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It is known that stochastic integral must satisfy the isometry property which is $$ \mathbb{E}\left[ \left( \int_0^T X_t~dB_t\right)^2 \right] = \mathbb{E} \left[ \int_0^T X^2_t~dt \right] . $$ I am trying to prove this property for a simple stochastic process. What I said so far that is $$ \mathbb{E}\left[\sum_{i=0}^{n-1} X_i \left(B(t_{i+1})-B(t_i)\right)\right]^2, $$ then I am stuck. I know that we should to write the square sum as double sum to continue the proof but I couldn't do it. Any help please!

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    $\begingroup$ I don't understand the sentence after "What I said so far is..." $\endgroup$ Jul 4, 2012 at 13:12

2 Answers 2

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I take it that a simple process is a process of the form $$ Y_t=\sum_{i=0}^{n-1} X_i 1_{]t_i,t_{i+1}]}(t),\quad t\geq 0, $$ where $0\leq t_1<t_2<\cdots < t_n$ and $X_i$ is a bounded $\mathcal{F}_{t_i}$-measureable variable. Now we need the following result.

For every simple process $Y=(Y_t)_{t\geq 0}$ the process
$$ \left(\left(\int_0^t Y_s\;\mathrm{d}B_s\right)^2 -\int_0^t Y_s^2\;\mathrm{d}s\right)_{t\geq 0} $$ is a martingale which is $0$ at $0$.

By taking expectation this shows the desired property.

If you don't know the above result you can just prove it. Here it is useful to note that when $Y=(Y_t)_{t\geq 0}$ is a simple process, then $Y^2=(Y_t^2)_{t\geq 0}$ is also a simple process which satisfies $$ Y_t^2=\sum_{i=0}^{n-1} X_i^2 1_{]t_{i},t_{i+1}]},\quad t\geq 0. $$

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A simple process $\Phi = (\Phi_t)_{t\geq 0}\in\mathcal{E}$ is a process of the form $$\Phi_t = \sum_{i=0}^{n-1}U_i1_{(t_i,t_{i+1}]}(t)$$ where $0=t_0<\dots<t_n<\infty$ and $U_i\in b\mathcal{F}_{t_i}$, i.e. $U_i$ is a bounded $\mathcal{F}_{t_i}$-measurable random variable. Since $t\mapsto \Phi_t$ is a step function, we get that the integral of $\Phi\in\mathcal{E}$ is given by:

\begin{align*} \int_0^\infty \Phi_s \mathrm{d}s &= \sum_{i=0}^{n-1} U_i \left(t_{i+1}-t_i\right)\\ \int_0^t \Phi_s \mathrm{d}s &= \sum_{i=0}^{n-1} U_i \left(\left(t_{i+1}\land t\right)-\left(t_i\land t\right)\right) \end{align*}

The stochastic integral of $\Phi\in\mathcal{E}$ with respect to the Brownian motion $B$ is defined by: $$\int_0^t \Phi_s \mathrm{d}B_s := \sum_{i=0}^{n-1} U_i \left(B_{t_{i+1}\land t} - B_{t_i\land t}\right)$$


Let $t\geq 0$ and let $\Phi\in\mathcal{E}$ be a simple process, i.e. $$\Phi_t = \sum_{i=0}^{n-1}U_i1_{(t_i,t_{i+1}]}(t)$$ where $0=t_0<\dots<t_n<\infty$ and $U_i\in b\mathcal{F}_{t_i}$. We start by calculating the integral of $\Phi^2$. Notice that the square of a summation is given by:

\begin{align*} \left(\sum_{i=0}^{m} Z_i\right)^2 = \left(\sum_{i=0}^{m} Z_i^2\right) + 2\left(\sum_{\substack{i,j=0\\i<j}}^{m} Z_iZ_j\right) \end{align*}

If $0\leq i < j \leq n-1$, then $(t_i,t_{i+1}]\cap(t_j,t_{j+1}]=\emptyset$. It follows that: \begin{align*} \Phi_t^2 = &\left(\sum_{i=0}^{n-1}U_i^2 1_{(t_i,t_{i+1}]}(t)^2\right) + 2\left(\sum_{\substack{i,j=0\\i<j}}^{n-1} U_iU_j 1_{(t_i,t_{i+1}]}(t) 1_{(t_j,t_{j+1}]}(t)\right)\\ = &\sum_{i=0}^{n-1}U_i^2 1_{(t_i,t_{i+1}]}(t)^2 = \sum_{i=0}^{n-1}U_i^2 1_{(t_i,t_{i+1}]}(t) \end{align*}

Since $t\mapsto \Phi_t^2$ is a step function, we get that the integral of $\Phi^2$ is given by:

\begin{align*} \int_0^t \Phi_s^2 \mathrm{d}s = \sum_{i=0}^{n-1} U_i^2 \left(\left(t_{i+1}\land t\right)-\left(t_i\land t\right)\right) \end{align*}

Now we are going to simplify the left hand side of Itô's isometry. From the definition of the stochastic integral follows:

\begin{align*} \mathbb{E}\left[\left(\int_{0}^{t}\Phi_s\mathrm{d}B_s\right)^2\right] = \mathbb{E}\left[\left(\sum_{i=0}^{n-1} U_i \left(B_{t_{i+1}\land t} - B_{t_i\land t}\right)\right)^2\right] \end{align*}

Recall the formula for computing the square of a summation. From $\mathbb{E}$ being linear follows:

\begin{align*} &\mathbb{E}\left[\left(\sum_{i=0}^{n-1} U_i \left(B_{t_{i+1}\land t} - B_{t_i\land t}\right)\right)^2\right]\\ =\ &\sum_{i=0}^{n-1} \mathbb{E}\left[U_i^2 \left(B_{t_{i+1}\land t} - B_{t_i\land t}\right)^2\right]\\ &+2\sum_{\substack{i,j=0\\i<j}}^{n-1} \mathbb{E}\left[U_iU_j\left(B_{t_{i+1}\land t} - B_{t_i\land t}\right)\left(B_{t_{j+1}\land t} - B_{t_j\land t}\right)\right] \end{align*}

Because of $0\leq i < j \leq n-1$ we have $i,i+1\leq j$ as well as $t_i, t_{i+1}\leq t_j$. It follows that $\mathcal{F}_{t_i},\mathcal{F}_{t_{i+1}}\subseteq \mathcal{F}_{t_j}$ and that $U_i$, $U_{j}$, $B_{t_{i}\land t}$, $B_{t_{i+1}\land t}$ are $\mathcal{F}_{t_j}$-measurable. Using the law of total expectation yields

\begin{align*} &\mathbb{E}\left[U_iU_j\left(B_{t_{i+1}\land t} - B_{t_i\land t}\right)\left(B_{t_{j+1}\land t} - B_{t_j\land t}\right)\right]\\ =\ &\mathbb{E}\left[\mathbb{E}\left[U_iU_j\left(B_{t_{i+1}\land t} - B_{t_i\land t}\right)\left(B_{t_{j+1}\land t} - B_{t_j\land t}\right)\vert \mathcal{F}_{t_j}\right]\right]\\ =\ &\mathbb{E}\left[U_iU_j\left(B_{t_{i+1}\land t} - B_{t_i\land t}\right)\mathbb{E}\left[ B_{t_{j+1}\land t} - B_{t_j\land t}\vert \mathcal{F}_{t_j}\right]\right]\\ =\ &\mathbb{E}\left[U_iU_j\left(B_{t_{i+1}\land t} - B_{t_i\land t}\right)0\right] = 0 \end{align*}

where we also used a fact about the expected value of a Brownian motion as well as:

\begin{align*} &\mathbb{E}\left[B_{t_{j+1}\land t} - B_{t_j\land t}\vert \mathcal{F}_{t_j}\right]\\ =\ &\begin{cases} \mathbb{E}\left[B_t - B_t\vert \mathcal{F}_{t_j}\right] = \mathbb{E}\left[0\vert \mathcal{F}_{t_j}\right]& (t \leq t_j < t_{j+1})\\ \mathbb{E}\left[ B_t - B_{t_j}\vert \mathcal{F}_{t_j}\right] & (t_j < t \leq t_{j+1})\\ \mathbb{E}\left[ B_{t_{j+1}} - B_{t_j}\vert \mathcal{F}_{t_j}\right] & (t_j < t_{j+1} < t)\\ \end{cases}\\ =\ &0 \end{align*}

Similiarly, by using that $U_i$ is $\mathcal{F}_{t_i}$-measurable together with the law of total expectation, we get

\begin{align*} &\mathbb{E}\left[U_i^2 \left(B_{t_{i+1}\land t} - B_{t_i\land t}\right)^2\right]\\ =\ &\mathbb{E}\left[\mathbb{E}\left[U_i^2 \left(B_{t_{i+1}\land t} - B_{t_i\land t}\right)^2\vert \mathcal{F}_{t_i}\right]\right]\\ =\ &\mathbb{E}\left[U_i^2\mathbb{E}\left[\left(B_{t_{i+1}\land t} - B_{t_i\land t}\right)^2\vert \mathcal{F}_{t_i}\right]\right]\\ =\ &\mathbb{E}\left[U_i^2\left(\left(t_{i+1}\land t\right)-\left(t_i\land t\right)\right)\right] \end{align*}

where we also used a fact about the expected value of a Brownian motion as well as:

\begin{align*} &\mathbb{E}\left[ \left(B_{t_{i+1}\land t} - B_{t_i\land t}\right)^2\vert \mathcal{F}_{t_i}\right]\\ =\ &\begin{cases} \mathbb{E}\left[\left(B_t - B_t\right)^2\vert \mathcal{F}_{t_i}\right] = \mathbb{E}\left[ 0 \vert \mathcal{F}_{t_i}\right] = 0 = t - t& (t \leq t_i < t_{i+1})\\ \mathbb{E}\left[\left(B_t - B_{t_i}\right)^2\vert \mathcal{F}_{t_i}\right] = t - t_i & (t_i < t \leq t_{i+1})\\ \mathbb{E}\left[\left(B_{t_{i+1}} - B_{t_i}\right)^2\vert \mathcal{F}_{t_i}\right] = t_{i+1} - t_i & (t_i < t_{i+1} < t)\\ \end{cases}\\ =\ & \left(t_{i+1}\land t\right)-\left(t_i\land t\right) \end{align*}

Putting everything together yields:

\begin{align*} &\mathbb{E}\left[\left(\int_{0}^{t}\Phi_s\mathrm{d}B_s\right)^2\right] = \sum_{i=0}^{n-1} \mathbb{E}\left[U_i^2\left(\left(t_{i+1}\land t\right)-\left(t_i\land t\right)\right)\right]\\ &= \mathbb{E}\left[\sum_{i=0}^{n-1} U_i^2\left(\left(t_{i+1}\land t\right)-\left(t_i\land t\right)\right)\right] = \mathbb{E}\left[\int_{0}^{t}\Phi_s^2\mathrm{d}s\right] \end{align*}

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