Conditional Expectation of a Brownian Motion Increment

Summary

The document asks whether the conditional expectation of an exponential Brownian increment, given information available at an earlier time, can equal its unconditional expectation. It assumes the earlier time is no later than the endpoint and questions whether the equality can hold except when the earlier time is zero.

The issue concerns independent increments and conditional expectation: a Brownian increment after time s is independent of the information generated up to s, so conditioning on that past information leaves its expectation unchanged. The post does not include the accepted answer’s reasoning, derive the expectation, or state assumptions about the filtration beyond its notation. It is therefore a focused conceptual question rather than a worked explanation, and readers would need additional material to see why the conditional expectation is a constant random variable.

Key ideas

• Brownian motion increments after time s are independent of the information available up to s under the usual filtration assumptions.
• The question distinguishes a conditional expectation as a random variable from its possible constant value.
• The post provides no derivation or detailed answer to resolve the confusion.

Tags

# The conditional expectation of a geometric brownian motion


# The conditional expectation of a geometric brownian motion












In this question it states that $$\mathbb{E}[e^{\sigma(W_t-W_s)}|\mathcal{F}_s] = \mathbb{E}[e^{\sigma(W_t-W_s)}],$$ and I assume that $0 \leq s \leq t$. The accepted answer states that this step is correct. However, how can the random variable (LHS) ever be equal to a constant (RHS) for all $0 \leq s \leq t$? The only situation in which it seems to be true is when $s=0$, since the LHS then collapses to a constant as well. Can anybody provide an explanation?

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