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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.

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  • $\begingroup$ So, what does the $n$ mean in your notation? Is it a $n$-dimensional Brownian motion? $\endgroup$
    – Olorun
    Commented Dec 10, 2015 at 2:19
  • $\begingroup$ @Olorun It's just a constant n $\endgroup$ Commented Dec 10, 2015 at 2:21
  • $\begingroup$ @Olorun Its a power for $W_t$ $\endgroup$ Commented Dec 10, 2015 at 2:26
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    $\begingroup$ Note that $$\int_0^T W_t \, dW_t$$ does NOT equal $0$. The expectation of the randm variable is zero, but not the random variable itself. $\endgroup$
    – saz
    Commented Dec 10, 2015 at 6:41
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    $\begingroup$ It turns out that the expectation of "essentially" any function integrated against $dW_t$ (in the sense of Ito) is zero. $\endgroup$
    – Ian
    Commented Dec 10, 2015 at 23:46

2 Answers 2

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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:

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

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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.

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