I am confused in applying expectation in denominator.


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


5 Answers 5


can it be 1/E(X)?

No, in general it can't; 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 by other means.


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.

  • $\begingroup$ oh yes yes...I am very sorry that I could not apprehend that fact...I have one more q...Is this applicable to any kind of function???actually I am stuck with $|x|$...How can the expectation of $|x|$ can be deduced in terms of $E(x)$ and $V(x)$ $\endgroup$ May 8, 2014 at 11:58
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    $\begingroup$ I don't think you can use it for $|X|$ as that function is not differentiable. I would rather divide the problem into the cases and say $E(|X|) = E(X|X > 0)p(X>0) + E(-X|X < 0)p(X<0)$, I guess. $\endgroup$ May 8, 2014 at 20:35
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    $\begingroup$ @MatteoFasiolo Can you please explain why the symmetry of the distribution of $X$ (or lack thereof) has an effect on the accuracy of the Taylor approximation? Do you have a source that you could point me to that explains why this is? $\endgroup$ Jul 31, 2017 at 11:03
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    $\begingroup$ @AaronHendrickson my reasoning is simply that the next term in the expansion is proportional to $E\{(X-E(X))^3\}$ which is related to the skewness of the distribution of $X$. Skewness is an asymmetry measure. However, zero skewness does not guarantee symmetry and I am not sure whether symmetry guarantees zero skewness. Hence, this is all heuristic and there might be plenty of counterexamples. $\endgroup$ Aug 1, 2017 at 12:50
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    $\begingroup$ I don't understand how this solution gets so many upvotes. For a single random variable $X$ there is no justificiation about the quality of this approximation. The third derivative $f(x)=1/x$ is not bounded. Moreover the remainder of the approx. is $1/6f'''(\xi)(X-\mu)^3$ where $\xi$ is itself a random variable between $X$ and $\mu$. The remainder won't vanish in general and may be very huge. Taylor approx. may only be useful if one has sequence of random variables $X_n -\mu = O_p(a_n)$ where $a_n \to 0$. Even then uniform integrability is needed additionally if interested in the expectation. $\endgroup$
    – BloXX
    Sep 20, 2017 at 10:19

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 $\frac1{\E X}=\frac11=1$.


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 $$ \int_0^\infty E\left[e^{-\lambda X}\right] d\lambda =E\left[\frac{1}{X}\right]. $$

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    $\begingroup$ The idea here is right, but the details wrong. Pleasecheck $\endgroup$ Aug 7, 2017 at 19:56
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    $\begingroup$ @Kjetil I don't see what the problem is: apart from the inconsequential differences of using $t X$ instead of $-t X$ in the definition of the MGF and naming the variable $t$ instead of $\lambda$, the answer you just posted is identical to this one. $\endgroup$
    – whuber
    Aug 8, 2017 at 19:48
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    $\begingroup$ You are right, the problems was less than I thought. Still this answer would be better withm some more details. I will upvote this tomorrow ( when I have new votes) $\endgroup$ Aug 8, 2017 at 20:19
  • $\begingroup$ This is not correct. For example, if $X$ has exponential distribution with mean $\frac1\mu$ then its moment-generating function is $$ \mathbb E[e^{-\lambda X}] = \int_0^\infty e^{\lambda x}\mu e^{-\mu x}\ \mathsf dx = \frac\mu{\mu-\lambda}, $$ but the integral only converges for $\lambda<\mu$. So the integral of the moment-generating function over $(0,\infty)$ cannot possible converge - and indeed, $\mathbb E\left[\frac 1X\right]$ does not exist. $\endgroup$
    – Math1000
    Nov 22, 2020 at 2:03

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

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


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