How to show that $\mathbb{E}(\int_0^T t\mathrm \, dW_t) = 0 $?

Question

I just want to know why $\mathbb{E}\left(\int_0^T t \,\mathrm dW_t\right)=0$.

I know it's got something to do with the Gaussian distribution but I don't really know what.

Answer 1

I'm assuming $W(t)$ is the Wiener process.

Hint: One way is to use the fact that the Itō integral $I(u)=\int^u_0 tdW(t)$ is a martingale, so $\mathbb{E}(I(t))=\mathbb{E}(I(0))$ for all $t \geq 0.$

In general, this works for those Itō integrals with (mean) square-integrability condition. Thanks for nullUser for reminding me of this condition.

Answer 2

I would go with Ehsan's answer for this one, but since you mentioned "something to do with the Gaussian distribution" I will add my answer.

What you are thinking of is that if $f_t$ is deterministic, then $\int_0^T f_tdW_t \sim N(0, \int_0^t f_t^2dt)$ follows a normal distribution.

While this is a useful fact to know, in your particular case it would be much easier to simply note that the Ito integral is a martingale.