Let $W_t$ be a standard brownian motion. I am trying to compute $E[(\int_0^t s^2 dW_S)^4]$. I applied Ito's formula and got $$t^2 W_t = \int_0^t s^2 dWs + \int_0^t 2s W_s ds$$.
This gives us
$$E[(\int_0^t s^2 dW_S)^4] = E[(t^2 W_t-\int_0^t 2s W_s ds)^4]$$.
Now I'm not sure how to proceed. Is there a good way to deal with the 4th power on the right hand side?
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Since the integrand is a deterministic function, the following theorem can be used:
Define $$X(t) = \int_{0}^{t} s^2 dW_s,$$ then $\{X(t) \ | \ t \geq 0\}$ is a Gaussian process with zero mean and covariance function $\rho(s,t) = \int_{0}^{\min \{s,t\}} u^4du$. Hence, you have to compute the fourth moment of a normally distributed variable with zero mean and variance $\rho(t,t)$.
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