Evalute integral of Brownian motion $E[(\int_0^tB_sds)^2]$

Question

Can someone explain me how to compute this expectation?

$E[(\int_0^tB_sds)^2]$

I know that the result is $t^3/3$

Answer 1

$$E(\int_0^{t}B_sds)^{2}=E(\int_0^{t}\int_0^{t} B_sB_ududs)$$ $$=(\int_0^{t}\int_0^{t} EB_sB_ududs)=E(\int_0^{t}\int_0^{t} (s \wedge u) duds).$$ Now split the integral w.r.t $u$ into integral from $0$ to $s$ and the integral from $s$ to $t$. Can you finish?

Answer 2

Rewrite $\int_0^tB_sds=tB_t -\int_0^tsdB_s$ to evaluate

$\space\space \space \space E[(\int_0^tB_sds)^2]$

$=E[t^2B^2_t]-2E[tB_t\int_0^tsdB_s]+E[(\int_0^tsdB_s)^2]$

$=t^2E[B^2_t]-2tE[\int_0^tdB_{\tau}\int_0^tsdB_s]+E[\int_0^t \tau dB_{\tau}\int_0^tsdB_s]$

$=t^3\cdot t-2t\int_0^tsds+\int_0^t s^2ds$

$=\frac13t^3$