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Given a R.V $X$ and that $\ E(X) = 0 $ and $\ E(X^2) = \sigma^2 $. Is there anyway to compute $\ E(X^3) $ without knowing the density function of $X$? Is it possible to tell from these if $\ E(X^3) > 0$?

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You could ask this at – Applied mathematician Nov 9 '12 at 9:51
up vote 1 down vote accepted

No - if $\sigma^2 \gt 0$ then $E[X^3]$ can take any finite value, or be positively or negatively infinite, or be undefined (e.g. the calculation might hit $\infty-\infty$).

For a distribution symmetric about $0$, the third moment will be $0$ or undefined.

In particular, if $\ E[X] = 0 $ and $\ E[X^2] = \sigma^2 $ and $\ E[X^3] \gt 0$ then $Y=-X$ has $\ E[Y] = 0 $ and $\ E[Y^2] = \sigma^2 $ and $\ E[Y^3] \lt 0$ and you do not seem to have any information to distinguish them.

You can say $E[X^4] \ge \sigma^4.$

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