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So I'm looking at the following:

$\operatorname{Cov}(X, X^2)$ where $X$ is the standard normal. So I know that $E[X] = 0$ and we have:

$$\operatorname{Cov}(X, X^2) = E(X \cdot X^2) - E[X] \cdot E[X^2] = E[X^3] - 0 \cdot E[X^2] = E[X^3]$$

From googling around, apparently this $= 0$, but I'm not sure why.

Any clarification here?

Thanks, Mariogs

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$\newcommand{\E}{\operatorname{E}}$Symmetry: $X$ has exactly the same distribution as $-X$, so $X^3$ has the same distribution as $-X^3$, and so $\E(X^3)=\E(-X^3)$. Hence $\E(X^3)=-\E(X^3)$, and only one number is minus itself.

(This won't work for things like the Cauchy distribution because the positive and negative parts in the expected value are both infinite; hence the expected value is undefined and things like the strong law of large numbers cannot be applied.)

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The standard normal distribution is symmetric about $0$, so $E[X^n] = 0$ for any odd positive integer $n$.

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  • $\begingroup$ Is there a proof of this / way to show it more rigorously? $\endgroup$ – bclayman Feb 16 '15 at 23:02
  • $\begingroup$ Write down the density, substitute $-x$ for $x$, ... $\endgroup$ – Robert Israel Feb 17 '15 at 0:29
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The standard normal distribution (probability distribution function $\frac{e^{-\frac{1}{2}x^2}}{\sqrt{2\pi}}$) is symmetric around $x=0$, since $x^3$ is antisymmetric, $E[X^3]=0$.

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