Constant Expectations Do Not Make a Process a Martingale

Summary

The document distinguishes a process having the same unconditional expectation at every time from being a martingale. A martingale requires the conditional expectation of a future value, given the information available now, to equal the current value. Brownian motion illustrates this conditional property.

A counterexample is constructed from a sign determined by Brownian motion at an earlier time, active only over a chosen interval. Its unconditional expectation remains zero, yet the process fails the martingale condition. The example shows why checking a constant mean alone is insufficient: martingale behavior depends on conditional expectations relative to the filtration. The note is a compact probability illustration and does not discuss financial applications or broader conditions such as integrability.

Key ideas

• A constant unconditional expectation over time does not establish the martingale property.
• A martingale is characterized by conditional expectations given the information available at the current time.
• Brownian motion is presented as an example satisfying the conditional expectation property.
• A process can have zero expectation at every time and still fail to be a martingale.

Tags

# The same expectation means martingale?


# The same expectation means martingale?












If a stochastic process has the same expectation value for all pisitive t, then is it a martingale? I don’t know how to show it whether that is right.

## Answer by Sebapi (score 2)

https://quant.stackexchange.com/a/42998

The Brownian motion $W_t$ is a martingale because for all $u>t$: $$ W_t = E(W_u | \mathcal{F}_t), $$

However, the process $X_t$ $$X_t = 1_{\{a<t<b\}} (2 . 1_{\{W_a < 0\}} -1) $$ has 0 expectation but 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.