Any help would be appreciated.
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$\begingroup$ What does this mean? "$E(|X|X)$" makes no sense: it looks like a typographical error for the expectation of $|X|$ or possibly for the expectation of $X$ itself, which you have specified as $0.$ $\endgroup$– whuber ♦Commented Oct 17, 2019 at 21:21
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$\begingroup$ It is the expectation of the absolute value of $X$ times $X$. $\endgroup$– econ86Commented Oct 17, 2019 at 21:25
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$\begingroup$ The reason that needs explanation is due to the natural interpretation of the second bar as meaning a conditional expectation. Perhaps a well-placed pair of parentheses would help. $\endgroup$– whuber ♦Commented Oct 17, 2019 at 21:32
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$\begingroup$ You are right. Thank you! $\endgroup$– econ86Commented Oct 17, 2019 at 21:34
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$\begingroup$ $|X|\cdot X$ is odd and its absolute value is simply $X^2$. Now given a standard normal is symmetric about 0 with finite second moment, the result should follow immediately. $\endgroup$– Glen_bCommented Oct 17, 2019 at 23:07
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Using the symmetry of the standard normal distribution about $0$, if $X \sim N(0, 1)$, then $-X \sim N(0, 1)$ as well, so $$ \begin{aligned} E\big[|X|X\big] &= P(X > 0) E\big[|X|X \,\big|\, X > 0\big] + P(X \leq 0) E\big[|X|X \,\big|\, X \leq 0\big] \\ &= \frac{1}{2} E\big[X^2 \,\big|\, X > 0\big] - \frac{1}{2} E\big[X^2 \,\big|\, X \leq 0\big] \\ &= \frac{1}{2} E\big[X^2 \,\big|\, X > 0\big] - \frac{1}{2} E\big[\left(-X\right)^2 \,\big|\, -X \geq 0\big] \\ &= \frac{1}{2} E\big[X^2 \,\big|\, X > 0\big] - \frac{1}{2} E\big[X^2 \,\big|\, X \geq 0\big] \\ &= 0. \end{aligned} $$
This generalizes to the fact that if $f : \mathbb{R} \to \mathbb{R}$ is an odd function (i.e., $f(-x) = -f(x)$ for all $x \in \mathbb{R}$) and $X$ is a random variable with a distribution that is symmetric about zero, then $E[f(X)] = 0$, provided the expectation exists.
answered Oct 17, 2019 at 20:46
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$\begingroup$ Thanks! Last, in your generalization, how is it $E[f(X)]=0$? E.g. take $E(X^2)$ which for my random variable is $1$. $\endgroup$– econ86Commented Oct 17, 2019 at 20:55
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1$\begingroup$ @econ86 the generalization is for odd functions; the function $x \mapsto x^2$ is not odd $\endgroup$ Commented Oct 17, 2019 at 20:56
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$\begingroup$ Artem, Could you please explain what "$E[|X|X]$" means in your post and how "$X^2$" magically appeared in the second line? $\endgroup$– whuber ♦Commented Oct 17, 2019 at 21:21
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$\begingroup$ @whuber I interpreted $|X|X$ in the question to mean the absolute value of $X$ multiplied by $X$; i.e., $|X|X = \text{sign}(X)X^2$ $\endgroup$ Commented Oct 17, 2019 at 21:23
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1$\begingroup$ Thank you: I was stuck on interpreting the second bar as referring to a conditional expectation. $\endgroup$– whuber ♦Commented Oct 17, 2019 at 21:33