$E[e_te_s\Delta B_t\Delta B_s]$ for $\Delta B_t$ Brownian motion increments and $e_t(\omega)$ a measurable function.

Question

Let $\Delta B_j=B_{t_{j+1}}-B_{t_j}$ where $B_t$ is Brownian motion, and $e_i(\omega)$ measurable with respect to $\sigma(B_{t_i})$. In Oksendal's 'Stochastic Differential Equations' he states:

$$ E[e_ie_j\Delta B_i\Delta B_j]= \begin{cases} 0 & i\ne j \\ E[e_j^2] & i=j \end{cases} $$

Justifying this because '$e_ie_j\Delta B_i$ and $\Delta B_j$ are independent if $i<j$'. I am trying to understand this.

I understand that Brownian motion is normally distributed with independent increments, so: $$E[\Delta B_i\Delta B_j]=0$$ However, how can we justify $e_ie_j\Delta B_i$ being independent to $\Delta B_j$? Even if they are, how can we treat $E[e_ie_j\Delta B_i]$? Basically I believe my confusion stems from how to treat products of measurable functions.

Answer 1

For every $i$, let $F_i=\sigma(B_s,s\leqslant t_i)$ then each $e_i$ is $F_i$-measurable and each $\Delta B_i$ is centered, independent of $F_i$, and $F_{i+1}$-measurable.

Now, fix $i\lt j$. Then $e_i$, $e_j$ and $\Delta B_i$ are all $F_j$-measurable and $\Delta B_j$ is independent of $F_j$ hence $e_ie_j\Delta B_i$ and $\Delta B_j$ are independent, a fact which implies $$E(e_ie_j\Delta B_i\Delta B_j)=E(e_ie_j\Delta B_i)\cdot E(\Delta B_j)=E(e_ie_j\Delta B_i)\cdot0=0.$$ Thus, the argument does not require to know the value of $E(e_ie_j\Delta B_i)$.

Likewise, for every $i$, $e_i$ is $F_i$-measurable and $\Delta B_i$ is independent of $F_i$ hence $$E((e_i\Delta B_i)^2)=E(e_i^2)\cdot E((\Delta B_i)^2)=E(e_i^2)\cdot(t_{i+1}-t_i).$$

Answer 2

You can see it using the "tower property" for instance,

Assume $i<j$, then

$$E[e_i e_j \Delta B_i \Delta B_j] = E\left[ E[e_i e_j \Delta B_i \Delta B_j |\mathcal{F}_{t_j}]\right].$$

Now observe that since $i<j$ then $\mathcal{F}_{t_i}\subset \mathcal{F}_{t_j}$ and by hypothesis $e_i$ is $\mathcal{F}_{t_i}$-measurable and hence also $\mathcal{F}_{t_j}$-measurable, also $\Delta B_i = B_{t_{i+1}}-B_{t_i}$ is $\mathcal{F}_{t_j}$-measurable. Hence $$E[e_i e_j \Delta B_i \Delta B_j] = E\left[ e_i e_j \Delta B_i E[\Delta B_j |\mathcal{F}_{t_j}]\right].$$

Finally observe that $\Delta B_j$ is independent of $\mathcal{F}_{t_j}$ so $E[\Delta B_j |\mathcal{F}_{t_j}]=E[\Delta B_j]=0$ and hence

$$E[e_i e_j \Delta B_i \Delta B_j] = 0.$$

This was maybe very detailed but useful to understand why.