Finding $E[W]$ and $E[W^2]$, where $W = \int_{t=0}^T B_s$ $ds$

Question

I'm trying to find a)$E[W]$ and b) $E[W^2]$, where $W_t = \int_{t=0}^T B_s$ $ds$ ($B_s$ denotes a Brownian motion).

In addition, I'd like to find $E[Z_sZ_t]$, where $Z_t = \int_0^t B_s^2$$ ds$

$E[W] = 0$ since $B_s$ is $\mathcal{N}(0,s)$

$E[W^2] = \int_0^t \int_0^t E[W_sW_t] ds$ $dt $

$= \int_0^t \int_0^t $min$(s,t)ds$ $dt =\int_0^t\int_0^t sdsdt + \int_0^t \int_0^t tdsdt = \int_0^t \frac{t^2}{2}dt + \int_0^t (t-t^2)dt = \frac{t^3}{3}$

As for finding $E[Z_sZ_t]$, I'm stuck.

Answer 1

You can turn $Z_s Z_t$ into a double integral just like you did with $W^2$:

$$Z_s Z_t = \int_0^s \int_0^t B_u^2 B_r^2 dr du.$$

After a Fubini theorem, you can follow the approach in Distribution of $\| W_t \|^2_{L^2([0,T])}$, you get the expectation of the integrand, and then the rest of the problem is calculus.