Is the following Itô-Integral not zero?

Question

is the following statement true:

$$\int_0^T t \, dW(t) \neq 0$$

I need it for a counter-example, that one can not change the order of integration between $dW$ and $dP(\omega)$. I thought of taking the following: $$ E\left[\int_0^T W^2(t) \, dW(t)\right]=0 \neq \int_0^T t \, dW(t)=\int_0^T E[W^2(t)] \, dW(t)$$ Intuitively it should be correct but how can I proof it? Applying the defintion of the Itô-Integral?

Answer 1

The random variable $$ X=\int_0^Tt\mathrm dW_t $$ is normal with mean zero and variance $$ \int_0^Tt^2\mathrm dt\ne0, $$ hence $$P(X=0)=0.$$

Answer 2

What we need:

$$\int_0^t s \, dW_s$$

Take Ito for $f(t,W_t)=tW_t$:

$$f=f(0,W_0)+\int_0^t\frac{\partial f}{\partial s}ds+\int_0^t\frac{\partial f}{\partial W_s}dW_s+\frac{1}{2}\int_0^t\frac{\partial^2 f}{\partial W_s^2}d[W_s,W_s]$$

$$tW_t=\int_0^t W_s ds + \int_0^t s dW_s $$

Then

$$\int_0^t s \, dW_s =tW_t-\int_0^t W_s ds$$

Where the later integral as you can see here is a normal with mean $0$ and variance $\frac{t^3}{3}$