If $X$ has a binomial distribution with parameters $n$ and $p$, calculate $\textbf{E}[(1+X)^{-1}]$

Question

If $X$ has a binomial distribution with parameters $n$ and $p$, show that

\begin{align*} \textbf{E}\left(\frac{1}{1+X}\right) = \frac{1-(1-p)^{n+1}}{(n+1)p} \end{align*}

MY ATTEMPT

Since $X\sim\text{Binomial}(n,p)$, we have \begin{align*} \textbf{E}\left(\frac{1}{1+X}\right) & = \sum_{x=0}^{n}{n\choose x}\frac{p^{x}(1-p)^{n-x}}{1+x} = (1-p)^{n}\sum_{x=0}^{n}{n\choose x}\frac{1}{1+x}\left(\frac{p}{1-p}\right)^{x}\\\\ & = \frac{(1-p)^{n+1}}{p}\sum_{x=0}^{n}{n\choose x}\frac{1}{1+x}\left(\frac{p}{1-p}\right)^{x+1} \end{align*}

This is as far as I can get. Could someone help me out? Thanks in advance!

Answer 1

You can also sidestep the summation by using the moment generating function (via the Laplace transform). For any $X>-1$,

$$\frac{1}{1+X} = \int_0^{+\infty} e^{-(1+X)t}dt,$$

so, taking the expectation,

$$\mathbb{E} \left(\frac{1}{1+X}\right) = \mathbb{E} \left( \int_0^{+\infty} e^{-(1+X)t}dt\right) = \int_0^{+\infty} e^{-t} \mathbb{E} \left( e^{-tX}\right) dt.$$

The moment generating function of the binomial distribution is $\mathbb{E} \left( e^{-tX}\right) = (1-p+pe^{-t})^n$. Insert this value in the integral, solve the integral (there is an explicit antiderivative), and get the result.

Answer 2

$X \sim B(n,p)$

$$\mathbb{E}\left( {1 \over 1+X}\right) = \sum_{k=0}^n {n \choose k}{1 \over 1+k} p^k(1-p)^{n-k} = (1-p)^n\sum_{k=0}^n {n \choose k}{1 \over k+1} \left( {p \over 1-p}\right)^k$$

We know that $$(1+y)^n = \sum_{k=0}^n {n \choose k} y^k$$

Integrating both sides with respect to $y$ from 0 to $x$ gives

$${(1+x)^{n+1} \over n+1} -{1 \over n+1} = \sum_{k=0}^n {n \choose k} {x^{k+1} \over k+1}$$

Setting $x = {p \over 1-p}$ gives

$${1 \over n+1} \left( {1 \over (1-p)^{n+1}} -1 \right) = {p \over 1-p}\sum_{k=0}^n {n \choose k} {1 \over k+1} \left( {p \over 1-p}\right)^k$$

Multiplying the left side by ${1-p \over p}$ and further by $(1-p)^n$ should give us our final answer.

$$\mathbb{E}\left( {1 \over 1+X}\right) = {1-p \over p(n+1)} \left( {1 \over 1-p} - (1-p)^n\right) = {1 - (1-p)^{n+1} \over p(n+1)}$$