Expectation of the reciprocal of a standard normal random variable

Question

If $\mathbf{X} \sim_{iid} \mathcal{N}(\mu, 1)$ then we know that the sample mean $\bar{X} \sim \mathcal{N}(\mu, 1/n)$, how would we show that $$\mathbf{E}\left(\frac{1}{\bar{X}}\right) = \infty $$ and more generally for any normal random variable $X$ how would we show $\mathbb{E}\left(\frac{1}{X}\right) = \infty$, I am quite stuck and have tried just using the definition of expectation but my integral has given me no luck so far.

Maybe relevant: https://stats.stackexchange.com/questions/70045/mean-and-variance-of-the-reciprocal-of-a-random-variable — kjetil b halvorsen, Nov 19 '23 at 22:58
You don't have to calculate any integrals: you only need to demonstrate that some portion of a relevant integral diverges. — whuber, Nov 20 '23 at 13:15
The duplicate at https://stats.stackexchange.com/questions/299722/ive-heard-that-ratios-or-inverses-of-random-variables-often-are-problematic-in proves the expectation is undefined for all $\mu.$ This follows from the simple fact that the value of any Normal density in any finite interval always has a positive lower bound. — whuber, Nov 21 '23 at 15:24

Matt F. · Answer 1 · 2023-11-20T11:52:48.333

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If $\frac12>X>\frac14$, then $2<\frac1X<4$, so in particular $\frac1X>2$.

Similarly

\begin{align} E\left[\frac1X\right] &> 2P\left[\frac12>X>\frac14\right]+4P\left[\frac14>X>\frac18\right]+\cdots\\ &>2a(\frac12-\frac14)+4a(\frac14-\frac18)+\cdots\\ &=\frac{a}2 + \frac{a}2 +\cdots \end{align} where $a=\min(f(0),f(\frac12))$ and $f$ is the pdf of the given normal distribution.

Since a normal distribution has positive probability everywhere, $a$ is positive. So by adding enough terms to the above expression we can show that $E[1/X]$ is bigger than any specified quantity, i.e. $E[1/X]=\infty$.

edited Nov 20 '23 at 11:52

answered Nov 19 '23 at 22:59

Matt F.

4,726

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You need to review your first line ... – kjetil b halvorsen Nov 19 '23 at 23:02
@kjetilbhalvorsen, I don’t see an error in the first line – Matt F. Nov 20 '23 at 07:21
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Seriously, please take another, more careful look at your first line. X cannot be both between 1/4 and a half and between 2 and 4 at the same time. – Glen_b Nov 20 '23 at 09:40
Thank you, @Glen_b, for helping me see through that fog. – Matt F. Nov 20 '23 at 12:01
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Note that $1/X$ does not have expectation is very different from $E(1/X) = \infty$. So your conclusion (as well as OP's) is wrong. To see why, the very fist inequality "$E(1/X) > 2P(...)$" is wrong because $X$ is not a non-negative r.v.! – Zhanxiong Nov 20 '23 at 12:42
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+1 for the idea, which is excellent and is easily patched up to address the (valid) objections in the comments. For instance, invoke definitions in Lebesgue integration to point out that the expectation is undefined when the positive part of a random variable has a divergent integral. You cannot conclude, however, that $E[1/X]$ is itself infinite! – whuber Nov 20 '23 at 13:18

Expectation of the reciprocal of a standard normal random variable

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