expectation of $\int_0^t W_s^2 dW_s $ (integral of square of brownian wrt to brownian)

Question

Looking for answers on the site I came across this answer with 3 upvotes where I am having trouble to understand an integral. Having not enough reputation (SE requires 50 reputation at least and I did not have access to chat until after editing) I cannot comment the post and ask Sasha, the answerer for clarification.

The said user stated: If $X := \int_1^2 W_s^2 \, dW_s$ then $E[X]=0$ where $W_s$ is a Brownian motion.

Yet I cannot understand how he came to this conclusion.

I followed the next steps and started from a more general question starting with $X=\int_0^t W_s^2dW_s$ : 1)Ito's isometry tells me that :$$ E\left[ \left(\int_0^t W_s \, dW_s \right)^2 \right] = E \left[\int_0^t W_s^2 \, ds \right]$$

2) Fubini's theorem (if I did unsterdand correctly from this another answer by Byron Schmuland) that i have: $$ E \left[\int_0^t W_s^2 \, ds \right] = \int_0^t E[W_s^2] \, ds = \int_0^t s \, ds = \frac{t^2}{2}$$

3) applying this to the particular case i would end up with this primitive evaluated in the following bounds : $$\left[\frac{s^2}{2}\right]^2_1=1.5$$

Which is different than $E[X] =0$

I cannot see what I am doing wrong. On the form on the question are the \left \right brackets necessary for readability?

Answer 1

What you are doing is to calculate $\mathbb{E}(X^{\color{red}{2}})$. Itô's isometry allows you to compute $\mathbb{E}(X^{\color{red}{2}})$; you are interested in $\mathbb{E}(X)$!

It is well-known the stochastic integral $$\int_0^t H_s \, dW_s$$ has expectation zero for any (nicely measurable) function $H$ such that $$\mathbb{E} \left( \int_0^t H_s^2 \, ds \right)<\infty$$ for any $t \geq 0$. To prove this, recall that

$$M_t := \int_0^t H_s \, dW_s$$

is a martingale which implies that $$\mathbb{E}(M_t)= \mathbb{E}(M_0)=0, \qquad t \geq 0$$ since martingales have constant expectation.