$\sup_{t \in [0, \infty)} \left|[(H^{(n)} - H) \cdot X, Y]_t \right| \overset{P}{\rightarrow} 0$

Question

1. Notation

We start with establishing some (standard, I think) notation.

Let $(\Omega, \mathcal{A}, P)$ be a given probability space.

For any filtration $\mathcal{G} = (\mathcal{G}_t)_{t \in [0,\infty)}$ of $\mathcal{A}$, denote by $CM^{\text{loc}}(\mathcal{G})$ the set of continuous local martingales $(Z_t)_{t \in [0,\infty)}$ w.r.t. the filtered probability space $(\Omega, \mathcal{A}, P; \mathcal{G})$, and denote by $CM^2(\mathcal{G})$ the set of continuous martingales $(Z_t)_{t \in [0, \infty)}$ w.r.t. said filtered probability space, such that $|||Z|||_2 := \sup_{t \in [0, \infty)} \|Z_t\|_2 = \sup_{t \in [0, \infty)} \sqrt{E(Z_t^2)} < \infty$.

Define $|||Z|||_\infty := \sup_{(t, \omega) \in [0, \infty) \times \Omega} |Z_t(\omega)|$ and denote by $T(\mathcal{G})$ the set of $\mathcal{G}$-adapted stochastic processes, whose paths are step-functions, and by $\overline{T}(\mathcal{G})$ the set of $\mathcal{G}$-adapted stochastic processes $H$, such that $\lim_{n \rightarrow \infty} |||H^{(n)} - H|||_\infty = 0$ for some sequence $H^{(1)}, H^{(2)}, \dots \in T(\mathcal{G})$.

Define $\mathcal{G}_\infty := \sigma\left(\bigcup_{t \in [0, \infty)} \mathcal{G}_t\right)$ and denote by $\overline{\mathcal{G}} = (\overline{\mathcal{G}}_t)_{t \in [0, \infty)}$ the completion of the filtration $\mathcal{G}$, i.e. $\overline{\mathcal{G}}_t := \sigma(\mathcal{G}_t \cup \{B \in \mathcal{G}_\infty \mid: P(B) = 0\})$.

Let $\mathcal{F} = (\mathcal{F}_t)_{t \in [0, \infty)}$ be some given filtration of the space $(\Omega, \mathcal{A}, P)$.

If $X, Y \in CM^{\text{loc}}(\mathcal{F})$, we denote by $X \cdot Y = (X \cdot Y)_{t \in [0, \infty)}$ the stochastic integral $$ (X \cdot Y)_t = \int_0^t X\ dY $$

We define the compensator of $X$ and $Y$ by $[X, Y]$, thus: $$ [X, Y] := XY - X_0Y_0 - X \cdot Y - Y \cdot X $$ and we define the quadratic variation of $X$ thus: $$ [X] := [X, X] $$

It can be shown that $[X, Y]$ is a continuous, $\overline{\mathcal{F}}$-adapted stochastic process with $[X, Y]_0 = 0$, and that $[X]$'s paths are non-decreasing. We may therefore define $$ [X]_\infty := \lim_{t \rightarrow \infty} [X]_t $$

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2. Statement of the problem

The following claim ($(*)$ below) came up in a proof of the so called "fundamental identity of stochastic integration" (i.e. "If $H \in \overline{T}(\mathcal{F})$ and if $X, Y \in CM^{\text{loc}}(\mathcal{F})$, then $[H \cdot X, Y] = H \cdot [X, Y]$.") found in Jochen Wengenroth's German textbook on stochastic processes, "Wahrscheinlichkeitstheorie", de Gruyter 2008, on p. 184.

Let $X, Y \in CM^2(\mathcal{F})$ be such that $X_0 = Y_0 = 0$. Let $H \in \overline{T}(\mathcal{F})$ and let $H^{(1)}, H^{(2)}, \dots \in T(\mathcal{F})$ be such that $\lim_{n \rightarrow \infty} |||H^{(n)} - H|||_\infty = 0$.

It is known that under these circumstances the sequence of stochastic integrals $(H^{(n)} \cdot X)_{n \in \{1, 2, \dots\}}$ converges in $CM^2(\overline{\mathcal{F}})$ (to $H \cdot X$, by definition).

Then

$$ \sup_{t \in [0, \infty)} \left|[(H^{(n)} - H) \cdot X, Y]_t \right| \overset{P}{\rightarrow} 0 \tag{*} $$

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3. Hints

Wengenroth suggests that $(*)$ can be justified using the following two facts, proved earlier in the book.

1. Whenever $X^{(1)}, X^{(2)}, \dots \in CM^{\text{loc}}(\mathcal{F})$, we have: $$ [X^{(n)}]_\infty \overset{P}{\rightarrow} 0 \iff \hat{X}^{(n)} := \sup_{t \in [0, \infty)} \left|X^{(n)}_t - X^{(n)}_0\right| \overset{P}{\rightarrow} 0 $$

2. The polarization identity: Whenever $X, Y \in CM^{\text{loc}}(\mathcal{F})$ we have $$ [X, Y] = \frac{1}{4} \left([X + Y] - [X - Y]\right) $$

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4. My attempts at a solution

I'm aware that it is frowned upon in this forum to present a question without indicating the attempts made at solving it. Unfortunately, I'm at a loss and have no clue how to go about solving this problem. If any of you would be kind enough to point me in the right direction, I'll make an attempt to solve the problem based on your suggestions, and, if successful, I'll publish my results in an answer below. Alternatively, if you wish to provide a detailed answer to this problem, I'll be very grateful.

Thanks.

Answer 1

Recall the following well-known theorem:

Let $B$ be a bilinear form (on some $\mathbb{R}$-vector space $X$) such that $B(x,y) = B(y,x)$ and $B(x,x) \geq 0$ for any $x,y \in X$. Then $$|B(x,y)| \leq \sqrt{B(x,x)} \sqrt{B(y,y)} \qquad \text{for all $x,y \in X$}.$$

Applying this to the bilinear form $B(X,Y) := [X,Y]_t$ gives

$$|[X,Y]_t| \leq \sqrt{[X]_t} \sqrt{[Y]_t}$$

for any two continous local martingales $X,Y$. This implies

$$\sup_t |[(H^n-H) \cdot X,Y]_t| \leq \sqrt{[(H^n-H) \cdot X]_{\infty}} \sqrt{[Y]_{\infty}} \tag{1} $$

On the other hand, it is well-known that

$$[(H^n-H) \cdot X]_{\infty} = \int_0^{\infty} (H^n-H)^2(s) \, d[X]_s. \tag{2}$$

Consequently,

$$\sup_t |[(H^n-H) \cdot X,Y]_t| \leq \sqrt{\|H^n-H\|^2_{\infty} [X]_{\infty}} \sqrt{[Y]_{\infty}}.$$

Since $\|H^n-H\|_{\infty} \to 0$ by assumption, the claim follows.

Edit: (Alternative argumentation which does not require $(2)$) By $(1)$, it suffices to show that

$$[(H_n-H) \cdot X]_{\infty} \stackrel{P}{\to} 0. $$

Using the second hint (mentioned in the OP), we find that this is equivalent to

$$\sup_{t \in [0,\infty)} |(H_n-H) \cdot X_t| \stackrel{P}{\to} 0. \tag{3}$$

In order to prove $(3)$, we note that, by Doob's inequality,

$$\mathbb{E} \left( \sup_{t \leq T} |(H_n-H) \cdot X_t|^2 \right) \leq 4 \sup_{t \leq T} \mathbb{E}(|(H_n-H) \cdot X_t|^2)$$

and therefore

$$\mathbb{E} \left( \sup_{t \geq 0} |(H_n-H) \cdot X_t|^2 \right) \leq 4 \|\|(H_n-H) \cdot X\|\|_2^2.$$

As $H^n \cdot X$ converges in $CM^2(\bar{\mathcal{F}})$ to $H \cdot X$, we get

$$\lim_{n \to \infty} \mathbb{E} \left( \sup_{t \geq 0} |(H_n-H) \cdot X_t|^2 \right) = 0.$$

Since $L^2$-convergence implies convergence in probability, $(3)$ follows.