Jensen Bounds for Reciprocal Expectations

Why Reciprocal Expectations Need Care

For a positive random variable, the reciprocal $1/X$ gives greater weight to values near zero than to values far from zero. Consequently, knowing the mean of $X$ does not mean that $E[1/X]$ is simply the reciprocal of that mean. Convexity instead gives a lower bound: under suitable conditions, the expected reciprocal is at least the reciprocal of the expectation.

The support condition is essential. The function $x\mapsto 1/x$ is convex on the positive half-line, but it is not defined at zero and is not convex across the whole real line. We will require $X>0$ with probability one, and we will distinguish the condition that makes the lower bound meaningful from the stronger condition that makes $E[1/X]$ finite.

Definition: For a random variable $X$ satisfying $P(X>0)=1$, its reciprocal $1/X$ is a positive random variable. Its expectation $E[1/X]$ is defined as a nonnegative extended expectation and may be $+\infty$. If $0<E[X]<\infty$, the reciprocal Jensen bound is $$ E\!\left[\frac{1}{X}\right]\geq \frac{1}{E[X]}. $$

Convexity and the Bound

Let $\phi(x)=1/x$ for $x>0$. Its first two derivatives are $$ \phi'(x)=-\frac{1}{x^2}, \qquad \phi''(x)=\frac{2}{x^3}>0. $$ Thus $\phi$ is strictly convex on its domain. Jensen's inequality for a convex function gives the reciprocal bound when the expectation is defined; the tangent-line argument below also handles the case in which $E[1/X]$ is infinite.

Theorem: Suppose $P(X>0)=1$ and $m=E[X]<\infty$. Then $m>0$ and $$ E\!\left[\frac{1}{X}\right]\geq \frac{1}{m}, $$ where the left side may be $+\infty$. Equality holds if and only if $P(X=m)=1$. If $X$ is not constant with probability one, the inequality is strict.

The assumptions have distinct roles. Positivity ensures that $1/X$ is defined and that the convexity argument applies. Finiteness of $E[X]$ ensures that $1/E[X]$ is a finite number. No assumption that $E[1/X]<\infty$ is needed for the inequality itself: if it is infinite, the inequality still holds, but it is not a finite numerical bound on that expectation.

Proof by a Tangent-Line Bound

The tangent line to $\phi(x)=1/x$ at $x=m>0$ is $\phi(m)+\phi'(m)(x-m)=2/m-x/m^2$. Convexity says the graph lies above this line. In this case the difference has a particularly useful exact form.

\frac{1}{x}-\left(\frac{2}{m}-\frac{x}{m^2}\right) =\frac{(x-m)^2}{xm^2}\geq 0 \qquad (x>0).

Because $X>0$ with probability one and $E[X]=m<\infty$, the tangent-line expression $2/m-X/m^2$ is integrable. The difference on the left is nonnegative, so its expectation is well-defined, possibly infinite. Taking expectations in the pointwise inequality gives

E\!\left[\frac{1}{X}\right] \geq E\!\left[\frac{2}{m}-\frac{X}{m^2}\right] =\frac{2}{m}-\frac{E[X]}{m^2} =\frac{1}{m}.

For the equality condition, the pointwise difference is zero exactly when $X=m$. If $P(X\ne m)>0$, the nonnegative difference is positive with positive probability, so its expectation is strictly positive. The bound is then strict. If $X=m$ with probability one, $E[1/X]=1/m$, so equality holds. Also, $m>0$: a nonnegative random variable that is positive with probability one cannot have expectation zero. This proves the theorem. $\square$

Worked Examples

Worked Example: A Two-Point Positive Distribution

Suppose $P(X=1)=P(X=4)=1/2$. Compute the reciprocal expectation and compare it with the reciprocal of the mean.

First calculate the mean: $$ E[X]=\frac{1}{2}(1)+\frac{1}{2}(4)=\frac{5}{2}. $$ Therefore, $$ \frac{1}{E[X]}=\frac{1}{5/2}=\frac{2}{5}. $$

Now calculate the expectation of the reciprocal: $$ E\!\left[\frac{1}{X}\right] =\frac{1}{2}\left(\frac{1}{1}\right) +\frac{1}{2}\left(\frac{1}{4}\right) =\frac{1}{2}+\frac{1}{8} =\frac{5}{8}. $$ Since $5/8>2/5$, the bound holds strictly. The variable takes two distinct values with positive probability, so it is not constant; strictness also follows from the equality condition in the theorem.

Worked Example: An Exponential Variable With Infinite Reciprocal Expectation

Let $X$ have an exponential distribution with rate $\lambda>0$, so its density is $f(x)=\lambda e^{-\lambda x}$ for $x>0$, and $E[X]=1/\lambda$. Determine whether the reciprocal bound is finite.

The right side is $$ \frac{1}{E[X]}=\lambda. $$ For the left side, the density formula gives $$ E\!\left[\frac{1}{X}\right] =\int_0^\infty \frac{1}{x}\lambda e^{-\lambda x}\,dx. $$ On $0<x<1$, $e^{-\lambda x}\geq e^{-\lambda}$, so $$ \int_0^\infty \frac{\lambda e^{-\lambda x}}{x}\,dx \geq \lambda e^{-\lambda}\int_0^1\frac{1}{x}\,dx =+\infty. $$ Thus $E[1/X]=+\infty$, even though $E[X]$ is finite. The reciprocal Jensen inequality is valid, but it only says $+\infty\geq\lambda$; it does not imply that the reciprocal expectation is finite. The divergence comes from the density assigning positive mass arbitrarily close to zero, where $1/x$ becomes unbounded.

Worked Example: Equality for a Constant

Suppose $X=3$ with probability one. Then $$ E[X]=3, \qquad \frac{1}{E[X]}=\frac{1}{3}, \qquad E\!\left[\frac{1}{X}\right]=\frac{1}{3}. $$ The two sides are equal because $X$ is constant. This checks the equality case directly and shows why the theorem allows equality, rather than asserting a strict lower bound in every case.

A Second Derivation When the Reciprocal Is Integrable

There is another way to see the bound using Cauchy–Schwarz. This derivation requires $E[1/X]<\infty$, so it does not cover the exponential example above. Under that additional assumption, both $\sqrt{X}$ and $1/\sqrt{X}$ have finite second moments: $E[(\sqrt{X})^2]=E[X]<\infty$ and $E[(1/\sqrt{X})^2]=E[1/X]<\infty$.

Derived inequality: If $P(X>0)=1$, $E[X]<\infty$, and $E[1/X]<\infty$, then $$ 1\leq E[X]E\!\left[\frac{1}{X}\right]. $$

Indeed, $\sqrt{X}(1/\sqrt{X})=1$ with probability one. Cauchy–Schwarz therefore gives $$ 1=\left(E\!\left[\sqrt{X}\frac{1}{\sqrt{X}}\right]\right)^2 \leq E[X]E\!\left[\frac{1}{X}\right]. $$ Since $E[X]>0$, dividing by it yields $E[1/X]\geq1/E[X]$. This proof confirms the same finite-expectation bound, while the tangent-line proof is more general because it also permits $E[1/X]=+\infty$. Equality in this Cauchy–Schwarz inequality requires $\sqrt{X}$ and $1/\sqrt{X}$ to be proportional with probability one, which forces $X$ to be constant.

Why Positivity Matters

The reciprocal bound is not a general statement about every random variable for which $1/X$ happens to be defined. Its justification uses convexity on $(0,\infty)$, so values of both signs cannot simply be included.

Worked Example: A Variable With Both Signs

Let $P(X=-1)=P(X=3)=1/2$. The variable is never zero, but it is not positive with probability one. Its mean is $$ E[X]=\frac{1}{2}(-1)+\frac{1}{2}(3)=1, \qquad \frac{1}{E[X]}=1. $$ Its reciprocal expectation is $$ E\!\left[\frac{1}{X}\right] =\frac{1}{2}(-1)+\frac{1}{2}\left(\frac{1}{3}\right) =-\frac{1}{2}+\frac{1}{6} =-\frac{1}{3}. $$ Here $-1/3<1$. This does not contradict the theorem: $X$ takes a negative value, so its support is not contained in the positive domain on which $x\mapsto1/x$ is convex.

Common Pitfalls and Practical Interpretation

Using the bound without checking support. Verify $P(X>0)=1$, not merely $P(X\ne0)=1$. A variable that takes negative values is outside the theorem's assumptions.
Assuming a finite mean makes the reciprocal expectation finite. It does not. The exponential example has a finite mean but an infinite reciprocal expectation because of behavior near zero.
Reversing the inequality. The reciprocal function is convex on the positive half-line, so Jensen gives $E[1/X]\geq1/E[X]$, not the reverse.
Assuming equality is typical. Equality requires $X$ to be constant with probability one. Any nonconstant positive variable with finite mean gives a strict inequality, including cases where the left side is infinite.

The bound is useful when a calculation involving $E[1/X]$ is difficult but $E[X]$ is known. It supplies a distribution-free lower bound, provided the variable is positive and has a finite mean. It does not provide an upper bound, and it does not guarantee that the reciprocal expectation is finite; those questions require further information about the distribution, especially its behavior near zero.

Check Your Understanding

Try these questions using the assumptions and results above.

State the support and moment conditions under which $E[1/X]\geq1/E[X]$ holds, allowing the left side to be infinite.
For a nonconstant positive random variable with finite mean, can equality hold? Explain using the tangent-line difference.
Why does a finite value of $E[X]$ not ensure that $E[1/X]$ is finite?
For $P(X=2)=P(X=6)=1/2$, calculate $E[X]$, $1/E[X]$, and $E[1/X]$, and check the inequality.
Does the reciprocal Jensen bound apply to a variable that takes values $-1$ and $3$? Identify the assumption that determines your answer.