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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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. $$ \lim_{a\to0+}\left(\int_{-\infty}^{-a}+\int_a^\infty\right) \frac1x e^{-x^2/2} \; dx = 0. $$ |
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