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.
1 Answer
$\begingroup$
$\endgroup$
6
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. $E[1/X]=\infty$.
-
1$\begingroup$ You need to review your first line ... $\endgroup$ Commented Nov 19, 2023 at 23:02
-
$\begingroup$ @kjetilbhalvorsen, I don’t see an error in the first line $\endgroup$– user225256Commented Nov 20, 2023 at 7:21
-
3$\begingroup$ 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. $\endgroup$– Glen_bCommented Nov 20, 2023 at 9:40
-
1$\begingroup$ 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.! $\endgroup$ Commented Nov 20, 2023 at 12:42
-
1$\begingroup$ +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! $\endgroup$– whuber ♦Commented Nov 20, 2023 at 13:18
self-studytag wiki info $\,$ 2. The premise of the question is false as framed. (If $\mu<0$, it doesn't come out to $\infty$ and if $\mu=0$ it's undefined in the sense that you get the limiting value of the difference of two things that each go to infinity) $\,$ 3. "My integral has given me no luck" is overly vague. What did you try, exactly? Did you consider what a value in the integral-limits of $\infty$ actually means in terms of limits? ($\infty$ is not a number) ... en.wikipedia.org/wiki/Improper_integral $\endgroup$