$\sum (B_{t_{i}^{n}}+B_{t_{i-1}^{n}})^{2} \left( \left(B_{t_{i}^{n}}-B_{t_{i-1}^{n}}\right)^2-[t^n_{i}-t^n_{i-1}]\right)$ converges in probability

Question

Let $T > 0$, $B$ be a Brownian motion and consider a sequence of partitions $\{0=t_{0}^{n}<...<t_{l(n)}^{n}=T\}, \; n \in \mathbb N$ such that the mesh tends to $0$ as $n \to \infty$.

How can I show that $\sum\limits_{i = 1}^{l(n)}(B_{t_{i}^{n}}+B_{t_{i-1}^{n}})^{2} \left(\left(B_{t_{i}^{n}}-B_{t_{i-1}^{n}}\right)^2-(t^n_{i}-t^n_{i-1})\right)$ converges in probability as $n \to \infty$?

Of course I know that $E[\left(B_{t_{i}^{n}}-B_{t_{i-1}^{n}}\right)^2-(t^n_{i}-t^n_{i-1})]=0$ and $B_{t}^2-t$ is a martingale but there is no independence or other notions that will help me prove $L^{2}$ convergence and by extension convergence in probability.

Any tips/hints appreciated.

Answer 1

Write $\Delta B_i = B_{t^n_i} - B_{t^n_{i-1}} $ and $\Delta t_i = t^n_i - t^n_{i-1}$. Then we may decompose the sum into three parts:

\begin{align*} \sum_{i=1}^{l(n)} (2B_{t^n_{i-1}} + \Delta B_i)^2 \bigl( (\Delta B_i)^2 - \Delta t_i \bigr) = S_{1,n} + S_{2,n} + S_{3,n}, \end{align*}

where

\begin{align*} S_{1,n} &= 4 \sum_{i=1}^{l(n)} B_{t^n_{i-1}}^2 \bigl( (\Delta B_i)^2 - \Delta t_i \bigr) \\ S_{2,n} &= \sum_{i=1}^{l(n)} 4B_{t^n_{i-1}}(\Delta B_i) \bigl( (\Delta B_i)^2 - \Delta t_i \bigr) \\ S_{3,n} &= \sum_{i=1}^{l(n)} (\Delta B_i)^2 \bigl( (\Delta B_i)^2 - \Delta t_i \bigr). \end{align*}

1. For the first sum, note that for $i < j$,

\begin{align*} &\mathbb{E}\bigl[ B_{t^n_{i-1}}^2 \bigl( (\Delta B_i)^2 - \Delta t_i \bigr) B_{t^n_{j-1}}^2 \bigl( (\Delta B_j)^2 - \Delta t_j \bigr) \bigr] \\ &= \mathbb{E}\bigl[ B_{t^n_{i-1}}^2 \bigl( (\Delta B_i)^2 - \Delta t_i \bigr) B_{t^n_{j-1}}^2 \mathbb{E}\bigl[ (\Delta B_j)^2 - \Delta t_j \bigm| \mathcal{F}_{t^n_{j-1}} \bigr] \bigr] \\ &= 0. \end{align*}

Also, by the independence,

\begin{align*} &\mathbb{E}\bigl[ \bigl( B_{t^n_{i-1}}^2 \bigl( (\Delta B_i)^2 - \Delta t_i \bigr) \bigr)^2 \bigr] = \mathbb{E}[ B_{t^n_{i-1}}^4 ] \mathbb{E}\bigl[ \bigl( (\Delta B_i)^2 - \Delta t_i \bigr)^2 \bigr] = C (t^n_{i-1})^2 \Delta t_i^2 \end{align*}

for some absolute constant $C>0$. (We may choose $C = \mathbb{E}[B_1^4]\mathbb{E}[(B_1^2-1)^2]$, although its value is not important here.) Combining altogether, it follows that

$$ \mathbb{E}\bigl[ S_{1,n}^2 \bigr] = C \sum_{i=1}^{l(n)} (t^n_{i-1})^2 \Delta t_i^2 \leq CT^3 \Bigl( \max_{1\leq i \leq l(n)} \Delta t_i \Bigr). $$

This tells that $S_{n,1} \to 0$ in $L^2$.

2. For the second sum, note that

$$ \mathbb{E} \bigl[ \bigl\lvert B_{t^n_{i-1}}(\Delta B_i) \bigl( (\Delta B_i)^2 - \Delta t_i \bigr) \bigr\rvert \bigr] \leq C_2 (t^n_{i-1})^{1/2} (\Delta t_i)^{3/2} $$

for another absolute constant $C_2 = \mathbb{E}[\lvert B_1\rvert]\mathbb{E}[\lvert B_1(B_1^2 - 1)\rvert] > 0$, and so,

$$ \mathbb{E} \bigl[ \bigl\lvert S_{2,n} \bigr\rvert \bigr] \leq C_2 T^{3/2} \Bigl( \max_{1\leq i \leq l(n)} \Delta t_i \Bigr). $$

This shows that $S_{2,n} \to 0$ in $L^1$.

3. A similar reasoning shows that $S_{3,n} \to 0$ in $L^1$ as well.

Therefore the original sum converges to $0$ in probability.