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My random variable $X$ is normally distributed with mean $b$ and variance $p$. I defined a new RV, $Y$, such that $Y=1/X$. Does anyone know how to find the mean of $Y$ and $Y^2$?

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  • $\begingroup$ Relevant: stats.stackexchange.com/questions/41896/… $\endgroup$ – Glen_b Sep 15 '13 at 2:08
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    $\begingroup$ ctd... for example, the Taylor expansion to first order suggests (assuming I didn't make an error) that $\text{Var}(1/X)$ should be approximately $\text{Var}(X)/\mu^4$. Taking $\sqrt p = .1$, and $b=10$ (CV=0.01), we get the approximate answer of $\text{Var}(1/X)\approx 1\times 10^{-6}$, and simulating in R (set.seed(239326);x=rnorm(1000,10,.1);y=1/x;var(y)) we obtain 1.05e-6. It works pretty well under a variety of values for $p$ and $b$, as long as the CV is small. If Taylor expansion doesn't work, we're left to account for its success under the conditions I suggested above. $\endgroup$ – Glen_b Sep 15 '13 at 23:22
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    $\begingroup$ It may be that we're dealing with a case where the expectation of the remainder isn't small but the remainder itself is nearly always small (e.g. where the middle 99.99...% of the distribution of the remainder is small); that could yield a situation where the low-order Taylor approximation is highly useful in practice (in that, essentially every time you try it with small coefficient of variation, it actually works quite well), even though it doesn't work 'on average' because of a extremely small chance of an even more extremely large result. $\endgroup$ – Glen_b Sep 15 '13 at 23:58
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    $\begingroup$ Dear @Glen_b: From a mathematical point of view, the Taylor series doesn't work, regardless of the coefficient of variation. Since this is a question about a mathematical property, I think it's important to point this out. Regarding "practicalities", I think that, in certain circumstances, one could easily argue the opposite---the outlined approach could be highly dangerous precisely because it appears like it might work the vast majority of the time. But, when it doesn't, the error is catastrophic. (The relevance of such an argument would depend on the application, of course.) $\endgroup$ – cardinal Sep 16 '13 at 0:46
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    $\begingroup$ @cardinal Certainly the required expectations don't exist (and the Taylor series for the expectations cannot work, as you say), which is all that need be said for the question at hand. My remaining questions about why the sample moments appear to work so well should go elsewhere (and will clarify how to modify at least one answer of mine elsewhere), but will have to wait until I can formulate a better question. I'm uncertain which comments should be deleted and which should remain (here and elsewhere); I'm happy to leave those where the explanatory responses would be relevant to others. $\endgroup$ – Glen_b Sep 16 '13 at 2:11
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This is equivalent to asking for $\int_{- \infty}^\infty \frac{1}{x} P(X=x) \, dx$ and $\int_{- \infty}^\infty \frac{1}{x^2} P(X=x) \, dx$.

Considering the second, let's choose some arbitrary $a > 0$: $$\eqalign { \int_{- \infty}^\infty \frac{1}{x^2} P(X=x) \, dx &= \int_{- \infty}^{-a} \frac{1}{x^2} P(X=x) \, dx\\ &+ \int_{-a}^a \frac{1}{x^2} P(X=x) \, dx + \int_a^\infty \frac{1}{x^2} P(X=x) \, dx \\ &\ge \int_{-a}^a \frac{1}{x^2} P(X=x) \, dx \\ &\ge \int_{-a}^a \frac{1}{x^2} \min\left(P(X=-a), P(X=a)\right) \, dx \\ &= \min\left(P(X=-a), P(X=a)\right) \int_{-a}^a \frac{1}{x^2} \, dx \to \infty }$$ where first we split up the integral and note that the tails are nonnegative, and then use the fact that the normal pdf is quasiconcave. So $\mathbb{E}\left[\frac{1}{X^2}\right]$ doesn't exist.

We can do exactly the same thing to show that $\mathbb{E}\left[\frac{1}{\lvert X \rvert}\right]$ doesn't exist (since $\int_{0}^a \frac{1}{X} dx$ also diverges). But $\frac{1}{X}$ is a little trickier, since one of the tail integrals is negative-valued. Still, $\mathbb{E}[\frac{1}{X} \mid \frac{1}{X} < a]$ doesn't exist, since the integral diverges. I think that, because $P(\frac{1}{X} < a) > 0$, using basically the same argument as the law of total expectation, this then implies that the overall expectation cannot exist. But I'm not 100% on that.

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    $\begingroup$ There is a cute application of quasiconcavity here. Some comments: (1) Essentially the same argument works with a much weaker hypothesis, though. As long as the density is continuous and nonzero at $x = 0$, that is already enough. This follows because a nonzero minimum is attained on $[-a,a]$ for some $a > 0$ and the rest falls out. (2) Because both $\mathbb E \max(X^{-1},0)$ and $\mathbb E -\min(X^{-1},0)$ are infinite, then $\mathbb E X^{-1}$ does not exist. (3) Strictly speaking, $\mathbb EX^{-2}$ does exist since the integrand is nonnegative, but it is not finite. (cont.) $\endgroup$ – cardinal Sep 15 '13 at 13:39
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    $\begingroup$ (cont.) (4) It is clear that you are using the notation $\mathbb P(X = x)$ to refer to the density here, but a little caution might be advised. (+1) :-) $\endgroup$ – cardinal Sep 15 '13 at 13:41
  • $\begingroup$ @cardinal 1) Yes, absolutely. 2-3) Okay; I've seen "does not exist" and "infinite" used interchangeably, e.g. I think in Feller, but the distinction makes sense. 4) Whoops, yes, that should have been a lowercase p -- it was very late at night when I wrote this. :) $\endgroup$ – Danica Sep 15 '13 at 19:30
  • $\begingroup$ In case I didn't state it very clearly, what I intended for point 2 was simply that you need not be less than 100% confident about the result you arrived at! Cheers. :-) $\endgroup$ – cardinal Sep 16 '13 at 0:48
  • $\begingroup$ To second @cardinal's argument, I also think it is enough to prove that $\mathbb{E}[|X|^{-1}]$ does not exist (or more precisely is infinite) to conclude that $\mathbb{E}[X^{-1}]$ does not exist, by using the $|X|^-1= \max(X^{-1},0)-\min(X^{-1},0)$ decomposition. $\endgroup$ – Xi'an Jun 30 '16 at 6:17
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Actually, I wrote a paper in 1991 addressing this issue, available on Research Gate. In short, the mean and variance do not exist.

More precisely, if one defines a generalised inverse normal distribution by its density ($\alpha>1,\sigma>0$) $$f(z|\mu,\sigma,\alpha) = \dfrac{K(\alpha,\mu,\sigma)}{|z|^\alpha}\exp\left\{-(z^{-1}-\mu)^2/2\sigma^2\right\}$$ then $$K(\alpha,\mu,\sigma)^{-1}=\sigma^{\alpha-1}e^{-\mu^2/2\sigma^2}2^{(\alpha-1)/2}\Gamma((\alpha-1)/2){}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)$$ where ${}_1F_1$ is a confluent hypergeometric function. From this derivation of the normalising constant, one deduces that the mean only exists for $\alpha>2$ (while the inverse normal distribution corresponds to $\alpha=2$) and the variance only exists for $\alpha>3$. The mean is given by $$\mathbb{E}[X]=\frac{\mu}{\sigma^2}\,\frac{{}_1F_1\left(\frac{1}{2}(\alpha-3);\frac{3}{2};\frac{\mu^2}{2\sigma^2}\right)}{{}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)}$$

A much simpler argument as to why the mean does not exist is that, if $f$ is the normal $\text{N}(b,p)$ density, $$x^{-1} f(x)\equiv \frac{e^{-b^2/2p}}{\sqrt{2\pi p}}x^{-1}$$ at zero, which is not integrable.

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  • $\begingroup$ @Xian Do you have a solution for the Cauchy principal value of $\mathbb{E}[X]$ when $\alpha=2$? $\endgroup$ – Aaron Hendrickson Jul 22 '17 at 19:55
  • $\begingroup$ @AaronHendrickson: I do not see how this is connected with this question or my answer. But more to the point, I do not have a solution! $\endgroup$ – Xi'an Jul 22 '17 at 20:35

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