2
$\begingroup$

In this post it is stated that due to Jensen's inequality the expected value of the reciprocal of a strictly postive random variable $X$ will satisfy:

$$\mathbb{E}\left[\frac{1}{X}\right] \geq \frac{1}{\mathbb{E}[X]}$$

My question is whether a similar inequality exists for the variance of $1/X$?

$\endgroup$
  • 2
    $\begingroup$ No, because $E\left[\frac 1X\right]$ can be unbounded and so the variance of $\frac 1X$ is kind of difficult to define. You need more restrictive conditions on $X$ than mere strict positivity. $\endgroup$ – Dilip Sarwate Jan 15 '18 at 17:09
4
$\begingroup$

Here is an answer (to a related question) which provides a valid generalization of $\mathbb{E}\left[\frac{1}{X}\right] \geq \frac{1}{\mathbb{E}[X]}$.

What is the relation between $\mathbb{E}X^r$ and $\mathbb{[E}X]^r$, for all possible values of $r$, when $X$ is a positive random variable? This can be answered by applying Jensens's inequality, based on the convexity or concavity of $x^r$ as a function of $x$, depending on the value of $r$.

The following presumes the relevant moments exist.

$\mathbb{E}X^r \ge \mathbb{[E}X]^r$, for $r \ge 1$ or $r \le 0$ ($x^r$ is convex in both cases)

$\mathbb{E}X^r \le \mathbb{[E}X]^r$, for $0 \le r \le 1$ ($x^r$ is concave, which includes the frequently used square root)

Note that equality holds for $r = 1$, which says $\mathbb{E}X = \mathbb{E}X$, and for $r = 0$, which says $1 = 1$. $\mathbb{E}\left[\frac{1}{X}\right] \geq \frac{1}{\mathbb{E}[X]}$ of course corresponds to $r = -1$.

Going back to the OP's original question, as shown by @kjetil b halvorsen , the analog for variance does not hold. But presuming the moments exist, we see by applying the above results with $r= -2$, that $\mathbb{E}\left[\frac{1}{X^2}\right] \geq [\frac{1}{\mathbb{E}X}]^2$

$\endgroup$
4
$\begingroup$

First, @Dilip Sarwate comments, for such ratio variables mean and variance often do not exist and then there is little to expect. For a detailed discussion of this see I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?. But, let us assume a case where expectation and variance exists.

Your question is My question is whether a similar inequality exists for the variance of $1/X$? It is not totally clear what you mean with similar but let us take it literally, that is, is it true that $$\DeclareMathOperator{\V}{\mathbb{V}} \V(\frac1{X})\cdot \V X \ge 1 \quad \text{?} $$ In that form it is clearly false, for instance take $X$ to have a uniform distribution on a very short interval close to 1, like $[0.9, 1.1]$. For that case the product of the two variances will be 0.1127601010e-4 falsifying the inequality. But maybe you where thinking of some other generalization?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.