Why the Cube of Brownian Motion Is Not a Martingale

Summary

The note tests whether cubing a Wiener process preserves the martingale property. It expands the later value as the sum of the earlier value and an independent Brownian increment, then takes a conditional expectation given the information available at the earlier time. The increment has mean zero and zero third moment, while its second moment is the elapsed time. These facts leave an extra term proportional to the earlier Brownian value, so the conditional expectation does not generally equal the current cube.

The calculation also identifies a corrected process: subtracting three times time multiplied by Brownian motion yields a martingale. This is a focused illustration of checking martingales from their defining conditional expectation, rather than assuming that a nonlinear transformation preserves the property. It relies on standard Wiener process increment properties and does not discuss broader stochastic processes or applications to trading.

Key ideas

• A martingale must have conditional expected future value equal to its current value.
• Independent Brownian increments allow the cube to be expanded and conditioned on current information.
• The squared increment contributes a term proportional to elapsed time and current Brownian motion.
• Subtracting three times time times Brownian motion produces a martingale.

Tags

# Why $W_{t}^3$ is not a martigale?(by Definition)


# Why $W_{t}^3$ is not a martigale?(by Definition)












If $W_t$ be a wiener process then,how can i show that $W_{t}^{3}$ is not a martingale by definition?

## Answer by Gordon (score 4, accepted)

https://quant.stackexchange.com/a/18552

Note that, for $0 \leq s < t$, \begin{align*} W_t^3 &= (W_t-W_s+W_s)^3\\ &= (W_t-W_s)^3 + 3(W_t-W_s)^2 W_s + 3 (W_t-W_s) W_s^2 + W_s^3. \end{align*} Moreover, \begin{align*} E\big( (W_t-W_s)^3 \mid \mathcal{F}_s\big) &= E\big( (W_t-W_s)^3\big)\\ &= 0,\\ E\big((W_t-W_s)^2 W_s \mid \mathcal{F}_s\big) &= W_s E\big( (W_t-W_s)^2\big)\\ &= (t-s)W_s, \end{align*} and \begin{align*} E\big( (W_t-W_s) W_s^2 \mid \mathcal{F}_s\big) &= W_s^2 E\big( (W_t-W_s)\big)\\ &=0. \end{align*} Then, \begin{align*} E\big(W_t^3\mid \mathcal{F}_s\big) &= 3(t-s)W_s+W_s^3. \end{align*} That is, $\{W_t^3 \mid t\geq 0\}$ is not a martingale. We note that, however, $\{W_t^3 -3tW_t \mid t\geq 0\}$ is a martingale. See Question Show that $E[B_t|\mathscr{F}_s] = B_s$

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