$E[W_{t/3}W_{t/2} \mid \mathcal{F}_{t/5}]$ where W is a Brownian Motion and $\mathcal{F}$ is the natural filtration?

Question

I am unsure how to go about finding this value.

$\mathrm{E}[W(t/3)*W(t/2)|$ $\mathrm{F}(t/5)]$

I assume the trick invovles an additional conditional expectation, but I am not sure how to go about it. Here we assume that the process W is adapted to the natural filtration ($\mathrm{F}(t)$ being the sigma algebra generated by W(s) for all 0 <= s <= t)

Answer 1

Hint Since $(W_t)_{t \geq 0}$ is a martingale, we find by the tower property

$$\begin{align*} \mathbb{E} \big( W_{t/3} \cdot W_{t/2} \mid \mathcal{F}_{t/5} \big) &=\mathbb{E} \big( \mathbb{E}(W_{t/3} \cdot W_{t/2} \mid \mathcal{F}_{t/3}) \mid \mathcal{F}_{t/5} \big) =\mathbb{E}\big(W_{t/3}^2 \mid \mathcal{F}_{t/5} \big) \end{align*}$$

Now write $$W_{t/3} = \big(W_{t/3}-W_{t/5} \big)+W_{t/5}$$ and use the independence of the increments.

Answer 2

Hint: Let $t\leqslant s\leqslant r$.

• First, $W_r=W_s+(W_r-W_s)$ with $W_r-W_s$ centered independent of $F_s$ hence $E[W_rW_s\mid F_s]=W_s^2$.
• Second, $W_s^2=W_t^2+2W_t(W_s-W_t)+(W_s-W_t)^2$ and $W_s-W_t$ is centered independent of $F_t$ with $E[(W_s-W_t)^2]=s-t$ hence $E[W_s^2\mid F_t]=W_t^2+s-t$.