I have the following Riemann sum approximation of a stochastic integral:
$$X = \sum_{i=0}^{n-1}W_{\frac in}^2(W_{\frac{i+1}n}-W_{\frac in})^3,$$
where $W_t$ is a Brownian motion for all $t$.
I am (attempting) to compute $\mathbb{E}(X^2)$.
Without the exponents above the terms in the summation notation, I recognize the equation to be the basic Itô integral:
$$X_t = \int_0^t W_sdW_s = \sum_i W_i(W_{i+1}-W_i).$$
With the two exponents, however, I'm not quite sure where to go. Any help or insight would be massively appreciated.
Let $\mu_n = \mathbb [Z^n]$ with $Z \sim \mathcal N (0,1)$ $$\mathbb E \left[X^2 \right] = \mathbb E \left[\sum_{i=0}^{n-1} \sum_{j=0}^{n-1} W_{\frac in}^2\left(W_{\frac{i+1}n}-W_{\frac in}\right)^3 W_{\frac jn}^2\left (W_{\frac{j+1}n}-W_{\frac jn}\right)^3\right] = \mathbb E \left[\sum_{i=0}^{n-1} \sum_{j=0}^{i-1} W_{\frac in}^2\left(W_{\frac{i+1}n}-W_{\frac in}\right)^3 W_{\frac jn}^2\left (W_{\frac{j+1}n}-W_{\frac jn}\right)^3 + \sum_{i=0}^{n-1} \sum_{j=i+1}^{n-1} W_{\frac in}^2\left(W_{\frac{i+1}n}-W_{\frac in}\right)^3 W_{\frac jn}^2\left (W_{\frac{j+1}n}-W_{\frac jn}\right)^3 + \sum_{i=0,~ j=i}^{n-1} W_{\frac in}^2\left(W_{\frac{i+1}n}-W_{\frac in}\right)^3 W_{\frac jn}^2\left (W_{\frac{j+1}n}-W_{\frac jn}\right)^3\right] = \mathbb E \left[\sum_{i=0}^{n-1} \sum_{j=0}^{i-1} W_{\frac in}^2\left(W_{\frac{i+1}n}-W_{\frac in}\right)^3 W_{\frac jn}^2\left (W_{\frac{j+1}n}-W_{\frac jn}\right)^3 + \sum_{j=0}^{n-1} \sum_{i=0}^{j-1} W_{\frac in}^2\left(W_{\frac{i+1}n}-W_{\frac in}\right)^3 W_{\frac jn}^2\left (W_{\frac{j+1}n}-W_{\frac jn}\right)^3 + \sum_{i=0,~ j=i}^{n-1} W_{\frac in}^2\left(W_{\frac{i+1}n}-W_{\frac in}\right)^3 W_{\frac jn}^2\left (W_{\frac{j+1}n}-W_{\frac jn}\right)^3\right] =2\mathbb E \left[\sum_{i=0}^{n-1} \sum_{j=0}^{i-1}\mathbb E\left[ W_{\frac in}^2\left(W_{\frac{i+1}n}-W_{\frac in}\right)^3 W_{\frac jn}^2\left (W_{\frac{j+1}n}-W_{\frac jn}\right)^3\Big | W_{\frac jn}, W_{\frac{j+1}n}, W_{\frac{i}n}\right]\right] + \sum_{i=0}^{n-1} \mathbb E\left[W_{\frac in}^4 \left(W_{\frac{i+1}n} - W_{\frac in}\right)^6\right] = \sum_{i=0}^{n-1}\frac{i^2}{n^2}\frac1{n^3} \mu_4\mu_6 + 2\mathbb E \left[\sum_{i=0}^{n-1} \sum_{j=0}^{i-1}W_{\frac jn}^2 W_{\frac in}^2\left (W_{\frac{j+1}n}-W_{\frac jn}\right)^3\mathbb E\left[\left(W_{\frac{i+1}n}-W_{\frac in}\right)^3\Big | W_{\frac jn}, W_\frac{j+1}n, W_{\frac in}\right] \right] = \mu_4\mu_6\frac{1}{n^5} \sum_{i=0}^{n-1} i^2$$
since $$\mathbb E\left[\left(W_{\frac{i+1}n}-W_{\frac in}\right)^3\Big | W_{\frac jn}, W_\frac{i+1}n, W_{\frac in}\right] = 0$$ and $W_{\frac in}, W_{\frac {i+1}n} - W_{\frac in}$ are independent and $$W_{\frac in} \sim \mathcal N(0, \frac in) = \sqrt\frac in \mathcal N(0,1)$$ and $$W_{\frac {i+1}n} - W_{\frac in} \sim \mathcal N (0, \frac 1n) = \sqrt \frac 1n \mathcal N(0,1)$$