Variation processes and strong solutions of stochastic differential equations Let


*

*$(\Omega,\mathcal{A},\operatorname{P})$ be a probability space

*$\mathbb{F}$ be a filtration on $(\Omega,\mathcal{A})$

*$\tau$ be a $\mathbb{F}$-stopping time


An $\mathbb{F}$-adapted, real-valued stochastic process $M=(M_t)_{t\ge 0}$ on $(\Omega,\mathcal{A},\operatorname{P})$ is called local $\mathbb{F}$-martingale until $\tau$ $:\Leftrightarrow$ There exists a sequence $(\tau_n)_{n\in\mathbb{N}}$ of $\mathbb{F}$-stopping times such that $$\tau_n\uparrow\tau\;\;\;\operatorname{P}\text{-almost surely}$$ and the stopped process $M^{\tau_n}:=(M_{\tau_n\wedge t})_{t\ge 0}$ is an uniformly integrable $\mathbb{F}$-martingale.

Now, let $t\mapsto H(t)=(H_{ij}(t))_{\stackrel{i=1,\ldots,n}{j=1,\ldots,m}}$ be $\mathbb{F}$-progressively measurable, $\left\|\;\cdot\;\right\|$ be the Frobenius norm, $B=(B^1,\ldots,B^m)$ be a $m$-dimensional Brownian motion and $$\operatorname{E}\left[\int_0^TH_{ij}^2\;dt\right]<\infty\;.$$ I want to show, that $$\operatorname{E}\left[\left\|\int_0^TH(t)\;dB_t\right\|^2\right]=\operatorname{E}\left[\int_0^T\left\|H(t)\right\|^2\;dt\right]\;.$$ Therefore, let $$I_i(t):=\sum_{j=1}^m\int_0^tH_{ij}(s)\;dW_s^j\;,$$ for $i\in\left\{1,\ldots,n\right\}$. By fundamental facts about the Itô integral, one knows, that each $I_i$ is a continuous martingale. However, I absolutely don't get, why the variation process of $I_i$ is given by $$\langle I_i\rangle_t:=\int_0^t\sum_{j=1}^mH_{ij}^2(s)\;ds$$ and why that implies $$\operatorname{E}\left[\left(I_i(T)\right)^2\right]=\operatorname{E}\left[\int_0^T\sum_{j=1}^mH_{ij}^2(s)\;ds\right]$$
 A: This answer takes a while. We will need several steps which I list here to guide our lines:
1)($Z$ simple) $I^M_t(Z)^2 - \int_0^t Z^2(s) \, d\langle M\rangle_s$ is a continuous local martingale
2) ($X$ $\Bbb{F}$- progressively measurable) $I^M_t(X)^2 - \int_0^t X^2(s) \, d\langle M\rangle_s$ is a continuous local martingale
3) $\langle I^M(X) + I^N(Y)\rangle_t = \langle I^M(X) \rangle_t + \langle I^N(Y)\rangle_t + 2 \langle I^M(X) , I^N(Y)\rangle_t$
4) $\langle I^M(X), I^N(Y)\rangle_t = \int_0^t X_uY_u \, d\langle M,N \rangle_u$
5) note that $I_i(H) = \sum_j I^{W_j}(H_{i,j})$ 
6) $$\langle I_i(H)\rangle = \langle \sum_j I^{W_j}(H_{i,j})\rangle = \sum_{j,k} \langle  I^{W_j}(H_{i,j}),I^{W_j}(H_{i,k}) \rangle \\
 = \sum_{j} \langle  I^{W_j}(H_{i,j}),I^{W_j}(H_{i,j}) \rangle = \int_0^t\sum_{j} H_{i,j}^2(u) d \langle W_j \rangle_u$$
each of this steps is classical and can be found in (Karatazas and Shreve- Brownian motion and stochastic calculus second edition), so I will give you the pages where you may find each result
1) pgs 132 and 137, 138
2) pg 139 
3) pg 32 
4) pg 142
5) just use the definition of $I_t^M(X) = \int_0^t X_s \,d M_s$
6) pg 32
