computation involving independent increments [closed]

One can rather easily show that E[$\sum_{i = 0}^{i = n - 1}W_{t_i}(W_{t_{i + 1}} - W_{t_i})]$ = -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[$\sum_{i = 0}^{i = n - 1}W_{t_i}(W_{t_{i + 1}} - W_{t_i})]$ = $\sum_{i = 0}^{i = n - 1}E[W_{t_{i}}]E[W_{t_{i + 1}} - W_{t_i}]$ = 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.

closed as off-topic by Ric, Bob Jansen♦Apr 23 '15 at 7:13

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• Question closed. Answer provided on math.stackexchange is sufficient. – user7348 Feb 21 '15 at 20:37
• Where on math.stackexchange can we find the answer? The question as posted here must be wrong as in the first line you have an ordinary expectation (thus a real number) on the lhs and a random variable on the rhs. – Ric Apr 23 '15 at 7:00
• The question is cross-posted here: math.stackexchange.com/questions/1159240/… user7348: please mind your manners and stop cross posting questions. – Bob Jansen Apr 23 '15 at 7:12
• I'm voting to close this question as off-topic because it's cross-posted here: math.stackexchange.com/questions/1159240/… – Bob Jansen Apr 23 '15 at 7:13

$W_{t_i}$ and $W_{t_i}$ are not independent (in fact $W_{t_i}=W_{t_i}$), e.g. $$E[W_t^2]=t^2\neq E[W_t]\cdot E[W_t]=0$$ So you cant separate the expectation in your equation.
• My comment said the question is closed, and I didn't want any additional answers.Furthermore, you're wrong. ($W_{t_{i}} - W_{t_{0}})(W_{t_{i + 1}} - W_{t_{i}})$ can be split because the increments are independent by definition of Brownian Motion. – user7348 Feb 22 '15 at 17:06