Show that $\lim_{|P|\to 0}\sum_{k=0}^{n-1}\frac{W(t_{k+1})+W(t_k)}{2}\left[W(t_{k+1})-W(t_k)\right]=\frac{W^2(T)}{2}$

Question

I have this problem which I am stuck in because it seems very obvious to me that the result is correct, but I don't know how $|P|\to 0$ can be used in the proof. Thanks a lot!

QUESTION: Show that $$\lim_{|P|\to 0}\sum_{k=0}^{n-1}\frac{W(t_{k+1})+W(t_k)}{2}\left[W(t_{k+1})-W(t_k)\right]=\frac{W^2(T)}{2}$$ where $$P=\{0=t_0<t_1<...<t_n=T\}$$ is a partition of $[0,T]$

ATTEMPT:

We have $$\sum_{k=0}^{n-1}\frac{W(t_{k+1})+W(t_k)}{2}\left[W(t_{k+1})-W(t_k)\right]=\frac{1}{2}\sum_{k=0}^{n-1}\left(W_{k+1}^2-W_k^2\right)$$

I know that $$\frac{1}{2}\mathbb{E}\sum_{k=0}^{n-1}\left(W_{k+1}^2-W_k^2\right)=\frac{1}{2}\sum_{k=0}^{n-1}\mathbb{E}\left(W^2(t_{k+1})-W^2(t_k)\right)\frac{1}{2}\sum_{k=0}^{n-1}\left(t_{k+1}-t_k\right)=\frac{1}{2}t_n=\frac{1}{2}\mathbb{E}W^2(t_n)=\frac{1}{2}\mathbb{E}W^2(T)$$ But how does $|P|\to 0$ affect this? Would it be correct to leave it like this?

COMMENT

I have attached the picture of a similar problem which I could solve, but I don't know how to use a similar technique to prove it in this case.

Answer 1

so we want to compute the Stratonovich $S$ integral, which is defined

$$ S:=\int_0^TW_t\circ dW_t:=\lim_{n\to\infty}\sum_{i=0}^{n-1}\frac{1}{2}(W_{t_{i+1}}+W_{t_{i}})(W_{t_{i+1}}-W_{t_{i}}) $$ while we have a partition $P$ of $[0,T]$ with stepsize $h=T/n$, so $$ t_0=0,t_1=h,\dots,t_k=kh,\dots,t_n=T $$ so as $n\to\infty$ we have $|P|\to 0$.

Now let's compute $S$

[EDIT: as pointed out by Did in the comments, the equality which we will show holds for every partition of $[0,T]$ since we will essentially make only use of the binomial formula $(a+b)(a-b)=a^2-b^2$ and the two fixed points of the partition $\{t_0=0,t_n=T\}$ and their relation to the Brownian motion.

So no limiting ever happens, although we should keep in mind, that the limiting process remains in general a part of the definition of the corresponding Stratonovich integral]

$$ S:=\int_0^TW_t\circ dW_t:=\lim_{n\to\infty}\sum_{i=0}^{n-1}\frac{1}{2}(W_{t_{i+1}}+W_{t_{i}})(W_{t_{i+1}}-W_{t_{i}})\\ =\lim_{n\to\infty}\sum_{i=0}^{n-1}\frac{1}{2}(W_{t_{i+1}}^2-W_{t_{i}}^2)=\lim_{n\to\infty}\sum_{i=0}^{n-1}\frac{1}{2}(W_{t_{n}}^2-W_{t_{0}}^2)=\frac{1}{2}(W_{T}^2-W_{0}^2)=\frac{1}{2}W_T $$ while we used the telescope sum and the fact that by definition we have $W_0=0$ and $W_{t_{n}}=W_T$.

bests