Prove that sum of squared Wiener increments tends to length of the interval.

Question

$$ S_{n} = \sum_{k=1}^{k_{n}} |W_{t_{k}} - W_{t_{k-1}}|^{2} \to b - a $$ in $L^{2}(\Omega)$, $max_{k}|t_{k} - t_{k-1}| \to 0.$

So I want to have $||S_{n} - (b-a)||_{L(\Omega)} = 0$.

$\mathbb{E}|S_{n} - (b-a)|^{2} = \mathbb{E}[(\sum_{k=1}^{k_{n}} (\Delta W_{k})^{2})^{2} - 2(b-a)(\sum_{k=1}^{k_{n}} (\Delta W_{k})^{2}) + (b-a)^{2}] = Var(\sum_{k=1}^{k_{n}} (\Delta W_{k})^{2}) + \mathbb{E}(\sum_{k=1}^{k_{n}} (\Delta W_{k})^{2}) - 2(b-a)\mathbb{E}(\sum_{k=1}^{k_{n}} (\Delta W_{k})^{2}) + (b-a)^{2} = \sum_{k=1}^{k_{n}} Var (\Delta W_{k})^{2} + \sum_{k=1}^{k_{n}} \Delta t_{k} - 2(b-a)(\sum_{k=1}^{k_{n}} (\Delta t_{k})) + (b-a)^{2}$

Is there somewhere a mistake? How to calculate this variance of $(\Delta W_{k})^{2}$? I cannot see that it will actually converge to zero... Could someone please give me a hint or suggest better solution?

Answer 1

So you want to prove the following assertion: $$S_{n} = \sum_{k=1}^{k_{n}} |W_{t_{k}} - W_{t_{k-1}}|^{2} \to b - a$$

Notice first (as you may already have) that $$\mathbb E(S_n)=\mathbb E\left(\sum_{k=1}^{k_{n}} |W_{t_{k}} - W_{t_{k-1}}|^{2}\right)=\sum_{k=1}^{k_{n}}\mathbb E\left(|W_{t_{k}} - W_{t_{k-1}}|^{2}\right)=\sum_{k=1}^{k_{n}}t_k-t_{k-1}=b-a$$

$$\mathbb E|S_n-(b-a)|^2=\mathbb V(S_n)=\mathbb V\left(\sum_{k=1}^{k_{n}} |W_{t_{k}} - W_{t_{k-1}}|^{2} \right)$$

By independence of the Brownian increments we have that the latter equals

$$\sum_{k=1}^{k_{n}} \mathbb V\left(|W_{t_{k}} - W_{t_{k-1}}|^{2} \right)=\sum_{k=1}^{k_{n}} \mathbb E\left(|W_{t_{k}} - W_{t_{k-1}}|^{4}\right)-\mathbb E\left(|W_{t_{k}} - W_{t_{k-1}}|^2\right)^2$$

Remember that the $4$th moment of a gaussian random variable is $3\sigma^4$ and hence $$=\sum_{k=1}^{k_{n}} 3(t_k-t_{k-1})^2-(t_k-t_{k-1})^2=2\sum_{k=1}^{k_{n}} (t_k-t_{k-1})^2\leq 2\max_{k}|t_{k} - t_{k-1}|\cdot \sum_{k=1}^{k_{n}}|t_{k} - t_{k-1}|=2\|\Pi_n\|(b-a) $$

Letting $\|\Pi_n\|$ (the mesh of the partition) go to $0$ you obtain the desired result.

Answer 2

maybe easier to see that $$ \mathbb{E}\left[|W_{t_k} - W_{t_{k-1}}|^2\right] = \mathbb{Var}[\Delta_k] = t_k - t_{k-1} $$ and $\mathbb{E}[S_n]$ can be computed with telescoping series.