Why the Time Integral of Brownian Motion Is Not a Martingale

Summary

The document asks whether a proposed proof correctly shows that the process formed by integrating Brownian motion over time is not a martingale. The argument observes that the process has mean zero because Brownian motion has mean zero, then incorrectly infers that its conditional expectation given the past is also zero. A martingale condition instead requires the conditional expectation of a future process value to equal the process’s current value.

The example is useful for distinguishing an unconditional expectation from a conditional one, a key point in stochastic-process reasoning. However, the document contains only the question and the attempted argument; it does not include a correction or a complete proof. Readers should treat the displayed conditional-expectation step as the issue under review, rather than as an established result.

Key ideas

• A process having zero unconditional mean does not establish its martingale property.
• The martingale condition compares a future value’s conditional expectation with the current process value.
• The proposed argument assumes a conditional expectation is zero based only on a zero unconditional mean.
• The document poses the proof question but does not provide its resolution.

Tags

# Proof that integral of Brownian motion wrt time is not a martingale


# Proof that integral of Brownian motion wrt time is not a martingale












Let $X_t=\int_0^t W_s ds$ where $W_s$ is Brownian motion, so $E[W_s]=0$.

Then $E[X_t]=\int_0^t E[W_s] ds=\int_0^t 0 ds=0$.

So $E[X_t|{\cal F}_s]=0\neq X_s$, almost everywhere. So by previous sentence, $X_t$ is not a martingale.

Question: Is the above a correctly stated and argued proof that $X_t$ is not a martingale?

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