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I am trying to show that the expectation of $\frac{1}{X^{2}}$ does not exist if $X$ is standard normal, but do not know how....could anyone help, please? The integration is $\int_{-\infty}^{\infty} \frac{1}{X^{2}}e^{-\frac{X^2}{2}}dx$, and obviously the value of $\frac{1}{X^2}$ goes to infinity around $0$. So the integral should be improper, but how to state the logic? Thanks!

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2 Answers 2

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Here is a hint: Show that on the interval $(0,1]$, $1/2 < e^{-x^2/2} < 1$, and consequently, $$0 < \frac{1}{2x^2} < \frac{e^{-x^2/2}}{x^2}.$$ Since the integrand $e^{-x^2/2}/x^2 > 0$ for all $x \in \mathbb R$, we can establish the inequalities $$\mathrm{E}[X^{-2}] > \int_{x=0}^1 \frac{e^{-x^2/2}}{x^2} \, dx > \int_{x=0}^1 \frac{1}{2x^2} \, dx.$$ What is your conclusion?

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  • $\begingroup$ did it! thanks a lot ! $\endgroup$
    – Amber Xue
    Sep 4, 2014 at 16:02
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Hint: Prove that the expectation is divergent. A necessary condition for the existence of $E(X)$ is the $\int |x|f(x)dx$ has to be convergent.

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