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

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  • $\begingroup$ what about showing that $E[W^3_t | \sigma(W_s)]$ is not $W^3_s$ ? $\endgroup$ – statquant Jun 28 '15 at 14:59
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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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