$\displaystyle\bm{p}^{\textnormal{VOL}}\preceq\left(\left|a_{1}\right|+\left|a_{2}\right|+\frac{1}{2}\left|a_{2}b_{2}K_{2}(0)\right|\right)({\mathcolor{NavyBlue}{\mathbf{1}}}+{\mathcolor{NavyBlue}{\mathbf{2}}})e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}M{\mathcolor{NavyBlue}{\mathbf{1}}}}+\left|y\right|{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}},$
(5.4)

By the sub-multiplicativity in Proposition 3.7 and lemma 3.10, we derive
$\bm{q}^{\textnormal{VOL}},\bm{p}^{\textnormal{VOL}}\in{\mathcal{A}_{h}}$. An
application of Theorem 3.9 yields that
$\bm{\ell}^{\textnormal{VOL}}\in{\mathcal{A}_{h}}$. In particular,
$\bm{\ell}^{\textnormal{VOL}}\in\mathcal{A}$ and the process

$\displaystyle\bar{Y}_{t}:=\left\langle\bm{\ell}^{\textnormal{VOL}},\widehat{\mathbb{W}}_{t}\right\rangle$
(5.5)

is well defined.

In the next step, we would like to apply Itô’s formula of Theorem 3.3 on (5.5)
to prove that $\bar{Y}$ solves the equation

$\displaystyle\begin{split}d\bar{Y}_{t}&=K_{1}(0)\left(a_{1}+b_{1}\bar{Y}_{t}\right)dt+\int_{0}^{t}K_{1}^{\prime}(t-s)\left(a_{1}+b_{1}\bar{Y}_{s}\right)dsdt\\\
&\quad+K_{2}(0)\left(a_{2}+b_{2}\bar{Y}_{t}\right)dW_{t}+\int_{0}^{t}K_{2}^{\prime}(t-s)\left(a_{2}+b_{2}\bar{Y}_{s}\right)dW_{s}dt,\\\
\bar{Y}_{0}&=y.\end{split}$ (5.6)

which is the differential version of (4.1) obtained from Itô’s lemma on (4.1)
since the kernels $K_{1},K_{2}$ are smooth. Then, we immediately deduce that
$Y_{t}=\bar{Y}_{t}=\left\langle\bm{\ell}^{\textnormal{VOL}},\widehat{\mathbb{W}}_{t}\right\rangle$
by uniqueness of the strong solution, which will end the proof of Theorem 4.2.

Therefore, it remains to

    1. (i)

Goal 1: argue that $\bm{\ell}^{\textnormal{VOL}}\in\mathcal{I}(\widehat{W})$
to apply Theorem 3.3 and get

$\displaystyle
d\left\langle\bm{\ell}^{\textnormal{VOL}},\widehat{\mathbb{W}}_{t}\right\rangle=\left\langle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}+\tfrac{1}{2}\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}},\widehat{\mathbb{W}}_{t}\right\rangle
dt+\left\langle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}},\widehat{\mathbb{W}}_{t}\right\rangle
dW_{t},$ (5.7)

    2. (ii)

Goal 2: identify the coefficients to obtain (5.6).

In order to achieve both goals, we start by computing the projections
$\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}},\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}},\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}}$.
First recall

$\displaystyle\bm{p}^{\textnormal{VOL}}$
$\displaystyle=a_{1}{\mathcolor{NavyBlue}{\mathbf{1}}}\int_{[0,\infty)}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{1}(dx)+a_{2}\left({\mathcolor{NavyBlue}{\mathbf{2}}}-\frac{1}{2}K_{2}(0)b_{2}{\mathcolor{NavyBlue}{\mathbf{1}}}\right)\int_{[0,\infty)}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{2}(dx)+y{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}},$
(5.8) $\displaystyle\bm{q}^{\textnormal{VOL}}$
$\displaystyle=b_{1}{\mathcolor{NavyBlue}{\mathbf{1}}}\int_{[0,\infty)}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{1}({dx})+b_{2}\left({\mathcolor{NavyBlue}{\mathbf{2}}}-\frac{1}{2}K_{2}(0)b_{2}{\mathcolor{NavyBlue}{\mathbf{1}}}\right)\int_{[0,\infty)}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{2}(dx).$
(5.9)

We compute that

$\displaystyle\begin{split}\bm{p}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}&=a_{1}\int_{[0,\infty)}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{1}(dx)-a_{2}\int_{[0,\infty)}\left(x{\mathcolor{NavyBlue}{\mathbf{2}}}+\frac{1}{2}K_{2}(0)b_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right)e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{2}(dx),\\\
\bm{q}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}&=b_{1}\int_{[0,\infty)}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{1}(dx)-b_{2}\int_{[0,\infty)}\left(x{\mathcolor{NavyBlue}{\mathbf{2}}}+\frac{1}{2}K_{2}(0)b_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right)e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{2}(dx),\\\
\bm{p}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}}&=\int_{[0,\infty)}\mu_{2}(dx)a_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}=K_{2}(0)a_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}},\qquad\bm{p}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}}=0,\\\
\bm{q}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}}&=\int_{[0,\infty)}\mu_{2}(dx)b_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}=K_{2}(0)b_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}},\qquad\bm{q}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}}=0.\end{split}$
(5.10)

Then, using Lemma A.2 and noticing
$(\bm{q}^{\textnormal{VOL}})^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}=0$,
it immediately follows that

$\displaystyle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}}$
$\displaystyle=\bm{p}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}}+\bm{\ell}^{\textnormal{VOL}}\left(\bm{q}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}}\right)=K_{2}(0)(a_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+b_{2}\bm{\ell}^{\textnormal{VOL}}),$
(5.11)
$\displaystyle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}}$
$\displaystyle=K_{2}(0)b_{2}\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}}=K_{2}^{2}(0)b_{2}(a_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+b_{2}\bm{\ell}^{\textnormal{VOL}}),$
(5.12)

and that

$\displaystyle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}$
$\displaystyle=\bm{p}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}+\bm{\ell}^{\textnormal{VOL}}\left(\bm{q}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}\right)$
(5.13)
$\displaystyle=\left(a_{1}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+b_{1}\bm{\ell}^{\textnormal{VOL}}\right)\int_{[0,\infty)}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{1}(dx)-\left(a_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+b_{2}\bm{\ell}^{\textnormal{VOL}}\right)\int_{[0,\infty)}\left(x{\mathcolor{NavyBlue}{\mathbf{2}}}+\frac{1}{2}K_{2}(0)b_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right)e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{2}(dx)$
(5.14)
$\displaystyle=K_{1}(0)\left(a_{1}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+b_{1}\bm{\ell}^{\textnormal{VOL}}\right)-\frac{1}{2}K_{2}^{2}(0)b_{2}\left(a_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+b_{2}\bm{\ell}^{\textnormal{VOL}}\right)$
(5.15)
$\displaystyle\quad+\left(a_{1}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+b_{1}\bm{\ell}^{\textnormal{VOL}}\right){\mathcolor{NavyBlue}{\mathbf{1}}}\int_{[0,\infty)}-xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{1}(dx)$
(5.16)
$\displaystyle\quad+\left(a_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+b_{2}\bm{\ell}^{\textnormal{VOL}}\right)\left({\mathcolor{NavyBlue}{\mathbf{2}}}-\frac{1}{2}K_{2}(0)b_{2}{\mathcolor{NavyBlue}{\mathbf{1}}}\right)\int_{[0,\infty)}-xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{2}(dx).$
(5.17)

The following proposition establishes our first goal:
$\bm{\ell}^{\textnormal{VOL}}\in\mathcal{I}(\widehat{W})$.

######  Lemma 5.1.

We have that
$\left\\{\bm{\ell}^{\textnormal{VOL}},\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}},\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}},\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}}\right\\}\subset{\mathcal{A}_{h}}$.
In particular, $\bm{\ell}^{\textnormal{VOL}}\in\mathcal{I}(\widehat{W}).$

###### Proof.

At the beginning of Section 5, we have shown that
$\bm{\ell}^{\textnormal{VOL}}\in{\mathcal{A}_{h}}$, then by (5.11)-(5.12), it
follows that
$\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}},\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}}\in{\mathcal{A}_{h}}$.
For $\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}$, we
remind that for $i=1,2$:

$\displaystyle\int_{[0,\infty)}e^{\shuffle-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{i}(dx)\preceq
e^{\shuffle
M{\mathcolor{NavyBlue}{\mathbf{1}}}},~{}~{}\int_{[0,\infty)}xe^{\shuffle-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{i}(dx)\preceq
Me^{\shuffle M{\mathcolor{NavyBlue}{\mathbf{1}}}}.$ (5.18)

Combined it with (5.10) and the sub-multiplicative property, we derive
$\bm{p}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}},\bm{q}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}\in{\mathcal{A}_{h}}$.
Recalling that
$\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}=\bm{p}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}+\bm{\ell}^{\textnormal{VOL}}(\bm{q}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}})$
in (5.15), an application of the sub-multiplicative property of the
${\mathcal{A}_{h}}$-norm in Proposition 3.7 yields that

$\left|\left|\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}\right|\right|_{T}^{\mathcal{A}_{h}}\leq\left|\left|\bm{p}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}\right|\right|_{T}^{\mathcal{A}_{h}}+\left|\left|\bm{\ell}^{\textnormal{VOL}}\right|\right|_{T}^{\mathcal{A}_{h}}\left|\left|\bm{q}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}\right|\right|_{T}^{\mathcal{A}_{h}}<\infty,$

thus proving that
$\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}\in{\mathcal{A}_{h}}$.
Finally, by Proposition 3.6, for every
$\bm{\ell}\in\left\\{\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}},\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}},\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}}\right\\}$,

$\mathbb{E}\left[\sup_{t\in[0,T]}\left|\left|\bm{\ell}\right|\right|_{t}^{\mathcal{A}}\right]\leq\left|\left|\bm{\ell}\right|\right|_{T}^{\mathcal{A}_{h}}<\infty.$

This proves that
$\sup_{t\in[0,T]}\left|\left|\bm{\ell}\right|\right|_{t}^{\mathcal{A}}<\infty$
a.s. and thus $\bm{\ell}^{\textnormal{VOL}}$ satisfies the integral assumption
in (3.11). This shows that
$\bm{\ell}^{\textnormal{VOL}}\in\mathcal{I}(\widehat{W})$ and completes the
proof. ∎

Thanks to Lemma 5.1,
$\bm{\ell}^{\textnormal{VOL}}\in\mathcal{I}(\widehat{W})$, so an application
of Itô’s formula in Theorem 3.3 and Remark 3.6 yields that the process
$\bar{Y}_{t}:=\left\langle\bm{\ell}^{\textnormal{VOL}},\widehat{\mathbb{W}}_{t}\right\rangle$
is an Itô process with decomposition

$\displaystyle\bar{Y}_{t}$
$\displaystyle=\left(\bm{\ell}^{\textnormal{VOL}}\right)^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+\int_{0}^{t}\left\langle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}+\tfrac{1}{2}\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}},\widehat{\mathbb{W}}_{s}\right\rangle
ds+\int_{0}^{t}\left\langle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}},\widehat{\mathbb{W}}_{s}\right\rangle
dW_{s}.$ (5.19)

It remains to achieve our second goal: identify the coefficients, which is the
object of the next lemma.

######  Lemma 5.2.

We have that

$\displaystyle\left\langle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}+\tfrac{1}{2}\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}},\widehat{\mathbb{W}}_{t}\right\rangle$
$\displaystyle=K_{1}(0)\left(a_{1}+b_{1}\bar{Y}_{t}\right)$ (5.20)
$\displaystyle\quad+\int_{0}^{t}K_{1}^{\prime}(t-s)\left(a_{1}+b_{1}\bar{Y}_{s}\right)ds+\int_{0}^{t}K_{2}^{\prime}(t-s)\left(a_{2}+b_{2}\bar{Y}_{s}\right)dW_{s},$
(5.21)
$\displaystyle\left\langle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}},\widehat{\mathbb{W}}_{t}\right\rangle$
$\displaystyle=K_{2}(0)(a_{2}+b_{2}\bar{Y}_{t}).$ (5.22)

###### Proof.

The proof is given in Subsection 5.1 below. ∎

We are now ready to complete the proof of Theorem 4.2.

###### Proof of Theorem 4.2.

Lemma 5.1 combined with Itô’s formula in (4.5) yields that $\bar{Y}$, given in
(5.5), is well-defined and satisfies (5.6), with
$\bm{\ell}^{\textnormal{VOL}}\in{\mathcal{A}_{h}}$. An application of Lemma
5.2, after observing that
$(\bm{\ell}^{\textnormal{VOL}})^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}=y$,
yields that $\bar{Y}_{t}$ solves (5.6) as needed. This shows that
$Y_{t}=\bar{Y}_{t}=\left\langle\bm{\ell}^{\textnormal{VOL}},\widehat{\mathbb{W}}_{t}\right\rangle$
by strong uniqueness of the solution to (4.1) and ends the proof. ∎

###  5.1 Proof of Lemma 5.2

We start by deriving a transformation formula, which is simple but crucial to
simplify the algebraic expressions.

######  Lemma 5.3.

Let $\bm{\ell}\in T((\mathbb{R}^{d}))$, ${\mathcolor{NavyBlue}{\mathbf{i}}}\in
A_{d}$ and $\bm{b}:=\sum_{{\mathcolor{NavyBlue}{\mathbf{j}}}\in
A_{d}}b^{\mathcolor{NavyBlue}{\mathbf{j}}}{\mathcolor{NavyBlue}{\mathbf{j}}}$
for $b^{\mathcolor{NavyBlue}{\mathbf{j}}}\in\mathbb{R}$, then

$\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{b}}=e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{b}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\left((e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-\bm{b}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}){\mathcolor{NavyBlue}{\mathbf{i}}}\right).$

###### Proof.

Let
$\bm{\gamma}:=\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{b}}-e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{b}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\left(\left(e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-\bm{b}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}\right){\mathcolor{NavyBlue}{\mathbf{i}}}\right)$,
then

$\displaystyle\bm{\gamma}$
$\displaystyle=\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{b}}\bm{b}\right)-\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{b}}\bm{b}\right)\mathrel{\sqcup\mkern-3.0mu\sqcup}\left(\left(e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-\bm{b}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}\right){\mathcolor{NavyBlue}{\mathbf{i}}}\right)$
$\displaystyle=\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}+\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{b}}\bm{b}-\left[{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\left(e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-\bm{b}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}\right)\right]{\mathcolor{NavyBlue}{\mathbf{i}}}-\left[(e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{b}}\bm{b})\mathrel{\sqcup\mkern-3.0mu\sqcup}\left(e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-\bm{b}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}\right)\right]{\mathcolor{NavyBlue}{\mathbf{i}}}-\left[e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{b}}\mathrel{\sqcup\mkern-3.0mu\sqcup}{\left(e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-\bm{b}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}\right){\mathcolor{NavyBlue}{\mathbf{i}}}}\right]\bm{b}$
$\displaystyle=\left[\bm{\ell}-\left(e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{b}}\mathrel{\sqcup\mkern-3.0mu\sqcup}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-\bm{b}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}\right)\right]{\mathcolor{NavyBlue}{\mathbf{i}}}+\left[\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{b}}-e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{b}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\left(\left(e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-\bm{b}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}\right){\mathcolor{NavyBlue}{\mathbf{i}}}\right)\right]\bm{b}$
$\displaystyle=0+\bm{\gamma}\bm{b},$

implying that $\bm{\gamma}=0$. ∎

Now, as we will be using integral of elements of $T((\mathbb{R}^{2}))$, recall
the Definition 4.1, we need to make sure its bracket with the signature and
concatenation make sense. This is the goal of the next two lemmas:

######  Lemma 5.4.

Let $\bm{p}\in L^{1}(T((\mathbb{R}^{2})),\mu)$, then
$\bm{\ell}\bm{p},\bm{p}\bm{\ell}\in L^{1}(T((\mathbb{R}^{2})),\mu)$ for every
$\bm{\ell}\in T((\mathbb{R}^{2}))$,

$\int_{[0,\infty)}\bm{\ell}\bm{p}(x)\mu(dx)=\bm{\ell}\left(\int_{[0,\infty)}\bm{p}(x)\mu(dx)\right)$

and

$\int_{[0,\infty)}\bm{p}(x)\bm{\ell}\mu(dx)=\left(\int_{[0,\infty)}\bm{p}(x)\mu(dx)\right)\bm{\ell}.$

###### Proof.

Obvious. ∎

######  Lemma 5.5.

Let $\bm{p}\in L^{1}(T((\mathbb{R}^{2})),\mu)$ such that
$\int_{[0,\infty)}\left|\left|\bm{p}(x)\right|\right|_{t}^{\mathcal{A}}\mu(dx)<\infty$
for every $t\in[0,T]$, then

$\left\langle\int_{[0,\infty)}\bm{p}(x)\mu(dx),\widehat{\mathbb{W}}_{t}\right\rangle=\int_{[0,\infty)}\left\langle\bm{p}(x),\widehat{\mathbb{W}}_{t}\right\rangle\mu(dx).$

###### Proof.

By assumption

$\displaystyle\int_{[0,\infty)}\left|\left|\bm{p}(x)\right|\right|_{t}^{\mathcal{A}}\mu(dx)$
$\displaystyle=\int_{[0,\infty)}\sum_{n=0}^{\infty}\left|\sum_{{\mathcolor{NavyBlue}{\mathbf{v}}}\in
V_{n}}\bm{p}^{\mathcolor{NavyBlue}{\mathbf{v}}}(x)\widehat{\mathbb{W}}_{t}^{\mathcolor{NavyBlue}{\mathbf{v}}}\right|\mu(dx)<\infty.$
(5.23)

This implies that $\bm{p}(x)\in\mathcal{A}$ $\mu$-a.e., so by the dominated
convergence theorem

$\displaystyle\left\langle\int_{[0,\infty)}\bm{p}(x)\mu(dx),\widehat{\mathbb{W}}_{t}\right\rangle$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{{\mathcolor{NavyBlue}{\mathbf{v}}}\in
V_{n}}\left(\int_{[0,\infty)}\bm{p}^{\mathcolor{NavyBlue}{\mathbf{v}}}(x)\mu(dx)\right)\widehat{\mathbb{W}}_{t}^{\mathcolor{NavyBlue}{\mathbf{v}}}$
$\displaystyle=\int_{[0,\infty)}\sum_{n=0}^{\infty}\sum_{{\mathcolor{NavyBlue}{\mathbf{v}}}\in
V_{n}}\bm{p}^{\mathcolor{NavyBlue}{\mathbf{v}}}(x)\widehat{\mathbb{W}}_{t}^{\mathcolor{NavyBlue}{\mathbf{v}}}\mu(dx)$
$\displaystyle=\int_{[0,\infty)}\left\langle\bm{p}(x),\widehat{\mathbb{W}}_{t}\right\rangle\mu(dx).$

∎

We now introduce two important propositions, which establish the commutativity
of the kernel integral and the bracket.

######  Proposition 5.6.

Let $\bm{\ell}\in{\mathcal{A}_{h}}$. Assume that $\mu$ is a measure on
$[0,\infty)$ satisfying $\int_{[0,\infty)}x^{n}\mu(dx)<M^{n}$ for every
$n\in\mathbb{N}$ and a positive constant $M$. We then have

$\displaystyle\left\langle\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}\left(\int_{[0,\infty)}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu(dx)\right),\widehat{\mathbb{W}}_{t}\right\rangle$
$\displaystyle=\left\langle\int_{[0,\infty)}\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu(dx),\widehat{\mathbb{W}}_{t}\right\rangle$
(5.24)
$\displaystyle=\int_{[0,\infty)}\left\langle\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle\mu(dx),\quad{\mathcolor{NavyBlue}{\mathbf{i}}}={\mathcolor{NavyBlue}{\mathbf{1}}},{\mathcolor{NavyBlue}{\mathbf{2}}}.$
(5.25)

###### Proof.

We start by proving the first equality (5.24). Fix
${\mathcolor{NavyBlue}{\mathbf{i}}}\in\left\\{{\mathcolor{NavyBlue}{\mathbf{1}}},{\mathcolor{NavyBlue}{\mathbf{2}}}\right\\}$.
By assumption of $\mu$, it is easy to show that
$\int_{[0,\infty)}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu(dx)\in{\mathcal{A}_{h}}$
by direct computation. In particular
$xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\in
L^{1}(\mathcal{A},\mu)$. A direct application of Lemma 5.4 shows that

$\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}\left(\int_{[0,\infty)}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu(dx)\right)=\int_{[0,\infty)}\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu(dx)\in
T((\mathbb{R}^{2})).$

Moreover, Proposition 3.7 ensures that it actually belongs to
${\mathcal{A}_{h}}$ so that the bracket is well defined, which proves the
first equality.

For the second equality (5.25), it suffices to verify that
$\int_{[0,\infty)}\left|\left|\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\right|\right|_{t}^{\mathcal{A}}\mu(dx)<\infty$
for every $t\in[0,T]$:

$\displaystyle\int_{[0,\infty)}\left|\left|\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\right|\right|_{t}^{\mathcal{A}}\mu(dx)$
$\displaystyle\leq\int_{[0,\infty)}\left|\left|\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\right|\right|_{t}^{\mathcal{A}_{h}}\mu(dx)\leq\int_{[0,\infty)}\left|\left|\bm{\ell}\right|\right|_{t}^{\mathcal{A}_{h}}\left|\left|{\mathcolor{NavyBlue}{\mathbf{i}}}\right|\right|_{t}^{\mathcal{A}_{h}}\left|\left|xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\right|\right|_{t}^{\mathcal{A}_{h}}\mu(dx),$

where the second equality comes from Proposition 3.7. It then follows

$\displaystyle\int_{[0,\infty)}\left|\left|\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\right|\right|_{t}^{\mathcal{A}}\mu(dx)$
$\displaystyle\leq\left|\left|\bm{\ell}\right|\right|_{t}^{\mathcal{A}_{h}}\left|\left|{\mathcolor{NavyBlue}{\mathbf{i}}}\right|\right|_{t}^{\mathcal{A}_{h}}\int_{[0,\infty)}\left|\left|xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\right|\right|_{t}^{\mathcal{A}_{h}}\mu(dx)$
$\displaystyle\leq
C\left|\left|\bm{\ell}\right|\right|_{t}^{\mathcal{A}_{h}}\left|\left|{\mathcolor{NavyBlue}{\mathbf{i}}}\right|\right|_{t}^{\mathcal{A}_{h}}\int_{[0,\infty)}\left(\sum_{n=0}^{\infty}x^{n+1}(n+1)^{-\frac{1}{4}}\frac{(2t)^{n}}{\sqrt{(2n-1)!}}\right)\mu(dx)$
$\displaystyle\leq
C\left|\left|\bm{\ell}\right|\right|_{t}^{\mathcal{A}_{h}}\left|\left|{\mathcolor{NavyBlue}{\mathbf{i}}}\right|\right|_{t}^{\mathcal{A}_{h}}\sum_{n=0}^{\infty}M^{n+1}(n+1)^{-\frac{1}{4}}\frac{(2t)^{n}}{\sqrt{(2n-1)!}}$
$\displaystyle\leq
C\left|\left|\bm{\ell}\right|\right|_{t}^{\mathcal{A}_{h}}\left|\left|{\mathcolor{NavyBlue}{\mathbf{i}}}\right|\right|_{t}^{\mathcal{A}_{h}}M\left(\sum_{n=1}^{\infty}M^{n}\frac{(2t)^{n}}{(n-1)!}+1\right)$
$\displaystyle=C\left|\left|\bm{\ell}\right|\right|_{t}^{\mathcal{A}_{h}}\left|\left|{\mathcolor{NavyBlue}{\mathbf{i}}}\right|\right|_{t}^{\mathcal{A}_{h}}M\left(2tMe^{2tM}+1\right)$
$\displaystyle<\infty.$

An application of Lemma 5.5 completes the proof. ∎

######  Proposition 5.7.

$\mu$ is a measure on $[0,\infty)$ satisfying
$\int_{[0,\infty)}x^{n}\mu(dx)<M^{n}$ for every $n\in\mathbb{N}$ and a
positive constant $M$, then

$\displaystyle\int_{[0,\infty)}\left\langle\bm{\ell}^{\textnormal{VOL}}{\mathcolor{NavyBlue}{\mathbf{1}}}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle\mu(dx)$
$\displaystyle=-\int_{0}^{t}J^{\prime}(t-s)\bar{Y}_{s}ds$ (5.26) and
$\displaystyle\int_{[0,\infty)}\left\langle\bm{\ell}^{\textnormal{VOL}}{\mathcolor{NavyBlue}{\mathbf{2}}}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle\mu(dx)$
$\displaystyle=-\int_{0}^{t}J^{\prime}(t-s)\bar{Y}_{s}dW_{s}-\frac{1}{2}K_{2}(0)\int_{0}^{t}J^{\prime}(t-s)(a_{2}+b_{2}\bar{Y}_{s})ds$
(5.27)

where $J(t)=\int_{[0,\infty)]}e^{-xt}\mu(dx)$.

###### Proof.

By Lemma 5.3 and the shuffle property, we derive

$\displaystyle\left\langle\bm{\ell}^{\textnormal{VOL}}{\mathcolor{NavyBlue}{\mathbf{1}}}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle$
$\displaystyle=x\left\langle
e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle\int_{0}^{t}\left\langle
e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}^{\textnormal{VOL}},\widehat{\mathbb{W}}_{s}\right\rangle
ds=x\int_{0}^{t}e^{-x(t-s)}\bar{Y}_{s}ds,$ (5.28)

Similarly, using Lemma 5.3 and the shuffle property,

$\displaystyle\left\langle\bm{\ell}^{\textnormal{VOL}}{\mathcolor{NavyBlue}{\mathbf{2}}}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle$
$\displaystyle=x\left\langle
e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle\left\langle\left(e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}^{\textnormal{VOL}}\right){\mathcolor{NavyBlue}{\mathbf{2}}},\widehat{\mathbb{W}}_{t}\right\rangle$
(5.29) $\displaystyle=x\left\langle
e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle\int_{0}^{t}\left\langle
e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}^{\textnormal{VOL}},\widehat{\mathbb{W}}_{s}\right\rangle
dW_{s}$ (5.30) $\displaystyle\quad+\frac{1}{2}x\left\langle
e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle\int_{0}^{t}\left\langle
e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}},\widehat{\mathbb{W}}_{s}\right\rangle
ds$ (5.31)
$\displaystyle=x\int_{0}^{t}e^{-x(t-s)}\bar{Y}_{s}dW_{s}+\frac{1}{2}K_{2}(0)x\int_{0}^{t}e^{-x(t-s)}(a_{2}+b_{2}\bar{Y}_{t})ds,$
(5.32)

where the second and third equalities come from Theorem 3.2 and (5.11)
respectively.

We will now show that we can apply the deterministic and stochastic Fubini
theorem. First, noticing that for all $t\in[0,T]$,

$\mathbb{E}\left|\bar{Y}_{t}\right|=\mathbb{E}\left|\left\langle\bm{\ell}^{\textnormal{VOL}},\widehat{\mathbb{W}}_{t}\right\rangle\right|\leq\left|\left|\bm{\ell}^{\textnormal{VOL}}\right|\right|_{t}^{\mathcal{A}_{h}},$

and thus

$\sup_{t\in[0,T]}\mathbb{E}\left|\bar{Y}_{t}\right|<\infty.$

We can now apply Tonelli’s theorem:

$\displaystyle\mathbb{E}\left[\int_{[0,\infty)}\int_{0}^{t}e^{-x(t-s)}\left|\bar{Y}_{s}\right|ds\mu(dx)\right]$
$\displaystyle=\int_{[0,\infty)}\int_{0}^{t}e^{-x(t-s)}\mathbb{E}\left|\bar{Y}_{s}\right|ds\mu(dx)$
$\displaystyle\leq\int_{[0,\infty)}\int_{0}^{t}\mathbb{E}\left|\bar{Y}_{s}\right|ds\mu(dx)$
$\displaystyle\leq\mu([0,\infty))\int_{0}^{t}\mathbb{E}\left|\bar{Y}_{s}\right|ds$
$\displaystyle\leq\mu([0,\infty))~{}T\sup_{t\in[0,T]}\mathbb{E}\left|\bar{Y}_{t}\right|<\infty.$

This leads to

$\int_{[0,\infty)}\int_{0}^{t}e^{-x(t-s)}\left|\bar{Y}_{s}\right|ds\mu(dx)<\infty\textnormal{
a.s.}.$ (5.33)

Now we can apply Fubini theorem on equation (5.28) and derive:

$\int_{[0,\infty)}\left\langle\bm{\ell}^{\textnormal{VOL}}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle\mu(dx)=\int_{0}^{t}\left(\int_{[0,\infty)}xe^{-x(t-s)}\mu(dx)\right)\bar{Y}_{s}ds=-\int_{0}^{t}J^{\prime}(t-s)\bar{Y}_{s}ds,$

thus proving (5.26). On the other hand, by Proposition 3.6 3(ii),

$\sup_{t\in[0,T]}\mathbb{E}\left[\bar{Y}_{t}^{2}\right]\leq\left(\left|\left|\bm{\ell}^{\textnormal{VOL}}\right|\right|_{T}^{\mathcal{A}_{h}}\right)^{2}<\infty$

and

$\displaystyle\mathbb{E}\left[\int_{[0,\infty)}\sqrt{\int_{0}^{t}e^{-2x(t-s)}\left|\bar{Y}_{s}\right|^{2}ds}\mu(dx)\right]$
$\displaystyle\leq\mu([0,\infty))\mathbb{E}\left[\sqrt{\int_{0}^{t}\left|\bar{Y}_{s}\right|^{2}ds}\right]$
$\displaystyle\leq\mu([0,\infty))\sqrt{\mathbb{E}\left[\int_{0}^{t}\left|\bar{Y}_{s}\right|^{2}ds\right]}$
$\displaystyle\leq\mu([0,\infty))\sqrt{T\sup_{t\in[0,T]}\mathbb{E}\left|\bar{Y}_{t}\right|^{2}}<\infty,$

which leads to

$\int_{[0,\infty)}\sqrt{\int_{0}^{t}e^{-2x(t-s)}\left|\bar{Y}_{s}\right|^{2}ds}\mu(dx)<\infty\textnormal{
a.s.}$ (5.34)

Note that (5.34) is actually the condition for stochastic Fubini theorem to
hold, see [30, Theorem 2.2]. Applying it, we derive

$\displaystyle\int_{[0,\infty)}\left\langle\bm{\ell}^{\textnormal{VOL}}{\mathcolor{NavyBlue}{\mathbf{2}}}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle\mu(dx)$
$\displaystyle=-\int_{0}^{t}J^{\prime}(t-s)\bar{Y}_{s}dW_{s}-\frac{1}{2}K_{2}(0)\int_{0}^{t}J^{\prime}(t-s)(a_{2}+b_{2}\bar{Y}_{s})ds,$

thus completing the proof. ∎

Now, taking $\mu=\mu_{1},\mu_{2}$, we combine Proposition 5.6 with Proposition
5.7 and derive, for $i=1,2$:

$\displaystyle\left\langle\bm{\ell}^{\textnormal{VOL}}{\mathcolor{NavyBlue}{\mathbf{1}}}\int_{[0,\infty)}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{i}(dx),\widehat{\mathbb{W}}_{t}\right\rangle$
$\displaystyle=-\int_{0}^{t}K_{i}^{\prime}(t-s)\bar{Y}_{s}ds.$ (5.35)
$\displaystyle\left\langle\bm{\ell}^{\textnormal{VOL}}{\mathcolor{NavyBlue}{\mathbf{2}}}\int_{[0,\infty)}xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{i}(dx),\widehat{\mathbb{W}}_{t}\right\rangle$
$\displaystyle=-\int_{0}^{t}K_{i}^{\prime}(t-s)\bar{Y}_{s}dW_{s}-\frac{1}{2}K_{2}(0)\int_{0}^{t}K_{i}^{\prime}(t-s)(a_{2}+b_{2}\bar{Y}_{s})ds.$
(5.36)

###### Proof of Lemma 5.2.

Following directly from (5.11)-(5.12), we get

$\displaystyle\left\langle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}},\widehat{\mathbb{W}}_{t}\right\rangle$
$\displaystyle=K_{2}(0)(a_{2}+b_{2}\bar{Y}_{t})\quad\text{and}\quad\left\langle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}},\widehat{\mathbb{W}}_{t}\right\rangle=K_{2}^{2}(0)b_{2}(a_{2}+b_{2}\bar{Y}_{t}).$
(5.37)

It now only remains to compute
$\left\langle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle$.
For this, recall (5.15)

$\displaystyle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}$
$\displaystyle=K_{1}(0)\left(a_{1}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+b_{1}\bm{\ell}^{\textnormal{VOL}}\right)-\frac{1}{2}K_{2}^{2}(0)b_{2}\left(a_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+b_{2}\bm{\ell}^{\textnormal{VOL}}\right)$
(5.38)
$\displaystyle\quad+\left(a_{1}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+b_{1}\bm{\ell}^{\textnormal{VOL}}\right){\mathcolor{NavyBlue}{\mathbf{1}}}\int_{[0,\infty)}-xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{1}(dx)$
(5.39)
$\displaystyle\quad+\left(a_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+b_{2}\bm{\ell}^{\textnormal{VOL}}\right)\left({\mathcolor{NavyBlue}{\mathbf{2}}}-\frac{1}{2}K_{2}(0)b_{2}{\mathcolor{NavyBlue}{\mathbf{1}}}\right)\int_{[0,\infty)}-xe^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-x{\mathcolor{NavyBlue}{\mathbf{1}}}}\mu_{2}(dx).$
(5.40)

A direct application of equation (5.35) and (5.36) yields:

$\displaystyle\left\langle\bm{\ell}^{\textnormal{VOL}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle$
$\displaystyle=K_{1}(0)\left(a_{1}+b_{1}\bar{Y}_{t}\right)-\frac{1}{2}K_{2}^{2}(0)b_{2}(a_{2}+b_{2}\bar{Y}_{t})$
(5.41)
$\displaystyle\quad+\int_{0}^{t}K_{1}^{\prime}(t-s)\left(a_{1}+b_{1}\bar{Y}_{s}\right)ds+\int_{0}^{t}K_{2}^{\prime}(t-s)\left(a_{2}+b_{2}\bar{Y}_{s}\right)dW_{s},$
(5.42)

which concludes the proof. ∎

##  6 Proof of Theorem 4.4

We first recall from Lemma A.2 that for all $\bm{p},\bm{q}\in
T((\mathbb{R}^{d}))$ such that
$\bm{q}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}=0$,

$\displaystyle\left(\bm{p}{\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}-\bm{q}\right)}^{-1}\right)|_{{\mathcolor{NavyBlue}{\mathbf{i}}}}$
$\displaystyle=\bm{p}|_{{\mathcolor{NavyBlue}{\mathbf{i}}}}+\bm{p}{\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}-\bm{q}\right)}^{-1}(\bm{q}|_{{\mathcolor{NavyBlue}{\mathbf{i}}}}).$
(6.1)

Returning to the definition of $\bm{q}^{\textnormal{DE}}$,
$\bm{p}^{\textnormal{DE}}$ and $\bm{\ell}^{\textnormal{DE}}$:

$\displaystyle\bm{\ell}^{\textnormal{DE}}$
$\displaystyle=\bm{p}^{\textnormal{DE}}{\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}-\bm{q}^{\textnormal{DE}}\right)}^{-1},$
(6.2) $\displaystyle\bm{p}^{\textnormal{DE}}$
$\displaystyle=\left(a_{1}-\tfrac{1}{2}b_{2}a_{2}\right){\mathcolor{NavyBlue}{\mathbf{1}}}+a_{2}{\mathcolor{NavyBlue}{\mathbf{2}}}+z{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}},$
(6.3) $\displaystyle\bm{q}^{\textnormal{DE}}$
$\displaystyle=\left(b_{1}-\tfrac{1}{2}b_{2}^{2}\right){\mathcolor{NavyBlue}{\mathbf{1}}}+b_{2}{\mathcolor{NavyBlue}{\mathbf{2}}}+\left(\sum_{m=1}^{k_{1}}c_{1}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\alpha_{1}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}}-\tfrac{1}{2}b_{2}\sum_{m=1}^{k_{2}}c_{2}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\alpha_{2}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}}\right){\mathcolor{NavyBlue}{\mathbf{1}}}+\left(\sum_{m=1}^{k_{2}}c_{2}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\alpha_{2}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}}\right){\mathcolor{NavyBlue}{\mathbf{2}}},$
(6.4)

it follows that

$\displaystyle\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}$
$\displaystyle=\bm{\ell}^{\textnormal{DE}}\left(\left(b_{1}-\tfrac{1}{2}b_{2}^{2}\right){\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+\sum_{m=1}^{k_{1}}c_{1}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\alpha_{1}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}}-\tfrac{1}{2}b_{2}\sum_{m=1}^{k_{2}}c_{2}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\alpha_{2}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}}\right)+\left(a_{1}-\tfrac{1}{2}b_{2}a_{2}\right){\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}},$
(6.5)
$\displaystyle\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}}$
$\displaystyle=\bm{\ell}^{\textnormal{DE}}\left(b_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+\sum_{m=1}^{k_{2}}c_{2}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\alpha_{2}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}}\right)+a_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}},$
(6.6)
$\displaystyle\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}}$
$\displaystyle=b_{2}\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}}.$
(6.7) Finally,
$\displaystyle\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}+\tfrac{1}{2}\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}}$
$\displaystyle=\bm{\ell}^{\textnormal{DE}}\left(b_{1}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+\sum_{m=1}^{k_{1}}c_{1}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\alpha_{1}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}}\right)+a_{1}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}.$
(6.8)

######  Lemma 6.1.

We have that
$\left\\{\bm{\ell}^{\textnormal{DE}},\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}},\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}},\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}}\right\\}\subset{\mathcal{A}_{h}}$.
In particular, $\bm{\ell}^{\textnormal{DE}}\in\mathcal{I}(\widehat{W}).$

###### Proof.

First, observe that
${\mathcolor{NavyBlue}{\mathbf{i}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\gamma{\mathcolor{NavyBlue}{\mathbf{i}}}}\preceq{\mathcolor{NavyBlue}{\mathbf{i}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\hat{\gamma}{\mathcolor{NavyBlue}{\mathbf{i}}}}=\frac{1}{\hat{\gamma}}\left(e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\hat{\gamma}{\mathcolor{NavyBlue}{\mathbf{i}}}}-{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right)$
where $\hat{\gamma}:=\max\left(\left|\gamma\right|,1\right)$ and $\preceq$ is
defined in Definition 3.8. Thus we obtain that

$\displaystyle\bm{q}^{\textnormal{DE}}$
$\displaystyle\preceq\left|b_{1}-\sum_{m=1}^{k_{1}}\frac{c_{1}^{m}}{\hat{\alpha}_{1}^{m}}-\frac{1}{2}b_{2}\left(b_{2}-\sum_{m=1}^{k_{2}}\frac{c_{2}^{m}}{\hat{\alpha}_{2}^{m}}\right)\right|{\mathcolor{NavyBlue}{\mathbf{1}}}+\left|b_{2}-\sum_{m=1}^{k_{2}}\frac{c_{2}^{m}}{\hat{\alpha}_{2}^{m}}\right|{\mathcolor{NavyBlue}{\mathbf{2}}}$
(6.9)
$\displaystyle\qquad+\sum_{m=1}^{k_{1}}\left|\frac{c_{1}^{m}}{\hat{\alpha}_{1}^{m}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\hat{\alpha}_{1}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}}-\frac{1}{2}b_{2}\frac{c_{2}^{m}}{\hat{\alpha}_{2}^{m}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\hat{\alpha}_{2}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}}\right|{\mathcolor{NavyBlue}{\mathbf{1}}}+\sum_{m=1}^{k_{2}}\left|\frac{c_{2}^{m}}{\hat{\alpha}_{2}^{m}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\hat{\alpha}_{2}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}}\right|{\mathcolor{NavyBlue}{\mathbf{2}}},$
(6.10)

where $\hat{\alpha}_{i}^{m}:=\max\left(\left|\alpha_{i}^{m}\right|,1\right)$,
$i=1,2$. We can then further dominate it with

$\bm{q}^{\textnormal{DE}}\preceq
Me^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\left(\sum_{m=1}^{k_{1}}\hat{\alpha}_{1}^{m}+\sum_{m=1}^{k_{2}}\hat{\alpha}_{2}^{m}\right){\mathcolor{NavyBlue}{\mathbf{1}}}}({\mathcolor{NavyBlue}{\mathbf{1}}}+{\mathcolor{NavyBlue}{\mathbf{2}}}),$

where

$\displaystyle M$
$\displaystyle=\left|b_{1}-\sum_{m=1}^{k_{1}}\frac{c_{1}^{m}}{\hat{\alpha}_{1}^{m}}-\frac{1}{2}b_{2}\left(b_{2}-\sum_{m=1}^{k_{2}}\frac{c_{2}^{m}}{\hat{\alpha}_{2}^{m}}\right)\right|+\left|b_{2}-\sum_{m=1}^{k_{2}}\frac{c_{2}^{m}}{\hat{\alpha}_{2}^{m}}\right|+\left|\sum_{m=1}^{k_{1}}\frac{c_{1}^{m}}{\hat{\alpha}_{1}^{m}}-\frac{1}{2}b_{2}\frac{c_{2}^{m}}{\hat{\alpha}_{2}^{m}}\right|+\left|\sum_{m=1}^{k_{2}}\frac{c_{2}^{m}}{\hat{\alpha}_{2}^{m}}\right|.$
(6.11)

Thus by Theorem 3.9

$\sup_{t\in[0,T]}\left|\left|\bm{\ell}^{\textnormal{DE}}\right|\right|_{t}^{\mathcal{A}_{h}}<\infty.$

With representation (6.5), we apply Proposition 3.7 and proving
$\left\\{\bm{\ell}^{\textnormal{DE}},\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}},\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}},\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}}\right\\}\subset{\mathcal{A}_{h}}$.
Similarly to lemma 5.1, we directly derive that
$\bm{\ell}^{\textnormal{DE}}\in\mathcal{I}(\widehat{W})$. ∎

We are now ready to apply Theorem 3.3 on $\bm{\ell}^{\textnormal{DE}}$ and get

$\displaystyle\left\langle\bm{\ell}^{\textnormal{DE}},\widehat{\mathbb{W}}_{t}\right\rangle$
$\displaystyle=\left(\bm{\ell}^{\textnormal{DE}}\right)^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+\int_{0}^{t}\left\langle\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{1}}}}+\tfrac{1}{2}\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{22}}}},\widehat{\mathbb{W}}_{s}\right\rangle
ds+\int_{0}^{t}\left\langle\bm{\ell}^{\textnormal{DE}}|_{{\mathcolor{NavyBlue}{\mathbf{2}}}},\widehat{\mathbb{W}}_{s}\right\rangle
dW_{s}$ (6.12)
$\displaystyle=z+\int_{0}^{t}\left\langle\bm{\ell}^{\textnormal{DE}}\left(b_{1}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+\sum_{m=1}^{k_{1}}c_{1}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\alpha_{1}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}}\right)+a_{1}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}},\widehat{\mathbb{W}}_{s}\right\rangle
ds$ (6.13)
$\displaystyle\qquad+\int_{0}^{t}\left\langle\bm{\ell}^{\textnormal{DE}}\left(b_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+\sum_{m=1}^{k_{2}}c_{2}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\alpha_{2}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}}\right)+a_{2}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}},\widehat{\mathbb{W}}_{s}\right\rangle
dW_{s}$ (6.14)
$\displaystyle=z+\int_{0}^{t}\left(a_{1}+b_{1}\left\langle\bm{\ell}^{\textnormal{DE}},\widehat{\mathbb{W}}_{s}\right\rangle+\sum_{m=1}^{k_{1}}c_{1}^{m}\left\langle\bm{\ell}^{\textnormal{DE}}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\alpha_{1}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{s}\right\rangle\right)ds$
(6.15)
$\displaystyle\qquad+\int_{0}^{t}\left(a_{2}+b_{2}\left\langle\bm{\ell}^{\textnormal{DE}},\widehat{\mathbb{W}}_{s}\right\rangle+\sum_{m=1}^{k_{2}}c_{2}^{m}\left\langle\bm{\ell}^{\textnormal{DE}}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\alpha_{2}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{s}\right\rangle\right)dW_{s}.$
(6.16)

Finally, using Lemma 5.3, i.e.
$\bm{\ell}{\mathcolor{NavyBlue}{\mathbf{i}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\gamma{\mathcolor{NavyBlue}{\mathbf{j}}}}=e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\gamma{\mathcolor{NavyBlue}{\mathbf{j}}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\left(\left(e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}-\gamma{\mathcolor{NavyBlue}{\mathbf{j}}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}\right){\mathcolor{NavyBlue}{\mathbf{i}}}\right)$
for all $\gamma\in\mathbb{R}$ and
${\mathcolor{NavyBlue}{\mathbf{i}}},{\mathcolor{NavyBlue}{\mathbf{j}}}\in
A_{d}$, and the fact that
$\bm{\ell}^{\textnormal{DE}}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\alpha_{i}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}}\in\mathcal{A}$
for all $m$ and $i=1,2$, it is then straightforward to see that

$\displaystyle\left\langle\bm{\ell}^{\textnormal{DE}}{\mathcolor{NavyBlue}{\mathbf{1}}}e^{\mathrel{\sqcup\mkern-3.0mu\sqcup}\alpha_{i}^{m}{\mathcolor{NavyBlue}{\mathbf{1}}}},\widehat{\mathbb{W}}_{t}\right\rangle$
$\displaystyle=\int_{0}^{t}e^{\alpha_{i}^{m}(t-s)}\left\langle\bm{\ell}^{\textnormal{DE}},\widehat{\mathbb{W}}_{s}\right\rangle
ds.$ (6.17)

This concludes the proof of Theorem 4.4 that
$\left\langle\bm{\ell}^{\textnormal{DE}},\widehat{\mathbb{W}}_{t}\right\rangle$
is solution to (4.21) and hence
$Z_{t}=\left\langle\bm{\ell}^{\textnormal{DE}},\widehat{\mathbb{W}}_{t}\right\rangle$
by strong uniqueness.

##  7 Proof of Theorem 4.5

We first illustrate that
$\left\langle\bm{\ell}_{t}^{\textnormal{GV}},\widehat{\mathbb{W}}_{t}\right\rangle$
with $\bm{\ell}_{t}^{\textnormal{GV}}$ defined in (4.33) is a well-defined
stochastic process.

$\displaystyle\mathbb{E}\bigg{[}\int_{0}^{T}\sum_{n=0}^{\infty}\left|K^{(n)}(t)\left\langle{\mathcolor{NavyBlue}{\mathbf{1}}}^{\otimes
n}{\mathcolor{NavyBlue}{\mathbf{2}}},\widehat{\mathbb{W}}_{t}\right\rangle\right|dt\bigg{]}$
$\displaystyle=\int_{0}^{T}\sum_{n=0}^{\infty}\left|K^{(n)}(t)\right|\mathbb{E}\left|\left\langle{\mathcolor{NavyBlue}{\mathbf{1}}}^{\otimes
n}{\mathcolor{NavyBlue}{\mathbf{2}}},\widehat{\mathbb{W}}_{t}\right\rangle\right|dt$
(7.1)
$\displaystyle=\int_{0}^{T}\sum_{n=0}^{\infty}\left|K^{(n)}(t)\right|\mathbb{E}\left|\int_{0}^{t}\frac{s^{n}}{n!}dW_{s}\right|dt$
(7.2)
$\displaystyle\leq\int_{0}^{T}\sum_{n=0}^{\infty}\left|K^{(n)}(t)\right|\frac{t^{n+\frac{1}{2}}}{n!\sqrt{2n+1}}~{}dt$
(7.3)

The last inequality is derived by Jensen inequality. Then by assumption we
know that

$\mathbb{E}\bigg{[}\int_{0}^{T}\sum_{n=0}^{\infty}\left|K^{(n)}(t)\left\langle{\mathcolor{NavyBlue}{\mathbf{1}}}^{\otimes
n}{\mathcolor{NavyBlue}{\mathbf{2}}},\widehat{\mathbb{W}}_{t}\right\rangle\right|dt\bigg{]}<\infty,$

thus

$\sum_{n=0}^{\infty}\left|K^{(n)}(t)\left\langle{\mathcolor{NavyBlue}{\mathbf{1}}}^{\otimes
n}{\mathcolor{NavyBlue}{\mathbf{2}}},\widehat{\mathbb{W}}_{t}\right\rangle\right|<\infty,~{}\mathbb{P}(d\omega)\otimes
dt-a.e.$

Consequently,
$\sum_{n=0}^{\infty}K^{(n)}(t)\left\langle{\mathcolor{NavyBlue}{\mathbf{1}}}^{\otimes
n}{\mathcolor{NavyBlue}{\mathbf{2}}},\widehat{\mathbb{W}}_{t}\right\rangle$
converges $\mathbb{P}(d\omega)\otimes dt-a.e.$, ensuring
$\left\langle\bm{\ell}_{t}^{\textnormal{GV}},\widehat{\mathbb{W}}_{t}\right\rangle$
to be a $\mathbb{P}(d\omega)\otimes dt-a.e.$-well-defined map. Notice that for
every $N\in\mathbb{N}$,
$\sum_{n=0}^{N}K^{(n)}(t)\left\langle{\mathcolor{NavyBlue}{\mathbf{1}}}^{\otimes
n}{\mathcolor{NavyBlue}{\mathbf{2}}},\widehat{\mathbb{W}}_{t}\right\rangle$ is
progressively measurable, so
$\left\langle\bm{\ell}_{t}^{\textnormal{GV}},\widehat{\mathbb{W}}_{t}\right\rangle$
is a also progressively measurable. It remains to prove that for every fixed
$t$,

$\displaystyle\left\langle\bm{\ell}_{t}^{\textnormal{GV}},\widehat{\mathbb{W}}_{t}\right\rangle=\int_{0}^{t}K(t-s)dW_{s}.$
(7.4)

For this we write

$\displaystyle\left\langle\bm{\ell}_{t}^{\textnormal{GV}},\widehat{\mathbb{W}}_{t}\right\rangle$
$\displaystyle=\sum_{n=0}^{\infty}K^{(n)}(t)(-1)^{n}\left\langle{\mathcolor{NavyBlue}{\mathbf{1}}}^{\otimes
n}{\mathcolor{NavyBlue}{\mathbf{2}}},\widehat{\mathbb{W}}_{t}\right\rangle=\sum_{n=0}^{\infty}K^{(n)}(t)\int_{0}^{t}\frac{(-s)^{n}}{n!}dW_{s}.$

Since
$\int_{0}^{t}\left(\sum_{n=0}^{\infty}\left|K^{(n)}(t)\right|\frac{s^{n}}{n!}\right)^{2}ds<\infty$
by assumption (LABEL:eq:assKgaussian2), the dominated convergence theorem of
stochastic integrals yields

$\sum_{n=0}^{N}K^{(n)}(t)\int_{0}^{t}\frac{(-s)^{n}}{n!}dW_{s}\longrightarrow\int_{0}^{t}\sum_{n=0}^{\infty}K^{(n)}(t)\frac{(-s)^{n}}{n!}dW_{s}=\int_{0}^{t}K(t-s)dW_{s},$

in probability, as $N\to\infty$. While the previous analysis ensures that

$\sum_{n=0}^{N}K^{(n)}(t)\int_{0}^{t}\frac{(-s)^{n}}{n!}dW_{s}\overset{\text{a.s.}}{\longrightarrow}\sum_{n=0}^{\infty}K^{(n)}(t)\int_{0}^{t}\frac{(-s)^{n}}{n!}dW_{s}=\left\langle\bm{\ell}_{t}^{\textnormal{GV}},\widehat{\mathbb{W}}_{t}\right\rangle,$

which proves (7.4) and completes the proof.

##  Appendix A Properties of the resolvent

######  Lemma A.1.

For $\bm{\ell}\in T((\mathbb{R}^{d}))$ such that
$\left|\bm{\ell}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right|<1$,
$\sum_{n=0}^{\infty}\bm{\ell}^{\otimes n}$ is well defined, i.e.
$\sum_{n=0}^{\infty}\left\langle\bm{\ell}^{\otimes
n},{\mathcolor{NavyBlue}{\mathbf{v}}}\right\rangle<\infty$ for every
${\mathcolor{NavyBlue}{\mathbf{v}}}\in V$.

###### Proof.

Let $m\in\mathbb{N}$ and ${\mathcolor{NavyBlue}{\mathbf{v}}}\in V_{m}$ a word
of length $m$. Then,

$\sup_{n\in\mathbb{N}}\sup_{\begin{subarray}{c}{\mathcolor{NavyBlue}{\mathbf{w_{1}}}},\dots,{\mathcolor{NavyBlue}{\mathbf{w_{n}}}}\in
V\setminus\left\\{{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right\\}\\\
{\mathcolor{NavyBlue}{\mathbf{w_{1}\cdots
w_{n}}}}={\mathcolor{NavyBlue}{\mathbf{v}}}\end{subarray}}\left|\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{w_{1}}}}\cdots\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{w_{n}}}}\right|<\infty,$

since the supremum is taken over a finite set. Consequently let us define

$M({\mathcolor{NavyBlue}{\mathbf{v}}}):=\sup_{n\in\mathbb{N}}\sum_{\begin{subarray}{c}{\mathcolor{NavyBlue}{\mathbf{w_{1}}}},\dots,{\mathcolor{NavyBlue}{\mathbf{w_{n}}}}\in
V\setminus\left\\{{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right\\}\\\
{\mathcolor{NavyBlue}{\mathbf{w_{1}\cdots
w_{n}}}}={\mathcolor{NavyBlue}{\mathbf{v}}}\end{subarray}}\left|\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{w_{1}}}}\cdots\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{w_{n}}}}\right|<\infty.$

We then have for all $n\in\mathbb{N}$ that

$\displaystyle\left|\left\langle\bm{\ell}^{\otimes
n},{\mathcolor{NavyBlue}{\mathbf{v}}}\right\rangle\right|$
$\displaystyle\leq\sum_{\begin{subarray}{c}{\mathcolor{NavyBlue}{\mathbf{w_{1}}}},\dots,{\mathcolor{NavyBlue}{\mathbf{w_{n}}}}\in
V\\\ {\mathcolor{NavyBlue}{\mathbf{w_{1}\cdots
w_{n}}}}={\mathcolor{NavyBlue}{\mathbf{v}}}\end{subarray}}\left|\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{w_{1}}}}\cdots\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{w_{n}}}}\right|$
(A.1)
$\displaystyle\leq\sum_{k=0}^{n}\binom{n}{k}\left|\bm{\ell}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right|^{k}\left(\sum_{\begin{subarray}{c}{\mathcolor{NavyBlue}{\mathbf{w_{1}}}},\dots,{\mathcolor{NavyBlue}{\mathbf{w_{n-k}}}}\in
V\setminus\left\\{{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right\\}\\\
{\mathcolor{NavyBlue}{\mathbf{w_{1}\cdots
w_{n-k}}}}={\mathcolor{NavyBlue}{\mathbf{v}}}\end{subarray}}\left|\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{w_{1}}}}\cdots\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{w_{n-k}}}}\right|\right)$
(A.2) $\displaystyle\leq
M({\mathcolor{NavyBlue}{\mathbf{v}}})\sum_{k=(n-m)^{+}}^{n}\binom{n}{k}\left|\bm{\ell}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right|^{k}.$
(A.3)

The last inequality follows from the fact that the set
$\left\\{{\mathcolor{NavyBlue}{\mathbf{w_{1}}}},\dots,{\mathcolor{NavyBlue}{\mathbf{w_{n}}}}\in
V\setminus\left\\{{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right\\}:{\mathcolor{NavyBlue}{\mathbf{w_{1}\cdots
w_{n}}}}={\mathcolor{NavyBlue}{\mathbf{v}}}\right\\}$ is empty when $n-k>m$.
Recall $m$ is the size of the word ${\mathcolor{NavyBlue}{\mathbf{v}}}$.
Therefore

$\displaystyle\sum_{n=0}^{\infty}\left|\left\langle\bm{\ell}^{\otimes
n},{\mathcolor{NavyBlue}{\mathbf{v}}}\right\rangle\right|\leq
M({\mathcolor{NavyBlue}{\mathbf{v}}})\sum_{n=0}^{\infty}\sum_{k=(n-m)^{+}}^{n}\binom{n}{k}\left|\bm{\ell}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right|^{k}=M({\mathcolor{NavyBlue}{\mathbf{v}}})\sum_{n=0}^{m-1}\sum_{k=0}^{n}\binom{n}{k}\left|\bm{\ell}^{{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}}\right|^{k}+M({\mathcolor{NavyBlue}{\mathbf{v}}})\sum_{n=m}^{\infty}\sum_{k=n-m}^{n}\binom{n}{k}\left|\bm{\ell}^{{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}}\right|^{k}.$
(A.4)

While the first part of the right-hand side is obviously finite, the second
needs more work, i.e. using Pascal’s rule and the upper bound for the binomial
coefficients we have

$\displaystyle\sum_{n=m}^{\infty}\sum_{k=n-m}^{n}\binom{n}{k}\left|\bm{\ell}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right|^{k}$
$\displaystyle=\sum_{n=m}^{\infty}\left|\bm{\ell}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right|^{n-m}\sum_{k=0}^{m}\binom{n}{n-m+k}\left|\bm{\ell}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right|^{k}$
(A.5)
$\displaystyle=\sum_{n=m}^{\infty}\left|\bm{\ell}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right|^{n-m}\sum_{k=0}^{m}\binom{n}{n-m+k}$
(A.6)
$\displaystyle\leq\left|\bm{\ell}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right|^{-m}\sum_{n=m}^{\infty}\left|\bm{\ell}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right|^{n}\binom{n+m}{n},$
(A.7)

where the first inequality follows from the assumption that
$\left|\bm{\ell}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right|<1$.
We can finally remark that the right-hand side has the same asymptotics as
$\sum_{n=0}^{\infty}n^{m}\left|\bm{\ell}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right|^{n}$,
which is finite, and conclude the proof. ∎

######  Lemma A.2.

Let $\bm{p},\bm{q}\in T((\mathbb{R}^{d}))$ such that
$\left|\bm{q}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\right|<1$
and ${\mathcolor{NavyBlue}{\mathbf{i}}}\in A_{d}$. Then

$\left(\bm{p}{\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}-\bm{q}\right)}^{-1}\right)|_{{\mathcolor{NavyBlue}{\mathbf{i}}}}=\frac{1}{1-\bm{q}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}}\left[\bm{p}|_{{\mathcolor{NavyBlue}{\mathbf{i}}}}+\bm{p}{\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}-\bm{q}\right)}^{-1}(\bm{q}|_{{\mathcolor{NavyBlue}{\mathbf{i}}}})\right].$

###### Proof.

First recall the decomposition in Remark 2.1, i.e.

$\bm{q}=\bm{q}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+\sum_{{\mathcolor{NavyBlue}{\mathbf{i}}}\in
A_{d}}\bm{q}|_{{\mathcolor{NavyBlue}{\mathbf{i}}}}{\mathcolor{NavyBlue}{\mathbf{i}}},$

and Proposition 2.2, i.e.

${\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}-\bm{q}\right)}^{-1}={\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+{\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}-\bm{q}\right)}^{-1}\bm{q}.$

One can then easily get that

$\displaystyle{\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}-\bm{q}\right)}^{-1}$
$\displaystyle={\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+{\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}-\bm{q}\right)}^{-1}\left(\bm{q}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+\sum_{{\mathcolor{NavyBlue}{\mathbf{i}}}\in
A_{d}}\bm{q}|_{{\mathcolor{NavyBlue}{\mathbf{i}}}}{\mathcolor{NavyBlue}{\mathbf{i}}}\right)=\frac{1}{1-\bm{q}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}}\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}+{\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}-\bm{q}\right)}^{-1}\sum_{{\mathcolor{NavyBlue}{\mathbf{i}}}\in
A_{d}}\bm{q}|_{{\mathcolor{NavyBlue}{\mathbf{i}}}}{\mathcolor{NavyBlue}{\mathbf{i}}}\right).$
(A.8)

It then follows that

$\displaystyle\bm{p}{\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}-\bm{q}\right)}^{-1}$
$\displaystyle=\frac{1}{1-\bm{q}^{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}}\left(\bm{p}+\bm{p}{\left({\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}-\bm{q}\right)}^{-1}\sum_{{\mathcolor{NavyBlue}{\mathbf{i}}}\in
A_{d}}\bm{q}|_{{\mathcolor{NavyBlue}{\mathbf{i}}}}{\mathcolor{NavyBlue}{\mathbf{i}}}\right).$
(A.9)

Taking the projection ends the proof. ∎

##  Appendix B Proof of Proposition 3.5

In this section, we are specifically interested in the case
$X_{t}=\widehat{W}_{t}=(t,W_{t}^{2},\dots,W_{t}^{d})$ where
$W=(W^{2},\dots,W^{d})$ is a (d-1)-dimensional Brownian motion. We will denote
$n({\mathcolor{NavyBlue}{\mathbf{v}}})$ the size of
${\mathcolor{NavyBlue}{\mathbf{v}}}$, i.e. its number of letters, and
$x({\mathcolor{NavyBlue}{\mathbf{v}}})$ the number of
${\mathcolor{NavyBlue}{\mathbf{1}}}$s specifically. If it does not cause
ambiguity, we will write $n$ for $n({\mathcolor{NavyBlue}{\mathbf{v}}})$ and
$x$ for $x({\mathcolor{NavyBlue}{\mathbf{v}}})$.

Let $g^{p}:[0,T]\times V\to\mathbb{R}$ be defined for all even
$p\in\mathbb{N}^{*}$ by

$\displaystyle
g_{t}^{p}({\mathcolor{NavyBlue}{\mathbf{v}}})=\left[\prod_{j=2}^{p}\binom{jn}{n}\right]\frac{t^{\frac{p}{2}(n+x)}}{\left(\frac{p}{2}(n+x)\right)!}2^{\frac{p}{2}(x-n)}$
(B.1)

and for all odd $p\in\mathbb{N}$ by

$\displaystyle
g_{t}^{p}({\mathcolor{NavyBlue}{\mathbf{v}}})=\left(g_{t}^{p+1}({\mathcolor{NavyBlue}{\mathbf{v}}})\right)^{\frac{p}{p+1}},$
(B.2)

for all $t\in[0,T]$ and all ${\mathcolor{NavyBlue}{\mathbf{v}}}\in V$. Notice
that $g_{t}^{p}({\mathcolor{NavyBlue}{\mathbf{v}}})$ only depends on
${\mathcolor{NavyBlue}{\mathbf{v}}}$ through its size and the number of its
${\mathcolor{NavyBlue}{\mathbf{1}}}$s, i.e.
$n({\mathcolor{NavyBlue}{\mathbf{v}}})$ and
$x({\mathcolor{NavyBlue}{\mathbf{v}}})$.

######  Proposition B.1.

Fix $t\in[0,T]$. We have

$\displaystyle\mathbb{E}\left[\left|\left\langle{\mathcolor{NavyBlue}{\mathbf{v}}},\widehat{\mathbb{W}}_{t}\right\rangle\right|^{p}\right]\leq
g_{t}^{p}({\mathcolor{NavyBlue}{\mathbf{v}}}),\quad{\mathcolor{NavyBlue}{\mathbf{v}}}\in
V,\quad p\in\mathbb{N}.$ (B.3)

###### Proof.

We first use Fawcett’s formula [25, Proposition 4.10] to write

$\mathbb{E}\left[\widehat{\mathbb{W}}_{t}\right]=\sum_{m=0}^{\infty}\frac{t^{m}}{m!}\left({\mathcolor{NavyBlue}{\mathbf{1}}}+\frac{1}{2}\sum_{{\mathcolor{NavyBlue}{\mathbf{j}}}\in\left\\{{\mathcolor{NavyBlue}{\mathbf{2}}},\dots,{\mathcolor{NavyBlue}{\mathbf{d}}}\right\\}}{\mathcolor{NavyBlue}{\mathbf{jj}}}\right)^{\otimes
m}.$

We then define for all $n\in\mathbb{N}$

$\widehat{V}_{n}:=\left\\{{\mathcolor{NavyBlue}{\mathbf{v}}}\in V_{n}:\exists
j\in\mathbb{N}\text{ such that
}{\mathcolor{NavyBlue}{\mathbf{v}}}={\mathcolor{NavyBlue}{\mathbf{u_{1}\cdots
u_{j}}}}\text{ for some
}{\mathcolor{NavyBlue}{\mathbf{u_{k}}}}\in\left\\{{\mathcolor{NavyBlue}{\mathbf{1}}},{\mathcolor{NavyBlue}{\mathbf{22}}},\dots,{\mathcolor{NavyBlue}{\mathbf{dd}}}\right\\}\text{
with }k=1,2,\dots,j\right\\}.$ (B.4)

This set corresponds to the coordinate of every signature element with non-
zero expectation, i.e. for every $n\in\mathbb{N}$ and
${\mathcolor{NavyBlue}{\mathbf{v}}}\in V_{n}$,

$\mathbb{E}\left[\left\langle{\mathcolor{NavyBlue}{\mathbf{v}}},\widehat{\mathbb{W}}_{t}\right\rangle\right]=\begin{cases}\frac{t^{\frac{1}{2}(n+x)}}{\left(\frac{1}{2}(n+x)\right)!}\cdot
2^{\frac{1}{2}(x-n)}&\text{if
}{\mathcolor{NavyBlue}{\mathbf{v}}}\in\widehat{V}_{n},\\\
0&\text{else}.\end{cases}$ (B.5)

We are now ready to compute even moments of every signature element. First,
remark that for all ${\mathcolor{NavyBlue}{\mathbf{u}}}\in V_{k}$ and
${\mathcolor{NavyBlue}{\mathbf{v}}}\in V_{m}$,
${\mathcolor{NavyBlue}{\mathbf{u}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}{\mathcolor{NavyBlue}{\mathbf{v}}}$
is a sum of $\binom{k+m}{k}$ elements ${\mathcolor{NavyBlue}{\mathbf{w}}}\in
V_{k+m}$ of length $n({\mathcolor{NavyBlue}{\mathbf{w}}})=k+m$ each, having
exactly
$x({\mathcolor{NavyBlue}{\mathbf{w}}})=x({\mathcolor{NavyBlue}{\mathbf{u}}})+x({\mathcolor{NavyBlue}{\mathbf{v}}})$
number of ${\mathcolor{NavyBlue}{\mathbf{1}}}$s, so that

$\mathbb{E}\left[\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}{\mathcolor{NavyBlue}{\mathbf{v}}},\widehat{\mathbb{W}}_{t}\right\rangle\right]\leq\binom{k+m}{k}\max_{\begin{subarray}{c}{\mathcolor{NavyBlue}{\mathbf{w}}}\in
V_{k+m}\\\ \textnormal{s.t.
}x({\mathcolor{NavyBlue}{\mathbf{w}}})=x({\mathcolor{NavyBlue}{\mathbf{u}}})+x({\mathcolor{NavyBlue}{\mathbf{v}}})\end{subarray}}\mathbb{E}\left\langle{\mathcolor{NavyBlue}{\mathbf{w}}},\widehat{\mathbb{W}}_{t}\right\rangle.$

Consequently, for every $p\in\mathbb{N}$, we can iteratively apply the shuffle
product $p$-times, and obtain the claimed upper bound for the even powers of
the signature:

$\displaystyle\mathbb{E}\left[\left\langle{\mathcolor{NavyBlue}{\mathbf{v}}},\widehat{\mathbb{W}}_{t}\right\rangle^{2p}\right]$
$\displaystyle=\mathbb{E}\left\langle{\mathcolor{NavyBlue}{\mathbf{v}}}^{\mathrel{\sqcup\mkern-3.0mu\sqcup}2p},\widehat{\mathbb{W}}_{t}\right\rangle\leq\left[\prod_{j=2}^{2p}\binom{jn}{n}\right]\max_{\begin{subarray}{c}{\mathcolor{NavyBlue}{\mathbf{w}}}\in
V_{2pn}\\\
s.t.\>x({\mathcolor{NavyBlue}{\mathbf{w}}})=2p\>x({\mathcolor{NavyBlue}{\mathbf{v}}})\end{subarray}}\mathbb{E}\left\langle{\mathcolor{NavyBlue}{\mathbf{w}}},\widehat{\mathbb{W}}_{t}\right\rangle=\left[\prod_{j=2}^{2p}\binom{jn}{n}\right]\frac{t^{p(n+x)}}{(p(n+x))!}2^{p(x-n)}.$
(B.6)

We can finally treat the odd moments by applying Jensen’s inequality on the
convex function $y\to y^{\frac{2p}{2p-1}}$ on $\mathbb{R}^{+}$:

$\displaystyle\mathbb{E}\left[\left|\left\langle{\mathcolor{NavyBlue}{\mathbf{v}}},\widehat{\mathbb{W}}_{t}\right\rangle\right|^{2p-1}\right]^{\frac{1}{2p-1}}\leq\mathbb{E}\left[\left\langle{\mathcolor{NavyBlue}{\mathbf{v}}},\widehat{\mathbb{W}}_{t}\right\rangle^{2p}\right]^{\frac{1}{2p}}\leq\left(h_{t}^{2p}({\mathcolor{NavyBlue}{\mathbf{v}}})\right)^{\frac{1}{2p}}=\left(h_{t}^{2p-1}({\mathcolor{NavyBlue}{\mathbf{v}}})\right)^{\frac{1}{2p-1}},$
(B.7)

concluding the proof of the proposition. ∎

We are now ready to prove Proposition 3.5:

###### Proof of Proposition 3.5.

Proposition 3.5-3(i) follows directly from an application of the above
Proposition B.1. In order to prove (3.16) we will consider two cases: either
when word ${\mathcolor{NavyBlue}{\mathbf{v}}}$ ends with a
${\mathcolor{NavyBlue}{\mathbf{1}}}$ or it ends with a
${\mathcolor{NavyBlue}{\mathbf{j}}}$ in
$\left\\{{\mathcolor{NavyBlue}{\mathbf{2}}},\dots,{\mathcolor{NavyBlue}{\mathbf{d}}}\right\\}$.

    1. (1)

Case ${\mathcolor{NavyBlue}{\mathbf{v}}}={\mathcolor{NavyBlue}{\mathbf{u1}}}$
for some word ${\mathcolor{NavyBlue}{\mathbf{u}}}\in V$. We apply Cauchy-
Schwarz inequality to get

$\displaystyle\mathbb{E}\left[\sup_{s\in[0,t]}\left\langle{\mathcolor{NavyBlue}{\mathbf{v}}},\widehat{\mathbb{W}}_{s}\right\rangle^{2}\right]$
$\displaystyle=\mathbb{E}\left[\sup_{s\in[0,t]}\left(\int_{0}^{s}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}},\widehat{\mathbb{W}}_{u}\right\rangle
du\right)^{2}\right]$ (B.8)
$\displaystyle\leq\mathbb{E}\left[\left(\int_{0}^{t}\left|\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}},\widehat{\mathbb{W}}_{u}\right\rangle\right|du\right)^{2}\right]$
(B.9)
$\displaystyle\leq\mathbb{E}\left[t\int_{0}^{t}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}},\widehat{\mathbb{W}}_{u}\right\rangle^{2}du\right]$
(B.10)
$\displaystyle=t\int_{0}^{t}\mathbb{E}\left[\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}},\widehat{\mathbb{W}}_{u}\right\rangle^{2}\right]du.$
(B.11)

Thus, using (3.15) with
$n=n({\mathcolor{NavyBlue}{\mathbf{v}}})=n({\mathcolor{NavyBlue}{\mathbf{u}}})+1$
and
$x=x({\mathcolor{NavyBlue}{\mathbf{v}}})=x({\mathcolor{NavyBlue}{\mathbf{u}}})+1$,
we have

$\displaystyle\mathbb{E}\left[\sup_{s\in[0,t]}\left\langle{\mathcolor{NavyBlue}{\mathbf{v}}},\widehat{\mathbb{W}}_{s}\right\rangle^{2}\right]$
$\displaystyle\leq\binom{2(n-1)}{n-1}\frac{t}{2^{n-x}(n+x-2)!}\int_{0}^{t}u^{n+x-2}du$
(B.12) $\displaystyle=\binom{2(n-1)}{n-1}\frac{t^{n+x}}{2^{n-x}(n+x-1)!}$
(B.13) $\displaystyle\leq\frac{2^{n+x}}{\sqrt{n}}\frac{t^{n+x}}{(n+x-1)!},$
(B.14)

where the last inequality comes from an application of Stirling’s
approximation that reads $\binom{2m}{m}\leq\frac{2^{2m}}{\sqrt{m+1}}$ for all
$m\in\mathbb{N}$.

    2. (2)

Case ${\mathcolor{NavyBlue}{\mathbf{v}}}={\mathcolor{NavyBlue}{\mathbf{uj}}}$
for some word ${\mathcolor{NavyBlue}{\mathbf{u}}}\in V$ and some letter
${\mathcolor{NavyBlue}{\mathbf{j}}}\in\left\\{{\mathcolor{NavyBlue}{\mathbf{2}}},\dots,{\mathcolor{NavyBlue}{\mathbf{d}}}\right\\}$.
First we apply Theorem 3.3 to get

$\displaystyle\left\langle{\mathcolor{NavyBlue}{\mathbf{v}}},\widehat{\mathbb{W}}_{s}\right\rangle^{2}$
$\displaystyle=\left(\int_{0}^{s}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}},\widehat{\mathbb{W}}_{u}\right\rangle
dW_{u}^{\mathcolor{NavyBlue}{\mathbf{j}}}+\frac{1}{2}\int_{0}^{s}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}}|_{{\mathcolor{NavyBlue}{\mathbf{j}}}},\widehat{\mathbb{W}}_{u}\right\rangle
du\right)^{2}$ (B.15) $\displaystyle\leq
2\left(\int_{0}^{s}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}},\widehat{\mathbb{W}}_{u}\right\rangle
dW_{u}^{\mathcolor{NavyBlue}{\mathbf{j}}}\right)^{2}+\frac{1}{2}\left(\int_{0}^{s}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}}|_{{\mathcolor{NavyBlue}{\mathbf{j}}}},\widehat{\mathbb{W}}_{u}\right\rangle
du\right)^{2}.$ (B.16)

Now, since
$\int_{0}^{s}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}},\widehat{\mathbb{W}}_{u}\right\rangle
dW_{u}^{\mathcolor{NavyBlue}{\mathbf{j}}}$ is a true martingale, we apply
Doob’s inequality followed by the previous result (B.14) with
$n=n({\mathcolor{NavyBlue}{\mathbf{v}}})=n({\mathcolor{NavyBlue}{\mathbf{u}}})+1$
and
$x=x({\mathcolor{NavyBlue}{\mathbf{v}}})=x({\mathcolor{NavyBlue}{\mathbf{u}}})$
and get

$\displaystyle\mathbb{E}\left[\sup_{s\in[0,t]}\left(\int_{0}^{s}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}},\widehat{\mathbb{W}}_{u}\right\rangle
dW_{u}^{\mathcolor{NavyBlue}{\mathbf{j}}}\right)^{2}\right]\leq
4~{}\mathbb{E}\left[\int_{0}^{t}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}},\widehat{\mathbb{W}}_{u}\right\rangle
dW_{u}^{\mathcolor{NavyBlue}{\mathbf{j}}}\right]^{2}=4\int_{0}^{t}\mathbb{E}\left[\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}},\widehat{\mathbb{W}}_{u}\right\rangle^{2}\right]du\leq
4\frac{2^{n+x}}{\sqrt{n}}\frac{t^{n+x}}{(n+x)!}.$ (B.17)

Then, again applying (B.14) to the right-most element in (B.16), we get

$\displaystyle\mathbb{E}\left[\sup_{s\in[0,t]}\left(\int_{0}^{s}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}}|_{{\mathcolor{NavyBlue}{\mathbf{j}}}},\widehat{\mathbb{W}}_{u}\right\rangle
du\right)^{2}\right]=\mathbb{E}\left[\sup_{s\in[0,t]}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}}|_{{\mathcolor{NavyBlue}{\mathbf{j}}}}{\mathcolor{NavyBlue}{\mathbf{1}}},\widehat{\mathbb{W}}_{s}\right\rangle^{2}\right]\leq\frac{2^{n+x}}{\sqrt{n}}\frac{t^{n+x}}{(n+x-1)!}.$
(B.18)

Note that if
${\mathcolor{NavyBlue}{\mathbf{u}}}|_{{\mathcolor{NavyBlue}{\mathbf{j}}}}=0$,
the left-hand side is 0 and the equation still holds.

Combining the above inequalities, we derive

$\displaystyle\mathbb{E}\left[\sup_{s\in[0,t]}\left\langle{\mathcolor{NavyBlue}{\mathbf{v}}},\widehat{\mathbb{W}}_{s}\right\rangle^{2}\right]$
$\displaystyle\leq C_{1}\frac{2^{n+x-1}}{\sqrt{n}}\frac{t^{n+x}}{(n+x-1)!},$
(B.19)

for some $C_{1}\in\mathbb{R}$ independent of $t$ and
${\mathcolor{NavyBlue}{\mathbf{v}}}$.

Combining the two equalities (B.14)-(B.19) above, we have that for every
${\mathcolor{NavyBlue}{\mathbf{v}}}\in V_{n}$, $n\geq 1$,

$\mathbb{E}\left[\sup_{s\in[0,t]}\left|\left\langle{\mathcolor{NavyBlue}{\mathbf{v}}},\widehat{\mathbb{W}}_{s}\right\rangle\right|^{2}\right]\leq
C_{2}\frac{2^{n+x-1}}{\sqrt{n}}\frac{t^{n+x}}{(n+x-1)!}.$ (B.20)

for some $C_{2}\in\mathbb{R}$ independent of $t$ and
${\mathcolor{NavyBlue}{\mathbf{v}}}$.

Moreover, as $n+1\sim n$, there exists a constant $C$ such that

$\mathbb{E}\left[\sup_{s\in[0,t]}\left|\left\langle{\mathcolor{NavyBlue}{\mathbf{v}}},\widehat{\mathbb{W}}_{s}\right\rangle\right|^{2}\right]\leq
C\frac{2^{n+x-1}}{\sqrt{n+1}}\frac{t^{n+x}}{(n+x-1)!}.$ (B.21)

This includes the case
${\mathcolor{NavyBlue}{\mathbf{v}}}={\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}\in
V_{0}$ and proves (3.16).

It remains to prove the sub-multiplicative property (3.17). First, in the case
${\mathcolor{NavyBlue}{\mathbf{u}}}={\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}$
or
${\mathcolor{NavyBlue}{\mathbf{w}}}={\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}$,
(3.17) holds trivially. Now we can easily see that

$\displaystyle\left(\frac{h_{t}({\mathcolor{NavyBlue}{\mathbf{u}}})h_{t}({\mathcolor{NavyBlue}{\mathbf{v}}})}{h_{t}({\mathcolor{NavyBlue}{\mathbf{u}}}{\mathcolor{NavyBlue}{\mathbf{v}}})}\right)^{2}$
$\displaystyle=C^{2}A({\mathcolor{NavyBlue}{\mathbf{u}}},{\mathcolor{NavyBlue}{\mathbf{v}}})B({\mathcolor{NavyBlue}{\mathbf{u}}},{\mathcolor{NavyBlue}{\mathbf{v}}}),$
(B.22)

where

$\displaystyle
A({\mathcolor{NavyBlue}{\mathbf{u}}},{\mathcolor{NavyBlue}{\mathbf{v}}}):=\frac{(n({\mathcolor{NavyBlue}{\mathbf{u}}})+x({\mathcolor{NavyBlue}{\mathbf{u}}})+n({\mathcolor{NavyBlue}{\mathbf{v}}})+x({\mathcolor{NavyBlue}{\mathbf{v}}})-1)!}{(n({\mathcolor{NavyBlue}{\mathbf{u}}})+x({\mathcolor{NavyBlue}{\mathbf{u}}})-1)!(n({\mathcolor{NavyBlue}{\mathbf{v}}})+x({\mathcolor{NavyBlue}{\mathbf{v}}})-1)!}\quad\text{and}\quad
B({\mathcolor{NavyBlue}{\mathbf{u}}},{\mathcolor{NavyBlue}{\mathbf{v}}}):=\sqrt{\frac{n({\mathcolor{NavyBlue}{\mathbf{u}}})+n({\mathcolor{NavyBlue}{\mathbf{v}}})+1}{(n({\mathcolor{NavyBlue}{\mathbf{u}}})+1)(n({\mathcolor{NavyBlue}{\mathbf{v}}})+1)}}.$
(B.23)

We now only need to show that
$C^{2}A({\mathcolor{NavyBlue}{\mathbf{u}}},{\mathcolor{NavyBlue}{\mathbf{v}}})B({\mathcolor{NavyBlue}{\mathbf{u}}},{\mathcolor{NavyBlue}{\mathbf{v}}})\geq
1$ for
${\mathcolor{NavyBlue}{\mathbf{u}}},{\mathcolor{NavyBlue}{\mathbf{v}}}\neq{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}$.
Without loss of generality, we assume
${\mathcolor{NavyBlue}{\mathbf{u}}}\neq{\color[rgb]{0.0600000000000001,0.46,1}\textup{{\o{}}}}$,
i.e. $n({\mathcolor{NavyBlue}{\mathbf{u}}})>0$, otherwise we could simply swap
${\mathcolor{NavyBlue}{\mathbf{u}}}$ and ${\mathcolor{NavyBlue}{\mathbf{v}}}$
in the following equations. Let us then remark that

$\displaystyle
A({\mathcolor{NavyBlue}{\mathbf{u}}},{\mathcolor{NavyBlue}{\mathbf{v}}})$
$\displaystyle=\binom{n({\mathcolor{NavyBlue}{\mathbf{u}}})+x({\mathcolor{NavyBlue}{\mathbf{u}}})+n({\mathcolor{NavyBlue}{\mathbf{v}}})+x({\mathcolor{NavyBlue}{\mathbf{v}}})-1}{n({\mathcolor{NavyBlue}{\mathbf{u}}})+x({\mathcolor{NavyBlue}{\mathbf{u}}})}(n({\mathcolor{NavyBlue}{\mathbf{u}}})+x({\mathcolor{NavyBlue}{\mathbf{u}}}))$
(B.24)

and that

$\displaystyle\sqrt{n({\mathcolor{NavyBlue}{\mathbf{u}}})+1}$
$\displaystyle\leq
C^{2}(n({\mathcolor{NavyBlue}{\mathbf{u}}})+x({\mathcolor{NavyBlue}{\mathbf{u}}}))$
(B.25)
$\displaystyle\sqrt{n({\mathcolor{NavyBlue}{\mathbf{v}}})+1}\sqrt{n({\mathcolor{NavyBlue}{\mathbf{u}}})+1}$
$\displaystyle\leq
C^{2}\sqrt{n({\mathcolor{NavyBlue}{\mathbf{v}}})+1}(n({\mathcolor{NavyBlue}{\mathbf{u}}})+x({\mathcolor{NavyBlue}{\mathbf{u}}}))\leq
C^{2}\sqrt{n({\mathcolor{NavyBlue}{\mathbf{u}}})+n({\mathcolor{NavyBlue}{\mathbf{v}}})+1}(n({\mathcolor{NavyBlue}{\mathbf{u}}})+x({\mathcolor{NavyBlue}{\mathbf{u}}})),$
(B.26)

which implies

$\displaystyle
B({\mathcolor{NavyBlue}{\mathbf{u}}},{\mathcolor{NavyBlue}{\mathbf{v}}})\geq\frac{1}{C^{2}(n({\mathcolor{NavyBlue}{\mathbf{u}}})+x({\mathcolor{NavyBlue}{\mathbf{u}}}))}$
(B.27)

and thus

$\displaystyle\left(\frac{h_{t}({\mathcolor{NavyBlue}{\mathbf{u}}})h_{t}({\mathcolor{NavyBlue}{\mathbf{v}}})}{h_{t}({\mathcolor{NavyBlue}{\mathbf{u}}}{\mathcolor{NavyBlue}{\mathbf{v}}})}\right)^{2}$
$\displaystyle\geq\binom{n({\mathcolor{NavyBlue}{\mathbf{u}}})+x({\mathcolor{NavyBlue}{\mathbf{u}}}))+n({\mathcolor{NavyBlue}{\mathbf{v}}})+x({\mathcolor{NavyBlue}{\mathbf{v}}})-1}{n({\mathcolor{NavyBlue}{\mathbf{u}}})+x({\mathcolor{NavyBlue}{\mathbf{u}}})}\geq
1.$ (B.28)

∎

##  Appendix C Proof of Proposition 3.1

Let $n\leq k\in\mathbb{N}$, then for all
${\mathcolor{NavyBlue}{\mathbf{u}}}\in V_{n}$,
${\mathcolor{NavyBlue}{\mathbf{v}}}\in V_{k-n}$, each element of
${\mathcolor{NavyBlue}{\mathbf{u}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}{\mathcolor{NavyBlue}{\mathbf{v}}}$
is of size $k$. This leads to the fact that
$\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}{\mathcolor{NavyBlue}{\mathbf{v}}},{\mathcolor{NavyBlue}{\mathbf{w}}}\right\rangle=0$
whenever ${\mathcolor{NavyBlue}{\mathbf{w}}}\notin V_{k}$, i.e.

$\displaystyle\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}{\mathcolor{NavyBlue}{\mathbf{v}}},\mathbb{X}_{t}\right\rangle=\sum_{{\mathcolor{NavyBlue}{\mathbf{w}}}\in
V_{k}}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}{\mathcolor{NavyBlue}{\mathbf{v}}},{\mathcolor{NavyBlue}{\mathbf{w}}}\right\rangle\mathbb{X}_{t}^{\mathcolor{NavyBlue}{\mathbf{w}}}.$
(C.1)

Moreover, by bilinearity of the shuffle product and for $\bm{p},\bm{\ell}\in
T((\mathbb{R}^{d}))$,

$\displaystyle\langle\bm{p}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell},{\mathcolor{NavyBlue}{\mathbf{w}}}\rangle$
$\displaystyle=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\sum_{{\mathcolor{NavyBlue}{\mathbf{u}}}\in
V_{n}}\sum_{{\mathcolor{NavyBlue}{\mathbf{v}}}\in
V_{m}}\bm{p}^{\mathcolor{NavyBlue}{\mathbf{u}}}\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{v}}}\langle{\mathcolor{NavyBlue}{\mathbf{u}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}{\mathcolor{NavyBlue}{\mathbf{v}}},{\mathcolor{NavyBlue}{\mathbf{w}}}\rangle=\sum_{n=0}^{k}\sum_{{\mathcolor{NavyBlue}{\mathbf{u}}}\in
V_{n}}\sum_{{\mathcolor{NavyBlue}{\mathbf{v}}}\in
V_{k-n}}\bm{p}^{\mathcolor{NavyBlue}{\mathbf{u}}}\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{v}}}\langle{\mathcolor{NavyBlue}{\mathbf{u}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}{\mathcolor{NavyBlue}{\mathbf{v}}},{\mathcolor{NavyBlue}{\mathbf{w}}}\rangle.$
(C.2)

Hence, it is easy to see that for all $k\in\mathbb{N}$

$\displaystyle\sum_{{\mathcolor{NavyBlue}{\mathbf{w}}}\in
V_{k}}\left\langle\bm{p}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell},{\mathcolor{NavyBlue}{\mathbf{w}}}\right\rangle\mathbb{X}_{t}^{\mathcolor{NavyBlue}{\mathbf{w}}}=\sum_{n=0}^{k}\sum_{{\mathcolor{NavyBlue}{\mathbf{u}}}\in
V_{n}}\sum_{{\mathcolor{NavyBlue}{\mathbf{v}}}\in
V_{k-n}}\bm{p}^{\mathcolor{NavyBlue}{\mathbf{u}}}\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{v}}}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}{\mathcolor{NavyBlue}{\mathbf{v}}},\mathbb{X}_{t}\right\rangle.$
(C.3)

It then follows that the semi-norm of
$\bm{p}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}$ is of the form

$\displaystyle\left|\left|\bm{p}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}\right|\right|_{t}^{\mathcal{A}}$
$\displaystyle=\sum_{k=0}^{\infty}\left|\sum_{{\mathcolor{NavyBlue}{\mathbf{w}}}\in
V_{k}}\left\langle\bm{p}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell},{\mathcolor{NavyBlue}{\mathbf{w}}}\right\rangle\mathbb{X}_{t}^{\mathcolor{NavyBlue}{\mathbf{w}}}\right|$
$\displaystyle=\sum_{k=0}^{\infty}\left|\sum_{n=0}^{k}\sum_{{\mathcolor{NavyBlue}{\mathbf{u}}}\in
V_{n}}\sum_{{\mathcolor{NavyBlue}{\mathbf{v}}}\in
V_{k-n}}\bm{p}^{\mathcolor{NavyBlue}{\mathbf{u}}}\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{v}}}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}{\mathcolor{NavyBlue}{\mathbf{v}}},\mathbb{X}_{t}\right\rangle\right|$
$\displaystyle\leq\sum_{k=0}^{\infty}\sum_{n=0}^{k}\left|\sum_{{\mathcolor{NavyBlue}{\mathbf{u}}}\in
V_{n}}\sum_{{\mathcolor{NavyBlue}{\mathbf{v}}}\in
V_{k-n}}\bm{p}^{\mathcolor{NavyBlue}{\mathbf{u}}}\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{v}}}\left\langle{\mathcolor{NavyBlue}{\mathbf{u}}}\mathrel{\sqcup\mkern-3.0mu\sqcup}{\mathcolor{NavyBlue}{\mathbf{v}}},\mathbb{X}_{t}\right\rangle\right|.$

Finally, using the shuffle product property on the right-hand side, recall
Lemma 2.5, the semi-norm can be bounded

$\displaystyle\left|\left|\bm{p}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell}\right|\right|_{t}^{\mathcal{A}}$
$\displaystyle\leq\sum_{k=0}^{\infty}\sum_{n=0}^{k}\left|\sum_{{\mathcolor{NavyBlue}{\mathbf{u}}}\in
V_{n}}\sum_{{\mathcolor{NavyBlue}{\mathbf{v}}}\in
V_{k-n}}\bm{p}^{\mathcolor{NavyBlue}{\mathbf{u}}}\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{v}}}\mathbb{X}_{t}^{\mathcolor{NavyBlue}{\mathbf{u}}}\mathbb{X}_{t}^{\mathcolor{NavyBlue}{\mathbf{v}}}\right|\leq\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\left|\sum_{{\mathcolor{NavyBlue}{\mathbf{u}}}\in
V_{n}}\bm{p}^{\mathcolor{NavyBlue}{\mathbf{u}}}\mathbb{X}_{t}^{\mathcolor{NavyBlue}{\mathbf{u}}}\sum_{{\mathcolor{NavyBlue}{\mathbf{v}}}\in
V_{m}}\bm{\ell}^{\mathcolor{NavyBlue}{\mathbf{v}}}\mathbb{X}_{t}^{\mathcolor{NavyBlue}{\mathbf{v}}}\right|\leq\left|\left|\bm{p}\right|\right|_{t}^{\mathcal{A}}\cdot\left|\left|\bm{\ell}\right|\right|_{t}^{\mathcal{A}}<\infty.$

It can then be argued very similarly that
$\left\langle\bm{p},\mathbb{X}_{t}\right\rangle\left\langle\bm{\ell},\mathbb{X}_{t}\right\rangle=\left\langle\bm{p}\mathrel{\sqcup\mkern-3.0mu\sqcup}\bm{\ell},\mathbb{X}_{t}\right\rangle$,
ending the proof.

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