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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.

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    Central limit theorem won't help, central limit theorem is for a different [type of convergence](http://en.wikipedia.org/wiki/Convergence_of_random_variables).2019-01-06

1 Answers 1

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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):

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

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    Th$a$nk you, I just copy-pasted the latex.2012-05-21