It's long time ago that I took the calculus class, so I dare to ask. If $X\sim N(0,1)$, what is $\mathbb{E}(1/X)$? $\mathbb{E}(1/X) = \int_{-\infty}^\infty \frac1x \cdot \frac1{\sqrt{2\pi}} \exp\left(-\frac{x^2}2\right) dx.$ Can I just claim $\mathbb{E}(1/X) = 0$ as $\frac1x \exp\left(-\frac{x^2}{2}\right)$ is an odd function even when it is not bounded?
What is $E[1/X]$ when $X$ is a standard normal random variable?
2
$\begingroup$
probability-distributions
-
0The is rather similar to asking for the expected value of the Cauchy distribution, which also gives 0 if we adopt the Cauchy PV; but that is not consider relevant/appropiate. In particular, if we use the Lebesgue integral (the sane thing to do in probability) it's clear that the integral does not exist. – 2011-12-10
1 Answers
3
You could speak of a Cauchy principal value. It's an integral whose positive and negative parts are both infinite, so the expected value does not exist.
$ \int_0^\infty \frac1x e^{-x^2/2}\;dx \ge \int_0^1 \frac1x e^{-1}\;dx = \infty, $ and similarly $ \int_{-\infty}^0 \frac1x e^{-x^2/2}\;dx = -\infty. $
The Cauchy principal value is $0$ because it's an odd function, i.e. $ \lim_{a\to0+}\left(\int_{-\infty}^{-a}+\int_a^\infty\right) \frac1x e^{-x^2/2} \; dx = 0. $
-
0Maybe I should add that limit theorems like laws of large numbers require actual integrals to exist; Cauchy principal values aren't good enough for those. – 2011-12-11