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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.

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  • $\begingroup$ Do you know ito's product rule and the condition for stochastic integrals to be martingales? $\endgroup$ – Henrik Nov 7 '13 at 10:42
  • $\begingroup$ Just briefly... but how does it help? $\endgroup$ – user106113 Nov 7 '13 at 11:16
  • $\begingroup$ Did's hint is way superior to mine, so just use his. $\endgroup$ – Henrik Nov 7 '13 at 17:49
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Use the three following facts: $$ (i)\int_0^Ts\mathrm dW_s=TW_T-\int_0^TW_s\mathrm ds,\ \ (ii)\ E[W_T^2]=T,\ \ (iii)\ E[W_TW_s]=\min(s,T).$$

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  • $\begingroup$ Thanks a lot! I could solve it now :) $\endgroup$ – user106113 Nov 7 '13 at 11:15

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