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Consider the following problem:

Let $X \sim N(\mu, \sigma^2)$ and assume that $|\mu| \gg \sigma^2$. Then, we construct a new random variable $Y = 1/X$ with pdf

$$f_Y (y) = \frac{1}{\sqrt{2 \pi} \sigma y^2} \exp\left(-\frac{1}{2} \frac{\left(1/y - \mu\right)^2}{\sigma^2}\right).$$

Unfortunately, the first and higher moments of $Y$ do not exist. I am trying to show that if $\mu$ is sufficiently large (compared to $\sigma^2$) we should be able to show that $var (Y) < \infty$. Does that make any sense? Any idea if this could be shown?

Thank you very much.

By the way, this is anything but a proof but if you run in R:

r x = rnorm(10000000,10,1) var(1/x) you should systematically get a value around 0.0001 and this result will always decrease if you increase $\mu$. This seems to suggest that the variance of $Y$ could be bounded in some cases, no?

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    $\begingroup$ You claim $Y$ has no moments but you want to show that $\text{Var}(Y) < \infty$ in some cases? $\endgroup$
    – dsaxton
    Commented Feb 25, 2016 at 0:52
  • $\begingroup$ Yes, I know how this sounds! :) But for example, what would happen if $\mu/\sigma \rightarrow \infty$? In this case, could we have $var(Y)$ to bounded? $\endgroup$
    – user304347
    Commented Feb 25, 2016 at 1:33
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    $\begingroup$ As long as there is a positive density at zero, the variance will diverge! But , if $P(Y \le 0) is really, really small, you couks make an approximation, which might be practically usefull, by truncating the integral. Thgis is a way of admitting that the normal model at the outset ius an approximation. $\endgroup$
    – kjetil b halvorsen ♦
    Commented Feb 25, 2016 at 10:35
  • $\begingroup$ By definition, when you take a limit $\mu/\sigma\to\infty$, you are always looking at finite values of the ratios. You have already asserted the variance is infinite for any such value. Thus, you are asking how to compute the limit of any sequence of infinite values. The answer--which I hope is now obvious--is that it is still $\infty$. $\endgroup$
    – whuber ♦
    Commented Feb 25, 2016 at 15:10
  • $\begingroup$ Thank you very much for your comments. I am sorry to insist but what you would think of the following. If $P(X > 0) \approx 1$ we could argue that $$P(Y \leq y) = P(1/(\mu + \sigma Z) \leq y) \approx P(1 \leq y(\mu + \sigma Z)) = \Phi \left(\frac{y \mu - 1}{y\sigma}\right),$$ where $Z$ denotes a standard normal distribution. The density of $Y$ is then easy to obtain and we verify that its variance is finite. Therefore, I think I could be reasonable to have $Var(Y)$ is (approximately) bounded when $\mu/\sigma$ is sufficiently large, no? Thanks again! $\endgroup$
    – user304347
    Commented Feb 25, 2016 at 23:53

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