For the stochastic integral, where $W_t$ is a Wiener process, I am trying to find the mean of $X_t = \int_0^t sW_sdW_s$. I have read before that any stochastic integral with $dWt$ has mean zero, but I dont know if it extends to cases where I have a random variable in the integrand as well. My approach is to decompose the integral $X_t$ into:
$$ \int_0^t sW_sdW_s = \lim_{n \to \infty}\sum_{j=0}^{n-1}t_jW_{t_i}(W_{t_{i+1}}-W_{t_i}) $$
Then,
$$ E\left(\int_0^t sW_sdW_s\right) = \lim_{n \to \infty}\sum_{j=0}^{n-1}E\left(t_jW_{t_i}(W_{t_{i+1}}-W_{t_i})\right) $$
I believe that I can separate the terms in the expectation into:
$$ E\left(t_jW_{t_i}(W_{t_{i+1}}-W_{t_i})\right) = E\left(t_jW_{t_i}\right)E\left(W_{t_{i+1}}-W_{t_i}\right) $$
I am not sure how to find $E\left(t_jW_{t_i}\right)$, although I know it is finite so it doesn't matter because $E\left(W_{t_{i+1}}-W_{t_i}\right) = 0$