Let $(W_t)_t$ a Brownian motion w.r.t $(F_t)_t$. Compute the following quantities:
- $E[(W_t^2 - t)(W_s^2 - s)]$, for $0<s < t$
Solution:
- With the usual trick of separating the increments I need to compute:
$$E[W_s^2((W_t - W_s)+ W_s)^2] - E[W_s^2 t] - s E[W_t^2] + st $$
I immediately note that $$-tE[W_s^2] - sE[W_t^2] + st = -t s -st + st= -st $$
Now I need to compute the more difficult part,i.e. $$E[W_s^2((W_t - W_s)+ W_s)^2] = E[W_s^2(W_t - W_s)^2]+ E[W_s^4] + 2E[W_s^3]E[W_t - W_s] = E[W_s^2]E[(W_t - W_s)^2] + E[W_s^4] + 2 E[W_s^3]E[W_t - W_s] = \\s(t-s)+3s^2 + 2\cdot0 = \\3s^2+s(t-s)$$
where I used that $E[X^4] = 3 \sigma^4$ for a normal random variable.
All in all I have $$3s^2 - s(t-s) -st = 2s^2$$
Here is another approach, maybe a little shorter.
$I=E[(W_t^2 - t)(W_s^2 - s)]=E[(W_t^2 - t)W_s^2]=E[W_t^2W_s^2]-ts$
Use $dW_t^2=2W_tdW_t+dt$ to express $W_t^2 = W_s^2+2\int_s^tW_{\tau}d{W_\tau}+(t-s)$. Then,
$E[W_t^2W_s^2]=E[W_s^4]+2E[\int_s^tW_{\tau}d{W_\tau}]E[W_s^2] +(t-s)E[W_s^2] =3s^2+0+(t-s)s$
and
$I=3s^2+(t-s)s-ts=2s^2$