Expectation of reciprocal of a variable

Question

I am confused in applying expectation in denominator.

$E(1/X)=\,?$

can it be $1/E(X)\,$?

Answer 1

can it be 1/E(X)?

No, in general it can't.

The first thing when pondering such things is to try a few simple cases/examples. To begin with, consider the simplest nondegenerate concrete example, a distribution with $\frac12$ probability on two distinct values, say $1$ and $2$.

$E(X) = \frac32$ so $1/E(X) = \frac23$ but $E(\frac{1}{X}) = \frac{\frac11 + \frac12}{2}=\frac{3}{4}$. Clearly the answer is "no" for that case. You might like to make a few additional ones of your own.

If you prefer to simulate, that takes but a moment. In R, we have:

> 1/mean(sample(1:2,1000000,replace=TRUE))
[1] 0.6664441
> mean(sample(1/(1:2),1000000,replace=TRUE))
[1] 0.7496865

Close to the above results. You can repeat this simulation several times to see that it's not some fluke. Additional examples are simple enough. For example, consider a uniform on (0.5,2):

> 1/mean(runif(1000000,0.5,2))
[1] 0.8000138
> mean(1/runif(1000000,0.5,2))
[1] 0.9242307

Again, repetition will convince you that it's not a fluke and that these sample averages from simulation are barely changing. By inspection we can see that in the first calculation the uniform has expected value (2.5)/2, so its reciprocal of expectation is 0.8, and some simple algebra establishes that the reciprocal has expected value $\frac23\log 4 \approx 0.9242$. We got nice agreement from the simulations.

Try a couple of cases of your own, either algebraically or via simulation.

Jensen's inequality tells us that if $X$ is a random variable and $\varphi$ is a convex function, then $\varphi(\text{E}[X]) \leq \text{E}\left[\varphi(X)\right]$. If $X$ is strictly positive, then $1/X$ is convex, so $\text{E}[1/X]\geq 1/\text{E}[X]$, and for a strictly convex function, equality only occurs if $X$ has zero variance ... so in cases we tend to be interested in, the two are generally unequal.

Assuming we're dealing with a positive variable, if it's clear to you that $X$ and $1/X$ will be inversely related ($\text{Cov}(X,1/X)\leq 0$) then this would imply $E(X \cdot 1/X) - E(X) E(1/X) \leq 0$ which implies $E(X) E(1/X) \geq 1$, so $E(1/X) \geq 1/E(X)$.

I am confused in applying expectation in denominator.

Use the law of the unconscious statistician

$$\text{E}[g(X)] = \int_{-\infty}^\infty g(x) f_X(x) dx$$

(in the continuous case)

so when $g(X) = \frac{1}{X}$, $\text{E}[\frac{1}{X}]=\int_{-\infty}^\infty \frac{f(x)}{x} dx$

In some cases the expectation can be evaluated by inspection (e.g. with gamma random variables), or by deriving the distribution of the inverse, or as some other answerers considered, by using the MGF, or indeed by other means.

Answer 2

As Glen_b says that's probably wrong, because the reciprocal is a non-linear function. If you want an approximation to $E(1/X)$ maybe you can use a Taylor expansion around $E(X)$:

$$ E \bigg( \frac{1}{X} \bigg) \approx E\bigg( \frac{1}{E(X)} - \frac{1}{E(X)^2}(X-E(X)) + \frac{1}{E(X)^3}(X - E(X))^2 \bigg) = \\ = \frac{1}{E(X)} + \frac{1}{E(X)^3}Var(X) $$ so you just need mean and variance of X, and if the distribution of $X$ is symmetric this approximation can be very accurate.

EDIT: the maybe above is quite critical, see the comment from BioXX below.

Answer 3

Others have already explained that the answer to the question is NO, except trivial cases. Below we give an approach to finding $\DeclareMathOperator{\E}{\mathbb{E}} \E \frac1{X}$ when $X>0$ with probability one, and the moment generating function $M_X(t) = \E e^{tX}$ do exist. An application of this method (and a generalization) is given in Expected value of $1/x$ when $x$ follows a Beta distribution, we will here also give a simpler example.

First, note that $\int_0^\infty e^{-t x}\; dt = \frac1{x}$ (simple calculus exercise). Then, write $$ \E \left(\frac1{X}\right) = \int_0^\infty x^{-1} f(x)\; dx = \int_0^\infty \left( \int_0^\infty e^{-tx}\; dt \right) f(x)\; dx =\\ \int_0^\infty \left( \int_0^\infty e^{-tx} f(x) \; dx \right) \; dt = \int_0^\infty M_X(-t) \; dt $$ A simple application: Let $X$ have the exponential distribution with rate 1, that is, with density $e^{-x}, x>0$ and moment generating function $M_X(t)=\frac1{1-t}, t<1$. Then $\int_0^\infty M_X(-t)\; dt = \int_0^\infty \frac1{1+t} \; dt= \ln(1+t) \bigg\rvert_0^\infty = \infty$, so definitely do not converge, and is very different from $\frac1{\E X}=\frac11=1$.

Answer 4

An alternative approach to calculating $E(1/X)​$ knowing X is a positive random variable is through its moment generating function $E\left[e^{-\lambda X}\right]​$. Since by elementary calculus $$ \int_0^\infty e^{-\lambda x} d\lambda =\frac{1}{x} $$ we have, by Fubini's theorem $$ \int_0^\infty E\left[e^{-\lambda X}\right] d\lambda =E\left[\frac{1}{X}\right]. $$

Answer 5

To first give an intuition, what about using the discrete case in finite sample to illustrate that $\text{E}(1/X)\neq 1/\text{E}(X)$ (putting aside cases such as $\text{E}(X)=0$)?

In finite sample, using the term average for expectation is not that abusive, thus if one has on the one hand

$\text{E}(X) = \frac{1}{N}\sum_{i=1}^N X_i$

and one has on the other hand

$\text{E}(1/X) = \frac{1}{N}\sum_{i=1}^N 1/X_i$

it becomes obvious that, with $N>1$,

$\text{E}(1/X) = \frac{1}{N}\sum_{i=1}^N 1/X_i \neq \frac{N}{\sum_{i=1}^N X_i} = 1/\text{E}(X)$

Which leads to say that, basically, $\text{E}(1/X)\neq 1/\text{E}(X)$ since the inverse of the (discrete) sum is not the (discrete) sum of inverses.

Analogously in the asymptotic $0$-centered continuous case, one has

$\text{E}(1/X)=\int_{-\infty}^\infty \frac{f(x)}{x} dx \neq 1/\int_{-\infty}^\infty xf(x) dx = 1/\text{E}(X)$.