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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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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