# Brownian motion and stochastic integration

How do I compute the following expectation?

$W$ is a standard Brownian motion (i.e.) $W(T)\sim N(0,T)$:

$E\left[ W(T)\int _{ 0 }^{ T }{ sdW(s) } \right]$.

I know that Brownian motion of disjoint time intervals are independent. Does this means that,

$E\left[ \left( W\left( T \right) -W\left( t \right) \right) \int _{ 0 }^{ t }{ sdW(s) } \right] =E\left[ \left( W\left( T \right) -W\left( t \right) \right) \right] E\left[ \int _{ 0 }^{ t }{ sdW(s) } \right] =0$

However, when I try to let $W(T) = W(T) - W(t) + W(t)$, the whole expectation becomes more complicated.

• Do you know ito's product rule and the condition for stochastic integrals to be martingales? – Henrik Nov 7 '13 at 10:42
• Just briefly... but how does it help? – user106113 Nov 7 '13 at 11:16
• Did's hint is way superior to mine, so just use his. – Henrik Nov 7 '13 at 17:49

Use the three following facts: $$(i)\int_0^Ts\mathrm dW_s=TW_T-\int_0^TW_s\mathrm ds,\ \ (ii)\ E[W_T^2]=T,\ \ (iii)\ E[W_TW_s]=\min(s,T).$$