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?
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probability-distributions
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2Isn't it an odd function? – 2011-12-10
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2The integral does not exist. There is no good reason to take existence in the PV sense as relevant. – 2011-12-10
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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