Let $W = \{W_t : t \geq 0\}$ be the standard Brownian motion and consider $$X_t=t W_t$$ Show that the quadratic variation of $X_t$ is $\frac{t^3}{3}$
I know this question has been answered here but I would like to do it by definition, ie, i would like to show that $$\lim_{n\to\infty}\sum_{i=1}^{n}(\Delta X_{t_i})^2\stackrel{L^2}{=}\frac{t^3}{3}, $$ $\text{ with } \ \Delta X_{t_i}=X_{t_i}-X_{t_{i-1}} \text{ and } t_{i}=\frac{t}{n}i$
My Attempt: I have shown that $\Delta X_{t_i} \sim \mathcal{N}(0,(\frac{t}{n})^3(i^2+i-1))$ and I was thinking about applying the LLN but I cannot do it since the variance has a term which is not constant in $i$.
Any ideas?
The limit is to be understood in mean-square or $L^2$ sense. Use the independence of the increments and properties of the Gaussian distribution in order to show that $E[(\sum_i(\Delta X_i)^2-t^3/3)^2]$ converges to zero.
Edit (additional details):
First show that $E[(\sum_i(\Delta X_i)^2]\to t^3/3$. Using your result that $\Delta X_i\sim\mathcal{N}(0,\sigma_i^2)$ with $\sigma_i^2=(t/n)^3(i^2+i-1)$, we obtain as $n\to\infty$
$$ E\left[\sum_{i=1}^n(\Delta X_i)^2\right]=\sum_{i=1}^n E\left[(\Delta X_i)^2\right] = \sum_{i=1}^n \sigma_i^2= \frac{t^3}{n^3}\frac{n^3+3n^2-n}{3}=\frac{t^3}{3}+O(1/n)\to \frac{t^3}{3}.$$ Therefore, because of independence, as $n\to\infty$ $$E\left[\left(\sum_{i=1}^n(\Delta X_i)^2-t^3/3\right)^2\right]=\text{Var}\left[\sum_{i=1}^n(\Delta X_i)^2\right]+O(1/n^2)=\sum_{i=1}^n\text{Var}\left[(\Delta X_i)^2\right]+O(1/n^2)\\=2\sum_{i=1}^n\sigma_i^4+O(1/n^2)=\frac{2t^6}{n^6}\frac{n^5+5n^4+5n^3-5n^2-n}{5}+O(1/n^2)=O(1/n)\to 0.$$ This shows that the $L^2$ limit of the sum of quadratic increments is equal to $t^3/3$.