4th moment of a Wiener stochastic integral?

Question

Suppose we have a Wiener stochastic integral

$$\int_0^T g(t) dW(t)$$

Then it is well known that the second moment of this integral is

$$E\left[ \left(\int_0^T g(t) \, dW(t) \right)^2\right] = \int_0^T g(t)^2 \, dt$$

My question is, by the same logic what is the 4th moment?

$$E\left[ \left(\int_0^T g(t) \, dW(t) \right)^4 \right] $$

My attempt goes like this:

$$\begin{align*} E \left[ \left( \int_0^T g(t) \, dW(t) \right)^4 \right] &= E \left(\lim_{n \to \infty} \sum_{i=1}^n g(t_i) \Delta W(t_i) \right)^4 \\ &= E \left(\lim_{n \to \infty} \sum_{i,j,k,z} g(t_i)g(t_j)g(t_k)g(t_z) \Delta W(t_i)\Delta W(t_j)\Delta W(t_k)\Delta W(t_z) \right) \end{align*}$$

Then, (being iffy with interchanging limit and expectation) since the expected value of increments will only be non-zero when $i=j=k=z$, we get

$$=\lim_{n \to \infty} \sum_{i} g(t_i)^4 E( \Delta W(t_i)^4)$$

$$=\lim_{n \to \infty} \sum_{i} g(t_i)^4 3 \Delta t_i^2$$

Now I'm unsure how to intepret this. If we just had $\Delta t_i$ it would be a Riemann integral. But with $\Delta t_i^2$, how does one intepret this?

Answer 1

Although you don't mention it explicitly, your calculations suggest that $g$ is a deterministic function; I'll assume this throughout my answer. If $g$ is not deterministic, then things are getting much more complicated.

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Set

$$X_t := \int_0^t g(s) \, dW_s.\tag{1}$$

Applying Itô's formula, we find

$$\begin{align*} X_t^4 &= 4 \int_0^t X_s^3 \, dX_s + 6 \int_0^t X_s^2 \, d\langle X \rangle_s \\ &= 4 \int_0^t X_s^3 g(s) \, dW_s +6 \int_0^t X_s^2 g(s)^2 \, ds. \end{align*}\tag{2}$$

Taking expectations on both sides we get

$$\mathbb{E}(X_t^4) = 6 \int_0^t \mathbb{E}(X_s^2) \, g(s)^2 \,ds.\tag{3}$$

By Itô isometry,

$$\mathbb{E}(X_s^2) = \int_0^s g(r)^2 \, dr\tag{4}\,.$$

This shows

\begin{align}\tag{5} \mathbb{E}(X_t^4) = 6 \int_0^t \int_0^s g(r)^2 g(s)^2 \, dr \, ds\,. \end{align} Writing $u(t)=\int_0^tg^2(r)\,dr$ we can switch to the new integration variable $u$ and obtain \begin{align} \mathbb{E}(X_t^4)&= 6\int_0^{u(t)} u\,du=3u^2(t) = 3 \left( \int_0^t g(r)^2 \, dr \right)^2.\tag{6} \end{align}

Alternative reasoning: The random variable

$$X_t := \int_0^t g(s) \, dW_s$$

is Gaussian with mean $0$ and variance $\sigma^2 = \int_0^t g(s)^2 \, ds$ (see here for more details). Since $\mathbb{E}(Y^4) = 3 \sigma^2$ for any $Y \sim N(0,\sigma^2)$, we get again

$$\mathbb{E}(X_t^4) = 3 \left( \int_0^t g(s)^2 \, ds \right)^2.\tag{7}$$