$\mathbb{E}(\int_{\rho}^{\tau}X(s)dW(s)\cdot\int_{\rho}^{\tau}Y(s)dW(s)|F_{\rho})=\mathbb{E}(\int_{\rho}^{\tau}X(s)Y(s)ds|F_{\rho})$

Question

Let $X,Y\in M_{[0,T]}^2=\left\{f:[0,T]\times\Omega\to\mathbb{R}:\text{f is adapted}, E\left(\int_0^Tf^2(t)dt\right)<\infty\right\}$ and $\rho, \tau$ - stopping times such that $0\le\rho\le \tau\le T$. Prove that: $\mathbb{E}(\int_{\rho}^{\tau}X(s)dW(s)\cdot\int_{\rho}^{\tau}Y(s)dW(s)|F_{\rho})=\mathbb{E}(\int_{\rho}^{\tau}X(s)Y(s)ds|F_{\rho})$

Can anyone prove it?

Answer 1

First we need to prove the statement for fixed times $a<b$ ,i.e., we need to prove that

$$ \mathbb{E}\left[\int_a^b X_s \, dW_s \cdot \int_a^b Y_s \, dW_s \right] = \int_a^b \mathbb{E}[X_s Y_s] \, dW_s$$ where $X=\{X_s : s \geq 0\}$ and $Y=\{Y_s: s \geq 0\}$ are Itô's diffusions.

Let $0=t_0 \leq t_1 \leq t_2 \leq \dots t_n=t$ be a partition of the interval $[0,t]$ and consider the following random variable $$A_n=\sum_{i=1}^{n}X_{t_{i-1}}\Delta W_{t_i}, \ B_n = \sum_{i=1}^{n}Y_{t_{i-1}}\Delta W_{t_i}, $$ with $$\Delta W_{t_i} = W_{t_i}-W_{t_{i-1}} \sim \mathcal{N}(0,t_i-t_{i-1})$$ and we know from the definition of Itô's integral that $\int_a^b X_s \, dW_s, \int_a^b Y_s \, dW_s$ are the random variables to which $A_n$ and $B_n$ converge in $L^2$ (respectively).

\begin{equation*} \begin{split} A_nB_n & =\sum_{i=1}^{n}X_{t_{i-1}}\Delta W_{t_i}\sum_{i=1}^{n}Y_{t_{i-1}}\Delta W_{t_i} \\ & = \sum_{i=1}^{n}X_{t_{i-1}}Y_{t_{i-1}}\Delta W_{t_i}^2 + \sum_{i\neq j}^{n}X_{t_{i-1}}Y_{t_{j-1}}\Delta W_{t_i}\Delta W_{t_j} \end{split} \end{equation*}

Note that because of the independent increments, for $i \neq j$, $\Delta W_{t_i}$ is independent of $\Delta W_{t_j}$. Also, since $X_{t_{i-1}} \in \mathcal{F}_{t_{i-1}}, \ Y_{t_{j-1}} \in \mathcal{F}_{t_{j-1}}$, it follows that \begin{equation*} \begin{split} \mathbb{E}\left[ \sum_{i\neq j}^{n}X_{t_{i-1}}Y_{t_{i-1}}\Delta W_{t_i}\Delta W_{t_j} \right] & = \sum_{i\neq j}^{n}\mathbb{E}\left[X_{t_{i-1}}Y_{t_{j-1}}\Delta W_{t_i}\Delta W_{t_j} \right] \\ & = \sum_{i\neq j}^{n}\mathbb{E}\left[X_{t_{i-1}}Y_{t_{j-1}}\right]\mathbb{E}\left[\Delta W_{t_i}\Delta W_{t_j} \right] \\ & = \sum_{i\neq j}^{n}\mathbb{E}\left[X_{t_{i-1}}Y_{t_{j-1}}\right]\mathbb{E}\left[\Delta W_{t_i}\right]\mathbb{E}\left[\Delta W_{t_j} \right] \\ & = 0 \\ \mathbb{E}\left[ \sum_{i=1}^{n}X_{t_{i-1}}Y_{t_{i-1}}\Delta W_{t_i}^2 \right] & = \sum_{i=1}^{n}\mathbb{E}\left[X_{t_{i-1}}Y_{t_{i-1}}\Delta W_{t_i}^2 \right] \\ & = \mathbb{E}\left[ \sum_{i=1}^{n}X_{t_{i-1}}Y_{t_{i-1}}\Delta W_{t_i}^2 \right] = \\ & = \sum_{i=1}^{n}\mathbb{E}\left[X_{t_{i-1}}Y_{t_{i-1}}\right]\mathbb{E}\left[\Delta W_{t_i}^2 \right] \\ & = \sum_{i=1}^{n}\mathbb{E}\left[X_{t_{i-1}}Y_{t_{i-1}}\right] (t_i-t_{i-1}) \end{split} \end{equation*} and consequently $$\mathbb{E}[A_nB_n]=\sum_{i=1}^{n}\mathbb{E}\left[X_{t_{i-1}}Y_{t_{i-1}}\right] (t_i-t_{i-1})$$ Thus (notice that $A_nB_n$ is a Riemann sum) $$\mathbb{E}[\lim_{n\to\infty}A_nB_n]\stackrel{L^2}{=}\int_a^b\mathbb{E}[X_s Y_s] \, ds$$ we finnaly get that \begin{equation*} \begin{split} \mathbb{E}\left[\int_a^b X_s \, dW_s \cdot \int_a^b Y_s \, dW_s \right] & = \int_a^b \mathbb{E}[X_s Y_s] \, dW_s \end{split} \end{equation*} Conditioning on $\mathbb{F}_a$, we have (also because of the independent increments) \begin{equation*} \begin{split} \mathbb{E}\left[\int_a^b X_s \, dW_s \cdot \int_a^b Y_s \, dW_s \Big| \mathcal{F}_a\right] & = \int_a^b \mathbb{E}[X_s Y_s] \, dW_s \end{split} \end{equation*} Now in order to prove this to stopping times is a little bit more complicated... The basic idea, is to discretize the stopping times $\rho$ and $\tau$ in a similar way to this Strong Markov property of Brownian motion and use the result proven here.

Answer 2

Another way you can prove the first identity is by leting $$A_t=\int_a^b X_s dW_s$$ $$B_t=\int_a^b Y_s dW_s$$ and since $A_t B_t = \frac12 (A_t+B_t)^2-\frac12 A_t^2-\frac12 B_t^2$ we get $$\mathbb{E}[A_t B_t]=\frac12 \mathbb{E}[(A_t + B_t)^2]-\frac12 \mathbb{E}[A_t^2]-\frac12 \mathbb{E}[B_t^2]$$ and recall Ito isomtry which says that $$\mathbb{E}[(\int_a^b Z_s dW_s)^2]=\int_a^b \mathbb{E}[Z_s^2] ds$$ and thus we get $$\mathbb{E}[A_t B_t]=\frac12 \int_a^b \mathbb{E}[(X_s+Y_s)^2] ds -\frac12 \int_a^b\mathbb{E}[X_s^2] ds-\frac12 \int_a^b\mathbb{E}[Y_s^2] ds = \int_a^b \mathbb{E}[X_s Y_s] ds$$