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Let $X$ be a random variable. Is it ever true that $$E\left({\frac 1 X}\right) \stackrel{?}{=} \frac 1 {E(X)} \text{ ?}$$

I'll assume $X$ never takes on the value of $0$.

I'll use the notation for discrete RV's, otherwise replace $\sum$ with $\int$. Let $f$ be the mass/density function.

$$E\left({\frac 1 X}\right) = \sum \frac 1 x f(x)$$

$$\frac 1 {E(X)} = \frac 1 {\sum x f(x)}$$

Assume they're equal. Then:

$$\sum \frac 1 x f(x) = \frac 1 {\sum x f(x)}$$

$$\implies \left(\sum \frac 1 x f(x)\right)\left(\sum xf(x)\right) = 1$$

This is a type of convolution, which would mean that $$\dfrac {f(x)}{x} = (x f(x))^{-1}$$

in regards to this convolution. Is that possible? Do functions have inverses under convolution?

GFauxPas
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    Do you want to assume $X>0$? – Chappers Jun 06 '17 at 15:00
  • @Chappers I'll assume $X \ne 0$ otherwise the question ends before it starts. I'll add that – GFauxPas Jun 06 '17 at 15:01
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    @GFauxPas : Note that I deleted the assertion about constant random variables from my answer. That would hold only if the constant is $1$ or $-1. \qquad$ – Michael Hardy Jun 06 '17 at 15:15
  • For every positive, non deterministic, random variable $X$, $$E(X)\cdot E\left(\frac1X\right)>1$$ For every nonzero deterministic random variable $X$, obviously, $$E(X)\cdot E\left(\frac1X\right)=1$$ hence this is the only equality case. – Did Jun 06 '17 at 15:31

1 Answers1

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By Jensen's inequality, for a positive-valued random variable $X$ we have

$$E\left(\frac{1}{X}\right)\ge \frac1{E(X)}$$

If $X$ is negative-valued, then the inequality is reversed.

This is because the map $x\mapsto \frac{1}{x}$ is convex for $x>0$ and concave for $x<0$. Equality holds precisely when $X$ is constant with probability $1$.

However equality can hold if we do not restrict ourselves to either positive-valued or negative-valued random variables. As a non-trivial example, consider a discrete random variable $X$ with $$P(X=-1)=\frac{1}{9}\,,\, P\left(X=\frac{1}{2}\right)=P(X=2)=\frac{4}{9}$$

Then indeed $$E\left(\frac{1}{X}\right)=\frac{1}{E(X)}=1$$

For details, see https://stats.stackexchange.com/q/305713/119261.

StubbornAtom
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  • And how do we know equality iff constant? – GFauxPas Jun 06 '17 at 16:12
  • That I think follows from the complete statement of Jensen's inequality itself: https://math.stackexchange.com/questions/1160095/convexity-and-equality-in-jensen-inequality, https://math.stackexchange.com/questions/628386/when-jensens-inequality-is-equality. – StubbornAtom Jun 06 '17 at 16:14
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    Assuming all expectations exist of course. – StubbornAtom May 15 '20 at 14:47