrt{\det(\boldsymbol{\Sigma}_{\boldsymbol{X}\boldsymbol{X}})}}\exp\left(-\frac{1}{2}(\boldsymbol{K}-E(\boldsymbol{X}))^{\prime}\boldsymbol{\Sigma}_{\boldsymbol{X}\boldsymbol{X}}^{-1}(\boldsymbol{K}-E(\boldsymbol{X}))\right)\\\
&=\frac{1}{2\pi\sqrt{Y_{1}Y_{3}-Y_{2}^{2}}}\exp\left(-\frac{K_{1}^{2}Y_{3}-2K_{1}K_{2}Y_{2}+K_{2}^{2}Y_{1}}{2(Y_{1}Y_{3}-Y_{2}^{2})}\right).\end{split}$
(36)

Hence, in accordance with (8), (35) and (36), the required result is proved.

##  3\. Reformulation of Theorem 2

If $\\{A_{j}\\}_{j=0}^{\infty}$ and $\\{B_{j}\\}_{j=0}^{\infty}$ are sequences
of i.i.d. real normal random variables for which the respective increments
$A_{j}-A_{j-1}$ and $B_{j}-B_{j-1}$ are independent for $j\geq 0$ and
$A_{-1}=B_{-1}=0$ by convention, then the increments

$\Delta_{j}=(A_{j}-A_{j-1})+i(B_{j}-B_{j-1})$

are independent complex normal random variables with mean zero and finite
$\operatorname{Var}(\Delta_{j})$ such that
$A_{j}+iB_{j}=\Delta_{0}+\Delta_{1}+\cdots+\Delta_{j}$ for $j\geq 0$. Then
$\\{A_{j}+iB_{j}\\}_{j=0}^{\infty}$ can be interpreted as a sequence of
successive observations of a Brownian motion. More precisely,
$A_{j}+iB_{j}=W(t_{j})$ for $j\geq 0$, where $t_{0}<t_{1}<\ldots$ and
$\\{W(t)\\}_{t=0}^{\infty}$ is the standard Brownian motion. It is plain that
$\operatorname{Var}(\Delta_{j})$ is the distance between the successive times
$t_{j-1}$ and $t_{j}$ for $j\geq 0$. Thus, the sum in (2) assumes the form

$S_{N}(z)=\sum_{j=0}^{N}(A_{j}+iB_{j})f_{j}(z)=\sum_{k=0}^{N}F_{k}(z)\Delta_{k},$
(37)

where

$F_{k}(z)=\sum_{j=k}^{N}u_{j}(x,y)+i\sum_{j=k}^{N}v_{j}(x,y)$ (38)

for $0\leq k\leq N$. Observe that $\\{F_{k}(z)\\}_{k=0}^{N}$ is a sequence of
holomorphic functions that are real-valued on $\mathds{R}$. Hence,
$\overline{F_{k}(z)}=F_{k}(\overline{z})$ for $0\leq k\leq N$ and all
$z\in\mathds{C}$. The covariance matrix of $\Delta_{k}$ is given by

$\Gamma_{k}=\begin{pmatrix}[c]\sigma_{a_{k}}^{2}&0\\\
0&\sigma_{b_{k}}^{2}\end{pmatrix}$

for $0\leq k\leq N$. Then, from Theorem 2 the following result is attained.

######  Theorem 5.

Provided all the conditions imposed on $S_{N}(z)$ in (37) and (38) and $T$ are
satisfied, then for all $N>1$ the formula for $h_{N,\boldsymbol{K}}(z)$ in
Theorem 2 now holds for

$D_{0}(z)=\sqrt{Y_{1}(z)Y_{3}(z)-Y_{2}^{2}(z)},$

where

$\begin{array}[]{c@{\qquad}c}\displaystyle
Y_{1}(z)=\sum_{k=0}^{N}\left(\sigma_{a_{k}}^{2}\left(\sum_{j=k}^{N}u_{j}\right)^{2}+\sigma_{b_{k}}^{2}\left(\sum_{j=k}^{N}v_{j}\right)^{2}\right),\\\\[21.52771pt]
\displaystyle
Y_{2}(z)=\sum_{k=0}^{N}\left((\sigma_{a_{k}}^{2}-\sigma_{b_{k}}^{2})\left(\sum_{j=k}^{N}u_{j}\right)\left(\sum_{j=k}^{N}v_{j}\right)\right),\\\\[21.52771pt]
\displaystyle
Y_{3}(z)=\sum_{k=0}^{N}\left(\sigma_{b_{k}}^{2}\left(\sum_{j=k}^{N}u_{j}\right)^{2}+\sigma_{a_{k}}^{2}\left(\sum_{j=k}^{N}v_{j}\right)^{2}\right),\end{array}$

and

$\begin{array}[]{c@{\qquad}c}\displaystyle
D_{1}(z)=\sum_{k=0}^{N}\left(\sigma_{a_{k}}^{2}\sum_{j=k}^{N}u_{j}-i\sigma_{b_{k}}^{2}\sum_{j=k}^{N}v_{j}\right)\left(\sum_{j=k}^{N}\frac{\partial
u_{j}}{\partial x}+i\sum_{j=k}^{N}\frac{\partial v_{j}}{\partial
x}\right),\\\\[21.52771pt] \displaystyle
D_{2}(z)=\sum_{k=0}^{N}\left(\sigma_{b_{k}}^{2}\sum_{j=k}^{N}u_{j}-i\sigma_{a_{k}}^{2}\sum_{j=k}^{N}v_{j}\right)\left(\sum_{j=k}^{N}\frac{\partial
u_{j}}{\partial x}+i\sum_{j=k}^{N}\frac{\partial v_{j}}{\partial
x}\right),\\\\[21.52771pt] \displaystyle
D_{3}(z)=\sum_{k=0}^{N}(\sigma_{a_{k}}^{2}+\sigma_{b_{k}}^{2})\left(\left(\sum_{j=k}^{N}\frac{\partial
u_{j}}{\partial x}\right)^{2}+\left(\sum_{j=k}^{N}\frac{\partial
v_{j}}{\partial x}\right)^{2}\right).\end{array}$

##  4\. Proof of Theorem 4

The proof of Theorem 4 mirrors that of Theorem 2. Below, only the differences
are highlighted. Direct computation leads to the equalities

$\begin{array}[]{c@{\qquad}c}\displaystyle
E(X_{1})=\sum_{j=0}^{N}(\mu_{a_{j}}u_{j}-\mu_{b_{j}}v_{j}),\qquad\displaystyle
E(X_{2})=\sum_{j=0}^{N}(\mu_{a_{j}}v_{j}+\mu_{b_{j}}u_{j}),\\\\[12.91663pt]
\displaystyle
E(X_{1}^{2})=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}u_{j}^{2}+\sigma_{b_{j}}^{2}v_{j}^{2}),\qquad\displaystyle
E(X_{2}^{2})=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}v_{j}^{2}+\sigma_{b_{j}}^{2}u_{j}^{2}).\end{array}$

Further,

$\displaystyle
E(X_{1}X_{2})=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}-\sigma_{b_{j}}^{2})u_{j}v_{j}.$

It follows that

$\boldsymbol{\Sigma}_{\boldsymbol{X}\boldsymbol{X}}=\begin{pmatrix}[l]Y_{1}^{\ast}&Y_{2}^{\ast}\\\
Y_{2}^{\ast}&Y_{3}^{\ast}\end{pmatrix},$

and hence

$\boldsymbol{\Sigma}_{\boldsymbol{X}\boldsymbol{X}}^{-1}=\frac{1}{Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2}}\begin{pmatrix}[r]Y_{3}^{\ast}&-Y_{2}^{\ast}\\\
-Y_{2}^{\ast}&Y_{1}^{\ast}\end{pmatrix}.$

On expanding our definitions, for the $j$th row and $k$th column

$\begin{split}\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{a}}&=\delta_{jk}\sigma_{a_{j}}^{2}-\mu_{a_{j}}\mu_{a_{k}},\\\
\boldsymbol{\Sigma}_{\boldsymbol{b}\boldsymbol{a}}&=-\mu_{b_{j}}\mu_{a_{k}},\end{split}\qquad\begin{split}\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{b}}&=-\mu_{a_{j}}\mu_{b_{k}},\\\
\boldsymbol{\Sigma}_{\boldsymbol{b}\boldsymbol{b}}&=\delta_{jk}\sigma_{b_{j}}^{2}-\mu_{b_{j}}\mu_{b_{k}}.\end{split}$

Since

$\begin{split}E((a_{j}-E(a_{j}))(X_{1}-E(X_{1})))&=A_{j,1},\\\
E((b_{j}-E(b_{j}))(X_{1}-E(X_{1})))&=-B_{j,1},\end{split}\qquad\begin{split}E((a_{j}-E(a_{j}))(X_{2}-E(X_{2})))&=A_{j,2},\\\
E((b_{j}-E(b_{j}))(X_{2}-E(X_{2})))&=B_{j,2},\end{split}$

we have

$\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{X}}=(\begin{matrix}[l]A_{j,1}&A_{j,2}\end{matrix}),\qquad\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{X}}=(\begin{matrix}[l]-B_{j,1}&B_{j,2}\end{matrix}).$

Using (10), simple algebra leads to

$\displaystyle\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{a},\boldsymbol{X}}$
$\displaystyle=\delta_{jk}\sigma_{a_{j}}^{2}-\mu_{a_{j}}\mu_{a_{k}}-\frac{A_{j,1}A_{k,1}Y_{3}^{\ast}+A_{j,2}A_{k,2}Y_{1}^{\ast}-(A_{j,1}A_{k,2}+A_{j,2}A_{k,1})Y_{2}^{\ast}}{Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2}},$
$\displaystyle\boldsymbol{\Sigma}_{\boldsymbol{b}\boldsymbol{b},\boldsymbol{X}}$
$\displaystyle=\delta_{jk}\sigma_{b_{j}}^{2}-\mu_{b_{j}}\mu_{b_{k}}-\frac{B_{j,1}B_{k,1}Y_{3}^{\ast}+B_{j,1}B_{k,2}Y_{1}^{\ast}+(B_{j,1}B_{k,2}+B_{j,2}B_{k,1})Y_{2}^{\ast}}{Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2}},$
$\displaystyle\boldsymbol{\Sigma}_{\boldsymbol{a}\boldsymbol{b},\boldsymbol{X}}$
$\displaystyle=-\mu_{a_{j}}\mu_{b_{k}}+\frac{A_{j,1}B_{k,1}Y_{3}^{\ast}-A_{j,2}B_{k,2}Y_{1}^{\ast}+(A_{j,1}B_{k,2}-A_{j,2}B_{k,1})Y_{2}^{\ast}}{Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2}},$
$\displaystyle\boldsymbol{\Sigma}_{\boldsymbol{b}\boldsymbol{a},\boldsymbol{X}}$
$\displaystyle=-\mu_{b_{j}}\mu_{a_{k}}+\frac{B_{j,1}A_{k,1}Y_{3}^{\ast}-B_{j,2}A_{k,2}Y_{1}^{\ast}+(B_{j,2}A_{k,1}-B_{j,1}A_{k,1})Y_{2}^{\ast}}{Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2}}.$

Using these in (26), we obtain

$\begin{split}&E(\boldsymbol{a}\mid\boldsymbol{X}=\boldsymbol{K})=\mu_{a_{j}}\\\
&\hskip
10.00002pt-\frac{((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})A_{j,1}+((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})A_{j,2}}{Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2}}\end{split}$

and, likewise,

$\begin{split}&E(\boldsymbol{b}\mid\boldsymbol{X}=\boldsymbol{K})=\mu_{b_{j}}\\\
&\hskip
10.00002pt+\frac{((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})B_{j,1}-((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})B_{j,2}}{Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2}}.\end{split}$

Thus, for the $j$th row and $k$th column

$\displaystyle
E(a_{j}a_{k}\mid\boldsymbol{X}=\boldsymbol{K})=\delta_{jk}\sigma_{a_{j}}^{2}-(A_{j,1}A_{k,1}Y_{3}^{\ast}+A_{j,2}A_{k,2}Y_{1}^{\ast}-(A_{j,1}A_{k,2}+A_{j,2}A_{k,1})Y_{2}^{\ast})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt-(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})(\mu_{a_{j}}A_{k,1}+\mu_{a_{k}}A_{j,1})$
$\displaystyle\hskip
10.00002pt+((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})(\mu_{a_{j}}A_{k,2}+\mu_{a_{k}}A_{j,2})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt+((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(K_{2}))Y_{2}^{\ast})((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))(Y_{2}^{\ast})^{\ast})$
$\displaystyle\hskip
10.00002pt\times(A_{j,1}A_{k,2}+A_{j,2}A_{k,1})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2}$
$\displaystyle\hskip
10.00002pt+(((K_{1}+E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})^{2}A_{j,1}A_{k,1}$
$\displaystyle\hskip
10.00002pt+((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))(Y_{2}^{\ast})^{2}A_{j,2}A_{k,2})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2},$
$\displaystyle
E(b_{j}b_{k}\mid\boldsymbol{X}=\boldsymbol{K})=\delta_{jk}\sigma_{b_{j}}^{2}-(B_{j,1}B_{k,1}Y_{3}^{\ast}+B_{j,2}B_{k,2}Y_{1}^{\ast}+(B_{j,1}B_{k,2}+B_{j,2}B_{k,1})Y_{2}^{\ast})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt+((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})(\mu_{b_{j}}B_{k,1}+\mu_{b_{k}}B_{j,1})$
$\displaystyle\hskip
10.00002pt-((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})(\mu_{b_{j}}B_{k,2}+\mu_{b_{k}}B_{j,2})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt-((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})$
$\displaystyle\hskip
10.00002pt\times(B_{j,1}B_{k,2}+B_{j,2}B_{k,1})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2}$
$\displaystyle\hskip
10.00002pt+(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2})Y_{2}^{\ast})^{2}B_{j,1}B_{k,1}$
$\displaystyle\hskip
10.00002pt+((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})^{2})B_{j,2}B_{k,2})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2},$
$\displaystyle
E(a_{j}b_{k}\mid\boldsymbol{X}=\boldsymbol{K})=(A_{j,1}B_{k,1}Y_{3}^{\ast}-A_{j,2}B_{k,2}Y_{1}^{\ast}+(A_{j,1}B_{k,2}-A_{j,2}B_{k,1})Y_{2}^{\ast})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt-(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})(\mu_{b_{k}}A_{j,1}-\mu_{a_{j}}B_{k,1})$
$\displaystyle\hskip
10.00002pt+((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})(\mu_{a_{j}}B_{k,2}+\mu_{b_{k}}A_{j,2}))/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt+((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})$
$\displaystyle\hskip
10.00002pt\times(A_{j,1}B_{k,2}-A_{j,2}B_{k,1})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2}$
$\displaystyle\hskip
10.00002pt+(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})^{2}A_{j,1}B_{k,1}$
$\displaystyle\hskip
10.00002pt+((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})^{2}A_{j,2}B_{k,2})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2},$
$\displaystyle
E(b_{j}a_{k}\mid\boldsymbol{X}=\boldsymbol{K})=(B_{j,1}A_{k,1}Y_{3}^{\ast}-B_{j,2}A_{k,2}Y_{1}^{\ast}+(B_{j,2}A_{k,1}-B_{j,1}A_{k,2})Y_{2}^{\ast})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt-(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})(\mu_{b_{j}}A_{k,1}-\mu_{a_{k}}B_{j,1})$
$\displaystyle\hskip
10.00002pt+((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})(\mu_{b_{j}}A_{k,2}+\mu_{a_{k}}B_{j,2}))/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt+((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})$
$\displaystyle\hskip
10.00002pt\times(B_{j,2}A_{k,1}-B_{j,1}A_{k,2})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2}$
$\displaystyle\hskip
10.00002pt-(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})^{2}B_{j,1}A_{k,1}$
$\displaystyle\hskip
10.00002pt-((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})^{2}B_{j,2}A_{k,2})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2}.$

Having obtained the four expectations above, the required expectations for
computing the value of
$E(\det(\nabla\boldsymbol{X})\mid\boldsymbol{X}=\boldsymbol{K})$ can now be
derived. Thus, for $j\neq k$

$\displaystyle
E(a_{j}a_{k}+b_{j}b_{k}\mid\boldsymbol{X}=\boldsymbol{K})=(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})^{2}(A_{j,1}A_{k,1}+B_{j,1}B_{k,1})$
$\displaystyle\hskip
10.00002pt+((K_{1}-E(X_{1}))Y_{1}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})^{2}(A_{j,2}A_{k,2}+B_{j,2}B_{k,2}))/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt-(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})(\mu_{a_{j}}A_{k,1}+\mu_{a_{k}}A_{j,1}-\mu_{b_{j}}B_{k,1}-\mu_{b_{k}}B_{j,1})$
$\displaystyle\hskip
10.00002pt+((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})$
$\displaystyle\hskip
10.00002pt\times(\mu_{a_{j}}A_{k,2}+\mu_{a_{k}}A_{j,2}+\mu_{b_{j}}B_{k,2}+\mu_{b_{k}}B_{j,2}))/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt+(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})$
$\displaystyle\hskip
10.00002pt\times(A_{j,1}A_{k,2}+A_{j,2}A_{k,1}-B_{j,1}B_{k,2}-B_{j,2}B_{k,1}))/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2}$
$\displaystyle\hskip
10.00002pt+(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})^{2}(A_{j,1}A_{k,1}+B_{j,1}B_{k,1})$
$\displaystyle\hskip
10.00002pt+((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})^{2}(A_{j,2}A_{k,2}+B_{j,2}B_{k,2}))/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2},$
$\displaystyle
E(a_{j}b_{k}-b_{j}a_{k}\mid\boldsymbol{X}=\boldsymbol{K})=-((B_{j,1}A_{k,1}-A_{j,1}B_{k,1})Y_{3}^{\ast}+(A_{j,2}B_{k,2}-B_{j,2}A_{k,2})Y_{1}^{\ast}$
$\displaystyle\hskip
10.00002pt-(A_{j,1}B_{k,2}+B_{j,1}A_{k,2}-A_{j,2}B_{k,1}-B_{j,2}A_{k,1})Y_{2}^{\ast})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt+(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}E(X_{2}))Y_{2}^{\ast})(\mu_{a_{j}}B_{k,1}+\mu_{b_{j}}A_{k,1}-\mu_{a_{k}}B_{j,1}-\mu_{b_{k}}A_{j,1})$
$\displaystyle\hskip
10.00002pt+((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})$
$\displaystyle\hskip
10.00002pt\times(\mu_{b_{j}}A_{k,2}+\mu_{a_{k}}B_{j,2}-\mu_{b_{k}}A_{j,2}-\mu_{a_{j}}B_{k,2}))/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt+(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})$
$\displaystyle\hskip
10.00002pt\times(A_{j,1}B_{k,2}+B_{j,1}A_{k,2}-A_{j,2}B_{k,1}-B_{j,2}A_{k,1}))/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2}$
$\displaystyle\hskip
10.00002pt+(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})^{2}(B_{j,1}A_{k,1}-A_{j,1}B_{k,1})$
$\displaystyle\hskip
10.00002pt+((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})^{2}(A_{j,2}B_{k,1}-B_{j,2}A_{k,2}))/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2},$
$\displaystyle
E(a_{j}^{2}+b_{j}^{2}\mid\boldsymbol{X}=\boldsymbol{K})=(\sigma_{a_{j}}^{2}+\sigma_{b_{j}}^{2})-((A_{j,1}^{2}+B_{j,1}^{2})Y_{3}^{\ast}+((A_{j,2}^{2}+B_{j,2}^{2})Y_{1}^{\ast}$
$\displaystyle\hskip
10.00002pt-2(A_{j,1}A_{j,2}-B_{j,1}B_{j,2})Y_{2}^{\ast})/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt-2(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{1}^{\ast})(\mu_{a{j}}A_{j,1}-\mu_{b_{j}}B_{j,1})$
$\displaystyle\hskip
10.00002pt+((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})(\mu_{a_{j}}A_{j,2}+\mu_{b_{j}}B_{j,2}))/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})$
$\displaystyle\hskip
10.00002pt+2(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})$
$\displaystyle\hskip
10.00002pt\times(A_{j,1}A_{j,2}-B_{j,1}B_{j,2}))/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2}$
$\displaystyle\hskip
10.00002pt+(((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})^{2}(A_{j,1}^{2}+B_{j,1}^{2})$
$\displaystyle\hskip
10.00002pt+((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})^{2}(A_{j,2}^{2}+B_{j,2}^{2}))/(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})^{2}.$

After all the necessary simplifications,

$\begin{split}E(\det(\nabla\boldsymbol{X})\mid\boldsymbol{X}=\boldsymbol{K})&=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}+\sigma_{b_{j}}^{2})\left(\left(\frac{\partial
u_{j}}{\partial x}\right)^{2}+\left(\frac{\partial v_{j}}{\partial
x}\right)^{2}\right)\\\ &\hskip
10.00002pt-\frac{I_{1}S_{1}-I_{2}S_{2}-I_{3}S_{3}-I_{4}S_{4}-I_{5}S_{5}}{Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2}},\end{split}$

where

$\displaystyle I_{1}$
$\displaystyle=Y_{3}^{\ast}-\frac{((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})^{2}}{Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2}},$
$\displaystyle I_{2}$
$\displaystyle=Y_{1}^{\ast}-\frac{((K_{1}-E(X_{1}))Y_{2}^{\ast}-(K_{2}-E(X_{2}))Y_{1}^{\ast})^{2}}{Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2}},$
$\displaystyle I_{3}$
$\displaystyle=Y_{2}^{\ast}+\frac{((K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast})((K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast})}{Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2}},$
$\displaystyle I_{4}$
$\displaystyle=(K_{1}-E(X_{1}))Y_{3}^{\ast}-(K_{2}-E(X_{2}))Y_{2}^{\ast},$
$\displaystyle I_{5}$
$\displaystyle=(K_{2}-E(X_{2}))Y_{1}^{\ast}-(K_{1}-E(X_{1}))Y_{2}^{\ast},$

and

$\displaystyle S_{1}$
$\displaystyle=\sum_{j=0}^{N}\sum_{k=0}^{N}\left((A_{j,1}A_{k,1}+B_{j,1}B_{k,1})\left(\frac{\partial
u_{j}}{\partial x}\frac{\partial u_{k}}{\partial x}+\frac{\partial
v_{j}}{\partial x}\frac{\partial v_{k}}{\partial x}\right)\right.$
$\displaystyle\hskip
10.00002pt\left.+(B_{j,1}A_{k,1}-A_{j,1}B_{k,1})\left(\frac{\partial
v_{j}}{\partial x}\frac{\partial u_{k}}{\partial x}-\frac{\partial
u_{j}}{\partial x}\frac{\partial v_{k}}{\partial x}\right)\right)$
$\displaystyle=\lvert D_{1}^{\ast}\rvert^{2},$ $\displaystyle S_{2}$
$\displaystyle=\sum_{j=0}^{N}\sum_{k=0}^{N}\left((A_{j,2}A_{k,2}+B_{j,2}B_{k,2})\left(\frac{\partial
u_{j}}{\partial x}\frac{\partial u_{k}}{\partial x}+\frac{\partial
v_{j}}{\partial x}\frac{\partial v_{k}}{\partial x}\right)\right.$
$\displaystyle\hskip
10.00002pt\left.+(A_{j,2}B_{k,2}-B_{j,2}A_{k,2})\left(\frac{\partial
v_{j}}{\partial x}\frac{\partial u_{k}}{\partial x}-\frac{\partial
u_{j}}{\partial x}\frac{\partial v_{k}}{\partial x}\right)\right)$
$\displaystyle=\lvert D_{2}^{\ast}\rvert^{2},$ $\displaystyle S_{3}$
$\displaystyle=\sum_{j=0}^{N}\sum_{k=0}^{N}\left((A_{j,1}A_{k,2}+A_{j,2}A_{k,1}-B_{j,1}B_{k,2}-B_{j,2}B_{k,1})\left(\frac{\partial
u_{j}}{\partial x}\frac{\partial u_{k}}{\partial x}+\frac{\partial
v_{j}}{\partial x}\frac{\partial v_{k}}{\partial x}\right)\right.$
$\displaystyle\hskip
10.00002pt\left.+(A_{j,1}B_{k,2}+B_{j,1}A_{k,2}-A_{j,2}B_{k,1}-B_{j,2}A_{k,2})\left(\frac{\partial
v_{j}}{\partial x}\frac{\partial u_{k}}{\partial x}-\frac{\partial
u_{j}}{\partial x}\frac{\partial v_{k}}{\partial x}\right)\right)$
$\displaystyle=\lvert D_{1}^{\ast}+iD_{2}^{\ast}\rvert^{2}-\lvert
D_{1}^{\ast}\rvert-\lvert D_{2}^{\ast}\rvert^{2},$ $\displaystyle S_{4}$
$\displaystyle=\sum_{j=0}^{N}\sum_{k=0}^{N}\left((\mu_{a_{j}}A_{k,1}+\mu_{a_{k}}A_{j,1}-\mu_{b_{j}}B_{k,1}-\mu_{b_{k}}B_{j,1})\left(\frac{\partial
u_{j}}{\partial x}\frac{\partial u_{k}}{\partial x}+\frac{\partial
v_{j}}{\partial x}\frac{\partial v_{k}}{\partial x}\right)\right.$
$\displaystyle\hskip
10.00002pt\left.+(\mu_{b_{k}}A_{j,2}+\mu_{a_{j}}B_{k,2}-\mu_{b_{j}}A_{k,2}-\mu_{a_{k}}B_{j,2})\left(\frac{\partial
v_{j}}{\partial x}\frac{\partial u_{k}}{\partial x}-\frac{\partial
u_{j}}{\partial x}\frac{\partial v_{k}}{\partial x}\right)\right),$
$\displaystyle=\lvert M+D_{1}^{\ast}\rvert^{2}-\lvert M\rvert^{2}-\lvert
D_{1}^{\ast}\rvert^{2},$ $\displaystyle S_{5}$
$\displaystyle=\sum_{j=0}^{N}\sum_{k=0}^{N}\left((\mu_{a_{j}}A_{k,2}+\mu_{a_{k}}A_{j,2}+\mu_{b_{j}}B_{k,2}+\mu_{b_{k}}B_{j,2})\left(\frac{\partial
u_{j}}{\partial x}\frac{\partial u_{k}}{\partial x}+\frac{\partial
v_{j}}{\partial x}\frac{\partial v_{k}}{\partial x}\right)\right.$
$\displaystyle\hskip
10.00002pt\left.+(\mu_{b_{k}}A_{j,2}+\mu_{a_{j}}B_{k,2}-\mu_{b_{j}}A_{k,2}-\mu_{a_{k}}B_{j,2})\left(\frac{\partial
v_{j}}{\partial x}\frac{\partial u_{k}}{\partial x}-\frac{\partial
u_{j}}{\partial x}\frac{\partial v_{k}}{\partial x}\right)\right)$
$\displaystyle=\lvert M+iD_{2}^{\ast}\rvert^{2}-\lvert M\rvert^{2}-\lvert
D_{2}^{\ast}\rvert^{2}.$

After regrouping the terms, with a little algebra we can write

$\begin{split}&E(\det(\nabla\boldsymbol{X})\mid\boldsymbol{X}=\boldsymbol{K})=D_{3}^{\ast}-\frac{1}{Y_{1}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2}}(\lvert
D_{1}^{\ast}\rvert^{2}I_{1}-\lvert D_{2}^{\ast}\rvert^{2}I_{2}+(\lvert
D_{1}^{\ast}+iD_{2}^{\ast}\rvert^{2}-\lvert D_{1}^{\ast}\rvert^{2}-\lvert
D_{2}^{\ast}\rvert^{2})I_{3}\\\ &\hskip 20.00003pt-(\lvert
M+D_{1}^{\ast}\rvert^{2}-\lvert M\rvert^{2}-\lvert
D_{1}^{\ast}\rvert^{2})I_{4}-(\lvert M+iD_{2}^{\ast}\rvert^{2}-\lvert
M\rvert^{2}-\lvert D_{2}^{\ast}\rvert^{2})I_{5}).\end{split}$

Hence, when combined with the joint density of two normal random variables
$X_{1}$ and $X_{2}$ as given by

$\begin{split}&p_{X_{1}X_{2}}(\boldsymbol{K}^{\prime})=\frac{1}{2\pi\sqrt{Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2}}}\\\
&\hskip
10.00002pt\times\exp\left(-\frac{(K_{1}-E(X_{1}))^{2}Y_{3}^{\ast}+(K_{2}-E(X_{2}))^{2}Y_{1}^{\ast}-2(K_{1}-E(X_{1}))(K_{2}-E(X_{2}))Y_{2}^{\ast}}{2(Y_{1}^{\ast}Y_{3}^{\ast}-(Y_{2}^{\ast})^{2})}\right),\end{split}$

the proof of the theorem is complete.

##  5\. Ramifications of Theorem 4

Theorem 4 has several important consequences. First, if
$\mu_{a_{j}}=\mu_{b_{j}}=\mu$ for $0\leq j\leq N$ in Theorem 4, then the
formula for $h_{N,\boldsymbol{K}}(z)$ holds with the following modifications
to the auxiliary functions:

$\displaystyle E(X_{1})$ $\displaystyle=\mu\sum_{j=0}^{N}(u_{j}-v_{j}),\qquad
E(X_{2})=\mu\sum_{j=0}^{N}(u_{j}+v_{j}),$ $\displaystyle Y_{1}^{\ast}(z)$
$\displaystyle=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}u_{j}^{2}+\sigma_{b_{j}}^{2}v_{j}^{2})-\mu^{2}\left(\sum_{j=0}^{N}(u_{j}-v_{j})\right)^{2},$
$\displaystyle Y_{2}^{\ast}(z)$
$\displaystyle=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}-\sigma_{b_{j}}^{2})u_{j}v_{j}-\mu^{2}\left(\sum_{j=0}^{N}(u_{j}-v_{j})\right)\left(\sum_{j=0}^{N}(u_{j}+v_{j})\right),$
$\displaystyle Y_{3}^{\ast}(z)$
$\displaystyle=\sum_{j=0}^{N}(\sigma_{a_{j}}^{2}v_{j}^{2}+\sigma_{b_{j}}^{2}u_{j}^{2})-\mu^{2}\left(\sum_{j=0}^{N}(u_{j}+v_{j})\right)^{2},$

and

$\begin{split}A_{j,1}(z)&=\sigma_{a_{j}}^{2}u_{j}-\mu^{2}\sum_{k=0}^{N}(u_{k}-v_{k}),\\\
B_{j,1}(z)&=\sigma_{b_{j}}^{2}v_{j}+\mu^{2}\sum_{k=0}^{N}(u_{k}-v_{k}),\end{split}\qquad\begin{split}A_{j,2}(z)&=\sigma_{a_{j}}^{2}v_{j}-\mu^{2}\sum_{k=0}^{N}(u_{k}+v_{k}),\\\
B_{j,2}(z)&=\sigma_{b_{j}}^{2}u_{j}-\mu^{2}\sum_{k=0}^{N}(u_{k}+v_{k}),\end{split}$

for $0\leq j\leq N$. Observe that $M(z)$ is changed implicitly by the change
in $E(\eta_{j})$. If, further, $\mu_{a_{j}}=\mu_{b_{j}}=0$ for $0\leq j\leq
N$, then $E(X_{1})=E(X_{2})=0$. Thus, $A_{j,1}=\sigma_{a_{j}}^{2}u_{j}$,
$A_{j,2}=\sigma_{a_{j}}^{2}v_{j}$, $B_{j,1}=\sigma_{b_{j}}^{2}v_{j}$,
$B_{j,2}=\sigma_{b_{j}}^{2}u_{j}$, and $M(z)=0$. Hence, Theorem 2 is
recovered. Further, if $\sigma_{a_{j}}^{2}=\sigma_{b_{j}}^{2}=1$ for $0\leq
j\leq N$, then Theorem 1 in [4] is recovered.

Second, if $\sigma_{a_{j}}^{2}=\sigma_{b_{j}}^{2}=\sigma^{2}$ for $0\leq j\leq
N$ in Theorem 4, then $Y_{2}^{\ast}(z)=-E(X_{1})E(X_{2})$ and

$Y_{1}^{\ast}(z)=\sigma^{2}\sum_{j=0}^{N}\lvert
f_{j}(z)\rvert^{2}-(E(X_{1}))^{2},\qquad
Y_{3}^{\ast}(z)=\sigma^{2}\sum_{j=0}^{N}\lvert
f_{j}(z)\rvert^{2}-(E(X_{2}))^{2}.$

Observe that

$\begin{split}A_{j,1}(z)-iB_{j,1}(z)&=\sigma^{2}\overline{f_{j}(z)}-E(X_{1})E(\eta_{j}),\\\
B_{j,2}(z)-iA_{j,2}(z)&=\sigma^{2}\overline{f_{j}(z)}+iE(X_{2})E(\eta_{j}).\end{split}$

Altogether, the formula for $h_{N,\boldsymbol{K}}(z)$ in Theorem 4 now holds
with

$\begin{split}D_{0}^{\ast}(z)&=\sqrt{\left(\sigma^{2}\sum_{j=0}^{N}\lvert
f_{j}(z)\rvert^{2}\right)^{2}-\sigma^{2}\sum_{j=0}^{N}\lvert
f_{j}(z)\rvert^{2}\left|\sum_{j=0}^{N}f_{j}(z)E(\eta_{j})\right|^{2}},\\\
D_{1}^{\ast}(z)&=\sigma^{2}\sum_{j=0}^{N}\overline{f_{j}(z)}f_{j}^{\prime}(z)+E(X_{1})\sum_{j=0}^{N}f_{j}^{\prime}(z)E(\eta_{j}),\\\
D_{2}^{\ast}(z)&=\sigma^{2}\sum_{j=0}^{N}\overline{f_{j}(z)}f_{j}^{\prime}(z)+iE(X_{2})\sum_{j=0}^{N}f_{j}^{\prime}(z)E(\eta_{j}),\\\
D_{3}^{\ast}(z)&=2\sigma^{2}\sum_{j=0}^{N}\lvert
f_{j}(z)\rvert^{2}.\end{split}$

The form of $M(z)$ remains unchanged. Further, if
$\mu_{a_{j}}=\mu_{b_{j}}=\mu$ for $0\leq j\leq N$,

$E(X_{1})=\mu\sum_{j=0}^{N}(u_{j}-v_{j}),\qquad
E(X_{2})=\mu\sum_{j=0}^{N}(u_{j}+v_{j}).$

Then the formula for $h_{N,\boldsymbol{K}}(z)$ in Theorem 4 now holds with

$\displaystyle D_{0}^{\ast}(z)$
$\displaystyle=\sqrt{\left(\sigma^{2}\sum_{j=0}^{N}\lvert
f_{j}(z)\rvert^{2}\right)^{2}-2\mu^{2}\sigma^{2}\sum_{j=0}^{N}\lvert
f_{j}(z)\rvert^{2}\left|\sum_{j=0}^{N}f_{j}(z)\right|^{2}},$ $\displaystyle
D_{1}^{\ast}(z)$
$\displaystyle=\sigma^{2}\sum_{j=0}^{N}\overline{f_{j}(z)}f_{j}^{\prime}(z)+\mu\sum_{j=0}^{N}(u_{j}-v_{j})\sum_{j=0}^{N}f_{j}^{\prime}(z)E(\eta_{j}),$
$\displaystyle D_{2}^{\ast}(z)$
$\displaystyle=\sigma^{2}\sum_{j=0}^{N}\overline{f_{j}(z)}f_{j}^{\prime}(z)+i\mu\sum_{j=0}^{N}(u_{j}+v_{j})\sum_{j=0}^{N}f_{j}^{\prime}(z)E(\eta_{j}),$
$\displaystyle D_{3}^{\ast}(z)$ $\displaystyle=2\sigma^{2}\sum_{j=0}^{N}\lvert
f_{j}(z)\rvert^{2}.$

Third, the following also follows from Theorem 4.

######  Corollary 3.

If the vector $\boldsymbol{K}$ is restricted to a circle of radius $K>0$, then
for all $N>1$ one has

$\begin{split}&h_{N,\boldsymbol{K}}(z)\\\ &\hskip 10.00002pt=\frac{1}{2\pi
D_{0}^{\ast}(z)}\exp\left(-\frac{(K-E(X_{1}))^{2}Y_{3}^{\ast}(z)+(K-E(X_{2}))^{2}Y_{1}^{\ast}(z)-2(K-E(X_{1}))(K-E(X_{2}))Y_{2}^{\ast}(z)}{2D_{0}^{\ast}(z)^{2}}\right)\\\
&\hskip 20.00003pt\times\left\\{D_{3}^{\ast}(z)-\frac{\lvert
D_{1}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\left(Y_{3}^{\ast}(z)-\frac{((K-E(X_{1}))Y_{3}^{\ast}(z)-(K-E(X_{2}))Y_{2}^{\ast}(z))^{2}}{D_{0}^{\ast}(z)^{2}}\right)\right.\\\
&\hskip 20.00003pt\left.-\frac{\lvert
D_{2}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\left(Y_{1}^{\ast}(z)-\frac{((K-E(X_{1}))Y_{2}^{\ast}(z)-(K-E(X_{2}))Y_{1}^{\ast}(z))^{2}}{D_{0}^{\ast}(z)^{2}}\right)\right.\\\
&\hskip 20.00003pt\left.+\left(\frac{\lvert
D_{1}^{\ast}(z)+iD_{2}^{\ast}(z)\rvert^{2}-\lvert
D_{1}^{\ast}(z)\rvert^{2}-\lvert
D_{2}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\right)\right.\\\ &\hskip
20.00003pt\left.\times\left(Y_{2}^{\ast}(z)-\frac{((K-E(X_{1}))Y_{3}^{\ast}(z)-(K-E(X_{2}))Y_{2}^{\ast}(z))((K-E(X_{1}))Y_{2}^{\ast}(z)-(K-E(X_{2}))Y_{1}^{\ast}(z))}{D_{0}^{\ast}(z)^{2}}\right)\right.\\\
&\hskip 20.00003pt\left.-\left(\frac{\lvert
M(z)+D_{1}^{\ast}(z)\rvert^{2}-\lvert M(z)\rvert^{2}-\lvert
D_{1}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\right)((K-E(X_{1}))Y_{3}^{\ast}(z)-(K-E(X_{2}))Y_{2}^{\ast}(z))\right.\\\
&\hskip 20.00003pt\left.+\left(\frac{\lvert
M(z)+iD_{2}^{\ast}(z)\rvert^{2}-\lvert M(z)\rvert^{2}-\lvert
D_{2}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\right)((K-E(X_{1}))Y_{2}^{\ast}(z)-(K-E(X_{2}))Y_{1}^{\ast}(z))\right\\}.\end{split}$

Then immediate by Corollary 3 is the following result.

######  Corollary 4.

If $\boldsymbol{K}$ is the zero vector, then for all $N>1$ one has

$\displaystyle h_{N,\boldsymbol{0}}(z)$ $\displaystyle=\frac{1}{2\pi
D_{0}^{\ast}(z)}\exp\left(-\frac{(E(X_{1}))^{2}Y_{3}^{\ast}(z)+(E(X_{2}))^{2}Y_{1}^{\ast}(z)-2E(X_{1})E(X_{2})Y_{2}^{\ast}(z)}{2D_{0}^{\ast}(z)^{2}}\right)$
$\displaystyle\hskip 10.00002pt\times\left\\{D_{3}^{\ast}(z)-\frac{\lvert
D_{1}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\left(Y_{3}^{\ast}(z)-\frac{(E(X_{1})Y_{3}^{\ast}(z)-E(X_{2})Y_{2}^{\ast}(z))^{2}}{D_{0}^{\ast}(z)^{2}}\right)\right.$
$\displaystyle\hskip 10.00002pt\left.-\frac{\lvert
D_{2}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\left(Y_{1}-\frac{(E(X_{1})Y_{2}^{\ast}(z)-E(X_{2})Y_{1}^{\ast}(z))^{2}}{D_{0}^{\ast}(z)^{2}}\right)+\left(\frac{\lvert
D_{1}^{\ast}(z)+iD_{2}^{\ast}(z)\rvert^{2}-\lvert
D_{1}^{\ast}(z)\rvert^{2}-\lvert
D_{2}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\right)\right.$
$\displaystyle\hskip
10.00002pt\left.\times\left(Y_{2}^{\ast}(z)-\frac{(E(X_{1})Y_{3}^{\ast}(z)-E(X_{2})Y_{2}^{\ast}(z))(E(X_{1})Y_{2}^{\ast}(z)-E(X_{2})Y_{1}^{\ast}(z))}{D_{0}^{\ast}(z)^{2}}\right)\right.$
$\displaystyle\hskip 10.00002pt\left.-\left(\frac{\lvert
M(z)+D_{1}^{\ast}(z)\rvert^{2}-\lvert M(z)\rvert^{2}-\lvert
D_{1}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\right)(E(X_{1})Y_{3}^{\ast}(z)-E(X_{2})Y_{2}^{\ast}(z))\right.$
$\displaystyle\hskip 10.00002pt\left.+\left(\frac{\lvert
M(z)+iD_{2}^{\ast}(z)\rvert^{2}-\lvert M(z)\rvert^{2}-\lvert
D_{2}^{\ast}(z)\rvert^{2}}{D_{0}^{\ast}(z)^{2}}\right)(E(X_{1})Y_{2}^{\ast}(z)-E(X_{2})Y_{1}^{\ast}(z))\right\\}.$

######  Acknowledgments.

During the preparation of this work, the first author received support from
the Office of the Vice Chancellor for Research and Dean of the Graduate School
at The University of Tennessee at Chattanooga. The second author received
support from the National Science Foundation under Grant DMS-1852288.

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