Expected Value of 1/(X + 1) for a Poisson Distribution

Assume the expression is $\mathbb{E}[1/(X+1)]$, where $X$ has a Poisson distribution with parameter $\lambda$. This expectation averages the reciprocal $1/(k+1)$ over all possible count values $k$.

For $\lambda>0$, the result is

The case $\lambda=0$ is handled separately: then $X=0$ with probability 1, and the expectation is 1.

Derivation from the Poisson probability mass function

For a Poisson random variable,

By the definition of expectation for a discrete random variable, multiply each possible value $1/(k+1)$ by its probability and sum:

In the second line, $\lambda^k/((k+1)k!)$ is rewritten as $\lambda^k/(k+1)!$, then a factor of $1/\lambda$ is taken out. The index shift $j=k+1$ makes the remaining sum the exponential series with its $j=0$ term removed.

Worked example: $\lambda=2$

If $X\sim\operatorname{Poisson}(2)$, then

This is an average over the possible counts, not the reciprocal evaluated at the average count. For instance, $\mathbb{E}[X]=2$, but substituting that mean into the expression gives $1/(2+1)=1/3\approx0.3333$, not the desired expectation. The terms in the expectation are weighted by the individual Poisson probabilities; values of $X$ near 2 are more likely than many other counts, but every nonnegative count contributes.

Why it is not the reciprocal of the mean

In general, a nonlinear function cannot be moved outside an expectation. Thus

For $\lambda>0$, the two expressions are strictly different. The function $f(x)=1/(x+1)$ is convex for $x\ge0$, so Jensen’s inequality gives

The numerical example confirms this: $0.4323>1/3$. Equality holds in the degenerate case $\lambda=0$, when $X$ is constant.

Common mistakes

• Dropping the +1: The formula above is for $1/(X+1)$, not $1/X$. The added 1 also ensures the expression is defined when $X=0$.
• Using only the mean: Replacing $X$ by $\mathbb{E}[X]=\lambda$ computes $1/(\lambda+1)$, which is generally not the expectation.
• Forgetting the zero-count probability: The Poisson sum starts at $k=0$. Since $\Pr(X=0)=e^{-\lambda}$, this term contributes $e^{-\lambda}$ to the expectation.
• Applying the formula at zero without a limit: The displayed fraction has denominator zero when $\lambda=0$. Use the distribution directly, or take the limit, which is 1.

If the intended expression was $\mathbb{E}[1/X]$, that is a different quantity and needs special care: an ordinary Poisson variable has positive probability of being zero, where $1/X$ is undefined. The shift by 1 avoids that problem.