$𝑋$ is a positive random variable (potentially unbounded) with $0 \le \mathbb{E}[X] = a < 1$.

Since $\phi(x) = \frac{1}{x}$ is a convex function, we can use Jensen's inequality to derive a lower bound: $\mathbb{E}[\frac{1}{1 + X}] \ge \frac{1}{1 + \mathbb{E}[X]} = \frac{1}{1 + a}$.

Is it possible to derive an upper bound?

A trivial upper bound would be 1 just because $X$ is positive. Can we derive an upper bound that involves $a$ or at least more information about $X$?


3 Answers 3


As shown by @jblood94, one can not improve on the trivial upper bound in the most general case. Yet the following inequality holds for $x \geq 0$ $$ \frac{1}{1 + x} \: \leqslant \:1 - x + x^2. $$ Then substituting $x$ by $X$ and taking the expectation $$ \mathbb{E}\left[\frac{1}{1 + X}\right] \leqslant 1 - a + \mathbb{E}[X^2], $$ so, if the variance of $X$ - or equivalently, $\mathbb{E}[X^2]$ - is known along with the expectation, an upper bound results.

  • 2
    $\begingroup$ Note, in addition, that $\mathbb{E}[X^2] \leq a$, as it is maximized for a Bernoulli distribution with probability parameter $p=a$. So your clever approach also gives $1$ as the least upper bound in the unknown variance case. $\endgroup$
    – jbowman
    Jan 5 at 20:36

$\mathbb{E}\big[\frac{1}{1+X}\big]\le1$ is the best you can do.

For example, consider the following PMF:

$$p_X(x) = \begin{cases} 1-\tau, & x=\tau \\ \tau, & x=\frac{\beta}{\tau}-1 \end{cases}$$

with $\tau<\beta<1+\tau^2$.



Now consider $\lim{\tau\to0}$:



and $0<\beta<1$. So the upper bound can approach 1 for any value of $a$ ($0<a<1$).


Suppose we assume that the second moment of $X$ exists, that is $E(X^2)<\infty$. Let $\sigma^2 = E(X^2) - [E(X)]^2$ be the variance of $X$. We can consider a concave function $g(x)=\dfrac{1}{1+x}-x^2$ as it is easy to see that $g''(x)=\dfrac{2} {(1+x)^3}-2\leq0$ for $x\geq0 $. By Jensen's inequality, we have

$$\begin{align*} E\left(\dfrac{1}{1+X}-X^2\right)&=E(g(X)) \\ &\leq g[E(X)] \\ &= \dfrac{1}{1+a}-a^2 \\ \implies E\left(\dfrac{1}{1+X}\right)&\leq \dfrac{1}{1+a}-a^2+(\sigma^2 + a^2)\\&=\dfrac{1}{1+a} + \sigma^2. \end{align*}$$


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