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 $Y^2$?
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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2$\begingroup$ 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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2$\begingroup$ (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
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$\begingroup$ @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
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$\begingroup$ 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
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
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$\begingroup$ @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
set.seed(239326);x=rnorm(1000,10,.1);y=1/x;var(y)
) we obtain1.05e-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