# How to compute $E[ (|X|) X]$ when $X \sim N(0,1)$?

Any help would be appreciated.

• 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 $0.$ – whuber Oct 17 '19 at 21:21
• It is the expectation of the absolute value of $X$ times $X$. – econ86 Oct 17 '19 at 21:25
• 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. – whuber Oct 17 '19 at 21:32
• You are right. Thank you! – econ86 Oct 17 '19 at 21:34
• $|X|\cdot X$ 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. – Glen_b Oct 17 '19 at 23:07

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.
• Thanks! Last, in your generalization, how is it $E[f(X)]=0$? E.g. take $E(X^2)$ which for my random variable is $1$. – econ86 Oct 17 '19 at 20:55
• @econ86 the generalization is for odd functions; the function $x \mapsto x^2$ is not odd – Artem Mavrin Oct 17 '19 at 20:56
• Artem, Could you please explain what "$E[|X|X]$" means in your post and how "$X^2$" magically appeared in the second line? – whuber Oct 17 '19 at 21:21
• @whuber I interpreted $|X|X$ in the question to mean the absolute value of $X$ multiplied by $X$; i.e., $|X|X = \text{sign}(X)X^2$ – Artem Mavrin Oct 17 '19 at 21:23