Calculate variance of stochastic integral $ E[(\int_t^T sW_s ds)^2]$

Question

Let $W_t$ be a Brownian motion, calculate

$$ E\left[\left(\int_t^T sW_s ds\right)^2\right]$$ I tried to use Ito' Lemma and break the integral but it didn't help. Do anyone have any ideas how to solve it? Thanks in advance.

Answer 1

Use Ito's rule $d(\frac12s^2dW_s)=\frac12s^2dW_s+sW_sds$ to express $$ \int_t^T sW_s ds = \frac12\left(T^2W_T-t^2W_t-\int_t^Ts^2dW_s\right)$$

Then, its expectation is

$$ I=E_t\left[\left(\int_t^T sW_s ds\right)^2\right]=\frac14 (I_1+I_2-2I_3)\tag{1}$$

where,

$$I_1=E[(T^2W_T-t^2W_t)^2]=E[((T^2-t^2)W_t+ T^2(W_T-W_t))^2]$$ $$=(T^2-t^2)^2W_t^2 +T^4(T-t) \tag{2}$$

$$I_2=E\left[\left(\int_t^Ts^2dW_s\right)^2\right] =\int_t^Ts^4ds=\frac15(T^5-t^5)\tag{3}$$

$$I_3=E\left[(T^2W_T-t^2W_t) \int_t^Ts^2dW_s\right] =T^2E\left[(W_T-W_t)\int_t^Ts^2dW_s\right]$$ $$=T^2\int_t^Ts^2ds=\frac13T^2(T^3-t^3)\tag{4}$$

Plug (2), (3) and (4) into (1) to obtain the variance,

$$ I=\frac14(T^2-t^2)^2W_t^2 +\frac14T^4(T-t)+ \frac1{20}(T^5-t^5)-\frac1{6}T^2(T^3-t^3)$$