# Itô integral of an elementary process

Let

• $(\Omega,\mathcal{A},\operatorname{P})$ be a probability space
• $\mathbb{F}=(\mathcal{F}_t,t\ge 0)$ be a filtration on $(\Omega,\mathcal{A})$
• $H=(H_t,t\ge 0)$ be a stochastic process on $(\Omega,\mathcal{A})$ with $$H_t(\omega)=\sum_{i=1}^nh_{i-1}(\omega)1_{(t_{i-1},t_i]}(t)$$ where $0=t_0<\ldots<t_n$ and $h_{i-1}$ is bounded and $\mathcal{F}_{t_{i-1}}$-measurable
• $\tau$ be a bounded $\mathbb{F}$-stopping time
• $(W_t,t\ge 0)$ be a Brownian motion on $(\Omega,\mathcal{A},\operatorname{P})$ and $$I^W_t(H):=\sum_{i=1}^nh_{i-1}\left(W_{t_i\wedge t}-W_{t_{i-1}\wedge t}\right)$$ where $x\wedge y:=\min(x,y)$ for all $x,y\in\mathbb{R}$

By definition, we've got $$\operatorname{E}\left[I_\tau^W(H)\right]=\sum_{i=1}^n\operatorname{E}\left[h_{i-1}\left(W_{t_i}^\tau-W_{t_{i-1}}^\tau\right)\right]\tag{1}$$ where $W_t^\tau:=W_{\tau\wedge t}$ is the so called stopped process. By definition of the conditional expectation, $$\operatorname{E}\left[W_{t_i}^\tau-W_{t_{i-1}}^\tau\right]=\operatorname{E}\left[\operatorname{E}\left[W_{t_i}^\tau-W_{t_{i-1}}^\tau\mid\mathcal{F}_{t_{i-1}}\right]\right]$$

However, I don't understand why we've got $$\sum_{i=1}^n\operatorname{E}\left[h_{i-1}\left(W_{t_i}^\tau-W_{t_{i-1}}^\tau\right)\right]=\sum_{i=1}^n\operatorname{E}\left[h_{i-1}\operatorname{E}\left[W_{t_i}^\tau-W_{t_{i-1}}^\tau\mid\mathcal{F}_{t_{i-1}}\right]\right]$$ That would be true iff $W_{t_i}^\tau-W_{t_{i-1}}^\tau$ and $h_{i-1}$ are uncorrelated; but why is that the case?

the answer to this is quite simple if you look carefully to the definition of H, you should notice that $h_{i-1}$ is an $\mathcal{F}_{t_{i-1}}$-measurable random variable. From this and the following elementary properties on conditional expectation we get the result :

If X is $\mathcal{F}$-measurable and for any bounded random variable $Y$, we have a.s. :

$$1-E[X.Y|\mathcal{F}]=X.E[Y|\mathcal{F}]$$

Property 1 allows you to get :

$$E\left[h_{i-1}.(W_{t_i}^\tau-W_{t_{i-1}}^\tau)\mid\mathcal{F}_{t_{i-1}}\right]= h_{i-1}E\left[(W_{t_i}^\tau-W_{t_{i-1}}^\tau)\mid\mathcal{F}_{t_{i-1}}\right]$$

Now let's remind the defining property of conditional expectation of a integrable random variable with respect to a $\sigma$-algebra $\mathcal{F}$ which is the only (modulo a.s. equivalence) $\mathcal{F}$-measurable random variable such that we have for any bounded $\mathcal{F}$-measurable random variable $Y$ the following property :
$$2-E[Y.X]=E[Y.E[X|\mathcal{F}]]$$

In our situation we are done because $h_{i-1}$ is bounded $\mathcal{F}_{t_{i-1}}$-meausarble and $(W_{t_i}^\tau-W_{t_{i-1}}^\tau)$ is an integrable random variable so we get by property 2 :

$$E[h_{i-1}E\left[(W_{t_i}^\tau-W_{t_{i-1}}^\tau)\mid\mathcal{F}_{t_{i-1}}\right]]=E[h_{i-1}.(W_{t_i}^\tau-W_{t_{i-1}}^\tau)]$$

Best regards