Problem
I need to calculate the limit of $X_n$ in mean squared when $n$ tends to infinity. $X_n$ is given as
$$X_n = \sum_{i=1}^n (W_{ti/n} - W_{t(i-1)/n})^3$$
where $W$ is Wiener process.
My attempt
If we prove that $E(X_n) = 0$ and $Var(X_n) \to 0$, we are done.
However, I get stuck at the very beginning when I try to calculate the expectation.
Issues
If we denote $W_{ti/n}$ as $Y_i$, I struggle evaluating $E(Y_i^2Y_{i-1})$ after rising the difference to cube. Even though I showed that $\sum_{i=1}^n Y_i \text{ and } \sum_{i=1}^n Y_{i-1}$ are almost the same with the exception of $Y_n$ and $Y_0$, I don't know how to get that $0$.
Use the fact that $Y_i-Y_{i-1}$ is independent of $Y_{i-1}$ and $\mathsf{E}Y_i^k=0$ for odd $k$, i.e., \begin{align} \mathsf{E}Y_i^2Y_{i-1}&=\mathsf{E}[(Y_i-Y_{i-1})^2Y_{i-1}]+2\mathsf{E}[(Y_i-Y_{i-1})Y_{i-1}^2]+\mathsf{E}Y_{i-1}^3 \\ &=\mathsf{E}[(Y_i-Y_{i-1})^2]\mathsf{E}Y_{i-1}+2\mathsf{E}[Y_i-Y_{i-1}]\mathsf{E}Y_{i-1}^2+\mathsf{E}Y_{i-1}^3=0. \end{align}
Alternatively, notice that $Z_i:=Y_i-Y_{i-1}\sim N(0,t/n)$ so that $\mathsf{E}Z_i^k=0$ for odd $k$, which implies that $\mathsf{E}X_n=\sum_{i=1}^n\mathsf{E}Z_i^3=0$. Next, since the increments of $(W_t)$ are independent, $$ \operatorname{Var}(X_n)=\sum_{i=1}^n \mathsf{E}Z_i^6=5!!\cdot\sum_{i=1}^n \left(\frac{t}{n}\right)^3=\frac{15 t^3}{n^2}. $$