Variance of stochastic integral

Question

How do I compute covariance $cov(X_t,X_s)$ of the process $X_t=\int_0^t (t-u)W_udu$ where $W_i$ is a Brownian Motion?

I only know that $EX_t=\int_0^t (t-u)E(W_u)du=0$ because $EW_i=0$.

Answer 1

WLOG assume that t>s,

$W_u$ is integrable because it is uniformly bounded ($E[|W_u|]\leq E[W_u^2]^{\frac 12} = \sqrt u$ ), we can use Fubini: $E[X_t] = \mathbb E[\int_0^t (t-u) W_u du] = \int_0^t (t-u) \mathbb E[W_u] du = 0$

Therefore, $cov(X_t,X_s) = \mathbb E[X_tX_s] $

With the same argument we use Fubini here: $$\mathbb E[X_tX_s]=\int_0^t\int_0^s(t-u)(s-v)\mathbb E[W_uW_v]\,\mathrm dv\,\mathrm du=\int_0^t\int_0^s (t-u)(s-v) \min(u,v)\,\mathrm dv\,\mathrm du$$

(My previous answer was wrong I misread the problem :/)