Ito integral for Brownian motion

Question

I know that because $W_t$ is a martingale, $$E\left[\int_{0}^{T} W_t dW_t\right] = 0$$ then what should the value for this equation be: $$E\left[\int_{0}^{T} W_t^{n}dW_t\right]?$$ $n$ is the power of $W_t$, a constant.

Answer 1

Another way to look at it is writing in differential form: $$ dX_t = 0 dt + W_t dW_t$$ It's easy to see $X_t$ has a zero drift term (i.e. $dt$ term is zero) and hence is a local martingale. Same argument holds if $W_t$ is raised to some power.

The expectation of both of your integrals will be zero.

Answer 2

If $f: (0,\infty) \times \Omega \to \mathbb{R}$ is a progressively measurable function such that

$$\mathbb{E} \left( \int_0^T f(t)^2 \, dt \right) < \infty \quad \text{for all $T>0$}, \tag{1}$$

then

$$M_T := \int_0^T f(t) \, dW_t$$

defines a martingale. In particular, we have $\mathbb{E}(M_T) = \mathbb{E}(M_0)=0$, i.e.

$$\mathbb{E} \left( \int_0^T f(t) \, dW_t \right) = 0.$$

Since the Brownian motion $(W_t)_{t \geq 0}$ is progressively measurable (because of the continuous sample paths), it suffices to check that $f(t,\omega) := W_t(\omega)^n$ satisfies $(1)$ in order to conclude

$$\mathbb{E} \left( \int_0^T W_t^n \, dW_t \right)=0.$$

To check $(1)$ we have to use that each $W_t$ has moments of arbitrary order. One possibility is using the scaling property, i.e. $W_t \stackrel{d}{\sim} t W_1$, and Tonelli's theorem:

$$\begin{align*} \mathbb{E} \left( \int_0^T f(t)^2 \, dt \right) &= \mathbb{E} \left( \int_0^T W_t^{2n} \, dt \right) \\ &= \int_0^T \mathbb{E}(W_t^{2n}) \, dt \\ &= \mathbb{E}(W_1^{2n}) \int_0^T t^{n} \, dt < \infty. \end{align*}$$