Independence of Brownian Increments and Product Expectations

Summary

The document asks whether a Brownian motion value at the start of an interval can be treated as independent of the increment over that interval when evaluating an expected product. Brownian motion has independent increments, so the value up to a partition time is independent of the subsequent increment. Since that increment has mean zero, the expected product is zero, and summing these expectations also gives zero.

The supplied answer appears to confuse the Brownian value with the increment: it says the two quantities are the same, then uses a second-moment example that does not address their independence. For Brownian motion, the second moment at time t is t, not t squared. Thus the answer as written is unreliable, even though the user’s stated independence reasoning for each interval is sound. The exchange provides no further derivation or financial application; its relevance is primarily to probability concepts used in stochastic modeling.

Key ideas

• A Brownian motion value at a partition time is independent of the following increment.
• Each Brownian increment has mean zero, so the expected product with the preceding value is zero.
• The supplied answer conflates the Brownian value and its increment and gives an incorrect second moment.
• The independence argument applies interval by interval to the partition sum.

Tags

# computation involving independent increments


# computation involving independent increments












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.

## Answer by emcor (score 0)

https://quant.stackexchange.com/a/16666

$W_{t_i}$ 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.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.