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One can rather easily show that $E\left[\sum\limits_{i = 0}^{i = n - 1}W_{t_i}(W_{t_{i + 1}} - W_{t_i})\right] = -T + W_T^2$.

What I'm confused about is why we can't simply say that for each $i$, $W_{t_{i}}$ is independent of $(W_{t_{i + 1}} - W_{t_i})$, so that upon interchanging sums and expectations, and using independence we have $E\left[\sum\limits_{i = 0}^{i = n - 1}W_{t_i}(W_{t_{i + 1}} - W_{t_i})\right] = \sum\limits_{i = 0}^{i = n - 1}E\left[W_{t_{i}}\right]E\left[W_{t_{i + 1}} - W_{t_i}\right] = 0$?

In this problem, we have naturally partitioned an interval as $0 = t_{0} < t_{1} < ... < t_{n} = T$, and $W_{t}$ is a Brownian motion.

rubik
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user7348
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2 Answers2

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You have an error: you find that the expectation is $-T+E[W_T^2]=-T+T=0$. That is, your observation is correct. We usually view this as arising from the fact that in the Ito convention, $\int_0^t W_s dW_s$ is a martingale. (So are the integrals of "most" functions against $W_t$.)

Ian
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    Oh my goodness, thanks so much. I realize what I've done wrong. I'm not taking an expectation at all. As the comment above suggests, I should not have a random variable in an expectation at all. Thanks! – user7348 Feb 21 '15 at 20:33
  • @user7348 If this answered your question, you should click the check mark next to it in order to accept it. – Ian Feb 21 '15 at 20:35
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Look here (it is in German, but you can translate the text): Stochastic integration

Note that there is set $B_i$ instead of $W_i$

kryomaxim
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