# $L^2$ convergence for variance of a Brownian motion

The Problem Given $$(B_s,s \in [0,t])$$ a Brownian motion on $$[0,t]$$ and $$0=t_0 < t_1 < ... < t_n \le t$$ a partition of $$[0,t]$$ such that $$\max_{j}|t_{j+1} - t_j| \to 0$$ as $$n \to \infty$$. Show that the random variable $$\sum_{j=0}^n (B_{t_j} - B_{t_{j-1}})^2$$ converges to $$t$$ in $$L^2$$ as $$n \to \infty$$.

I am taking this class in stochastic calculus and I am very dry on probability theory. However, I think I understand key concepts of Brownian Motion, so I need some help putting together the pieces. I know I want to show that $$\lim_{n \to \infty} \left\Vert \sum_{j=0}^n (B_{t_j} - B_{t_{j-1}})^2- t \right\Vert_2 = \lim_{n \to \infty} \mathbb{E}\left[ \left| \sum_{j=0}^n (B_{t_j} - B_{t_{j-1}})^2- t \right|^2 \right]=0$$

I also realize that $$\mathbb{E} \left[ \sum_{j=0}^n (B_{t_j} - B_{t_{j-1}})^2 \right ]= \sum_{j=0}^n \mathbb{E} \left[ (B_{t_j} - B_{t_{j-1}})^2 \right ]$$ $$=\sum_{j=0}^n t_j - t_{j-1}=t$$ by the property of the variance of Brownian motion and because the sum of the $$t_i$$ is $$t$$. I guess by biggest problem is I don't know how to mathematically break into this $$L^2$$ norm. Any help is greatly appreciated.

Hint

In your partition, you took $$t_n\leq t$$, but you must take $$t_n=t$$. Then,

\begin{align*} &\mathbb E\left[\left(\sum_{j=0}^{n-1}(B_{t_{j+1}}-B_{t_j})^2-t\right)^2\right]\\ &=\mathbb E\left[\left(\sum_{j=1}^{n-1}(B_{t_{j+1}}-B_{t_j})^2-(t_{j+1}-t_j)\right)^2\right]\\ &=\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\mathbb E\left[\Big((B_{t_{j+1}}-B_{t_j})^2-(t_{j+1}-t_j)\Big)\Big((B_{t_{i+1}}-B_{t_i})^2-(t_{i+1}-t_i)\Big)\right]. \end{align*}

I let you continue.

• OK that's great, I was a little surprised to see those summation symbols come outside the expectation, but I see what you did there. Feb 14, 2021 at 15:58
• I feel like now the next step is to break this expression into 2 cases: when $i=j$ and when $i \ne j$. If $i \ne j$ then we have independent intervals of Brownian motion, so we can say the expectation of the product is the product of the expectation, and for those we easily get zero. For the case where $i=j$ we can't assume independence, we would have $E[(B_{t_{j+1}} - B_{t_j})^4 - 2(t_{j+1} - t_j)(B_{t_{j+1}} - B_{t_j})^2 + (t_{j+1} - t_j)^2] = E[(B_{t_{j+1}} - B_{t_j})^4] - (t_{j+1} - t_j)^2$. This looks very promising but I don't know about the fourth moment?! Feb 14, 2021 at 16:08
• @jeffery_the_wind: Of course it won't be $0$. Nevertheless $$\sum_{j=0}^{n-1}(t_{j+1}-t_j)^2\leq t\cdot \max_{j=0,...,n-1}|t_{j+1}-t_j|\underset{n\to \infty }{\longrightarrow }0.$$
– Surb
Feb 14, 2021 at 20:05
• No specific property... just the fact that $|t_{i+1}-t_i|\leq \max_{j=0,...,n-1}|t_{j+1}-t_j|$ for all $i$.
– Surb
Feb 14, 2021 at 20:20
• Thank you so much! I got it! Feb 14, 2021 at 23:49