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Let $X_{n}$ be a martingale with respect to a filtration $\mathbb{P}_{n}$. Define:

$Y_{n}$ := $X_{n}^{3}$

Is $Y_{n}$ a martingale? Supermartingale?

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This smells like homework and is lacking indications about what the OP knows or what they tried. –  Did Feb 26 '13 at 16:29
    
@AlexanderShamov I suspect you did not read carefully the question (Itô formula? good grief! what for?). –  Did Feb 26 '13 at 16:30
    
@Did: Ah, right, of course I didn't. :) –  Alexander Shamov Feb 27 '13 at 18:50

1 Answer 1

In general it's neither martingale nor supermartingale. Consider for example $$X_n := \prod_{j=1}^n \xi_j$$ where $(\xi_j)_j$ are independent identically distributed random variables such that $\mathbb{E}\xi_1 = 1$, $\mathbb{E}|\xi_1|^3 < \infty$. Then $(X_n)_n$ is a martingale with respect to the filtration $\mathcal{F}_n := \sigma(\xi_1,\ldots,\xi_n)$ and we have $$\mathbb{E}(Y_n \mid \mathcal{F}_{n-1}) = Y_{n-1} \cdot \mathbb{E}\xi_1^3$$

  • $\mathbb{E}\xi_1^3 = 1$: $Y_n$ is a martingale.
  • $\mathbb{E}\xi_1^3 \geq 1$, $\xi_j \geq 0$ for all $j$: $Y_n$ is a sub-martingale.
  • $\mathbb{E}\xi_1^3 \not= 1$, $\xi_j \not \geq 0$ for all $j$: $Y_n$ is in general neither sub-martingale nor super-martingale.

There are some special cases which allow conclusion about $Y_n$:

  • If $(X_n)_n$ is a negative martingale, then $Y_n := X_n^3$ defines a super-martingale. This follows directly from Jensen inequality and the fact that $(-\infty,0] \ni x \mapsto x^3$ is concave.
  • If $(X_n)_n$ is a positive martingale, then we have that $(Y_n)_n$ is a sub-martingale.
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@did Thanks, you are right of course. $\mathbb{E}\xi_1=1$ implies $\mathbb{E}\xi_1^3 \geq 1$ anyway... –  saz Feb 26 '13 at 16:43
    
Indeed (but only if $\xi_1\geqslant0$ almost surely). +1. –  Did Feb 27 '13 at 7:07

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