Suppose $X$~$N(\mu,\sigma^2)$. Is the following statement correct: $\mathbb{E}\left(\frac{X}{X^2}\right)$ $\ne$ $\frac{\mathbb{E}(X)}{\mathbb{E}(X^2)}$?

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    $\begingroup$ Many threads here have discussed the distribution of $X/X^2=1/X$: because (no matter what $\mu$ and $\sigma\gt 0$ may be) $X$ has positive density around $X=0$, $1/X$ has an undefined expectation. That takes care of the left hand side. Can you evaluate the right hand side? $\endgroup$ – whuber Nov 8 '17 at 23:01
  • $\begingroup$ Thanks @whuber. Possibility of having undefined expectation is something I did not think through before asking the question. The right hand side is: $\frac{\mu}{\sigma^2+\mu^2}$ . My concern was whether the expectation of the ratio of a random variable can be written as the ratio of expectations of that random variable. In order to exclude the possibility of undefined expectation in above example, suppose that r.v. X takes only strictly positive values $x \in \mathbb{Z}^+$ (e.g. Gamma distributed). The inequality would be correct? $\endgroup$ – AlexMe Nov 9 '17 at 11:49
  • $\begingroup$ I found a nice relevant answer - See here: $\endgroup$ – AlexMe Nov 9 '17 at 11:58

The right hand side of your equality is a finite number (assuming $\sigma > 0$), while the left hand side simplifies to $\mathbb{E} \frac1X$, so do not exist, see for example I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?


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