Expected Value of 1/(1 + X) for a Geometric Random Variable

The expression is most naturally read as $\mathbb{E}[1/(1+X)]$ , where $X$ has a geometric distribution and $p$ is the probability of success on each independent trial. The answer depends on whether the geometric variable counts failures before the first success or counts the trial on which the first success occurs.

For the convention with support $0,1,2,\ldots$ , the expectation is $-p\ln(p)/(1-p)$ . The formula for support $1,2,3,\ldots$ is different; both are derived below.

If X counts failures

Suppose $X$ is the number of failures before the first success. Then

\Pr(X=k)=p(1-p)^k,\qquad k=0,1,2,\ldots

Write $q=1-p$ . The expectation is the probability-weighted sum

\mathbb{E}\left[\frac{1}{1+X}\right] = p\sum_{k=0}^{\infty}\frac{q^k}{k+1}.

Use the geometric-series identity $\sum_{k=0}^{\infty}q^k/(k+1)=-\ln(1-q)/q$ for $0<q<1$ . Since $1-q=p$ , this gives

\boxed{\mathbb{E}\left[\frac{1}{1+X}\right] = -\frac{p\ln p}{1-p}},\qquad 0<p<1.

At $p=1$ , the variable is always zero, so the expectation is 1. The displayed expression approaches 1 as $p$ approaches 1, so it can also be extended to that endpoint by continuity. A success probability of $p=0$ does not produce a proper geometric distribution: a success never occurs.

An integral derivation

The identity $1/(k+1)=\int_0^1 t^k\,dt$ turns the sum into an integral. Because the terms are nonnegative, the sum and integral can be interchanged:

\begin{aligned} \mathbb{E}\left[\frac{1}{1+X}\right] &=p\int_0^1\sum_{k=0}^{\infty}(qt)^k\,dt\\ &=p\int_0^1\frac{1}{1-qt}\,dt\\ &=-\frac{p}{q}\ln(1-q) =-\frac{p\ln p}{1-p}. \end{aligned}

This approach is useful for other expectations involving reciprocal expressions in an integer-valued random variable.

If X counts trials

Some books define a geometric random variable as the number of trials up to and including the first success. In that convention, $X$ takes values $1,2,3,\ldots$ , with

\Pr(X=k)=p(1-p)^{k-1}.

Now the expectation is

\begin{aligned} \mathbb{E}\left[\frac{1}{1+X}\right] &=p\sum_{k=1}^{\infty}\frac{q^{k-1}}{k+1}\\ &=p\int_0^1\frac{t}{1-qt}\,dt\\ &=\frac{p(-\ln p-q)}{q^2}\\ &=\boxed{\frac{p(-\ln p-1+p)}{(1-p)^2}},\qquad 0<p<1. \end{aligned}

At $p=1$ , $X=1$ with certainty, so the expectation is $1/2$ . Again, the formula has this limit as $p$ approaches 1.

Worked example

Let $p=1/2$ . If $X$ counts failures, then

\mathbb{E}\left[\frac{1}{1+X}\right] =-\frac{(1/2)\ln(1/2)}{1/2} =\ln 2\approx 0.6931.

If $X$ instead counts trials, then

\mathbb{E}\left[\frac{1}{1+X}\right] =\frac{(1/2)(\ln 2-1/2)}{(1/2)^2} =2\ln 2-1\approx 0.3863.

The different values make clear why you should identify the support before applying a formula.

Common mistakes

Ignoring the support convention. Check whether the problem defines $X$ on $0,1,2,\ldots$ or on $1,2,3,\ldots$ .
Using the reciprocal of the mean. In general, $\mathbb{E}[1/(1+X)]$ is not $1/(1+\mathbb{E}[X])$ . For example, with $p=1/2$ and support starting at zero, the expectation is $\ln 2$ , not $1/2$ .
Shifting the denominator incorrectly. When $X$ counts trials, it already starts at 1, but the requested quantity is still $1/(1+X)$ ; its first possible value is therefore $1/2$ .

References

A First Course in Probability — Sheldon Ross

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