Expectation of product of stochastic integral and brownian motion

Question

Find the covariance:

$$ COV((\int_t^T(T-s)dW_s), W_t) $$

I used the covariance formula: COV(X,Y) = E(XY) - E(X)E(Y) = E(XY) as E(X)=E(Y)=0

But I am stuck on figuring out the expectation of the product:

$$ E(\int_t^T(T-s)W_tdW_s) $$

Cheers

Answer 1

Rewrite these two factors as $$\int_t^T(T-s)dW_s=\int_0^Tf(s)dW_s,\qquad W_t=\int_0^Tg(s)dW_s, $$ for some suitable functions $f$ and $g$, then use the general result that $$ E\left(\int_0^Tf(s)dW_s\cdot\int_0^Tg(s)dW_s\right)=E\left(\int_0^Tf(s)g(s)ds\right). $$ The answer should be $$ 0. $$ Of course, more direct approaches exist, for example the fact that $$\int_t^T(T-s)dW_s$$ depends only on the process $\overline W$ defined by $\overline W_s=W_{t+s}-W_t$ for every $s\geqslant0$ since it coincides with $$ \int_0^{T-t}(T-t-s)d\overline W_{s}. $$ This yields the stronger result that the two factors are in fact independent.