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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?

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    Isn't it an odd function?2011-12-10
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    The 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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    The 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

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