Limit of Wiener processes

Question

Let $W_t$ be Wiener process. I am trying to evaluate the following limit $$\lim\limits_{n \to \infty}~{\sum\limits_{i=1}^{n}W_{\frac{i-1}{n}+\frac{1}{2n}}\left( W_{\frac{i}{n}} - W_{\frac{i-1}{n}} \right)}$$

I've expanded parethesis and got $$ \lim\limits_{n \to \infty}~ \left[ -W_0W_\frac{1}{2n} - W_\frac{1}{n}\left( W_\frac{3}{2n} - W_\frac{1}{2n} \right) - \cdots - W_\frac{n-1}{n}\left( W_\frac{2n-1}{2n} - W_\frac{2n-3}{2n} \right) + W_\frac{2n-1}{2n}W_1 \right] $$

I think I should apply central limit theorem here, but I don't understand how.

Central limit theorem won't help, central limit theorem is for a different type of convergence. — Tom Artiom Fiodorov, May 21 '12 at 20:34

score 1 · Accepted Answer · answered May 21 '12 at 21:28

1

HINT: Let $t_i = \frac{i}{n}$ be a partition of a unit interval. Then the sum can be rewritten as: $$ \sum_{i=1}^n W\left( \frac{t_{i-1}+t_i}{2}\right) \left( W(t_i) - W(t_{i-1}) \right) $$ Compare this with the definition of Stratonovich integral in terms of Riemann sum.

Thus the answer is (hover over):

$\lim_{n\to\infty} \sum\limits_{i=1}^n W\left(\frac{i-1}{n} + \frac{1}{2n}\right) \left( W\left(\frac{i}{n}\right) - W\left(\frac{i-1}{n}\right)\right) = \int_0^1 W(s) \circ\! \mathrm{d} W(s) = \frac{1}{2} W(1)^2$

answered May 21 '12 at 21:28

Sasha

70,631

My notes say that Stratonovich would be $$\displaystyle \lim_{n->\infty}\sum_{k=0}^{[2^n t]-1}\big(\frac{W_{(k+1)2^{-n}}+W_{k2^{-n}}}{2}\big)(W_{(k+1)2^{-n}}-W_{k2^{-n}}),$$ what am I missing? – Tom Artiom Fiodorov May 21 '12 at 21:45
There is a sign typo in your notes, it should be $W_{(k+1) 2^{-n}} - W_{k 2^{-n}}$ – Sasha May 21 '12 at 21:48
Thank you, I just copy-pasted the latex. – Tom Artiom Fiodorov May 21 '12 at 21:49

Limit of Wiener processes

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