Find the expected value of $E[\frac{1}{\left(X+1\right)^2}]$ where X is binomial

Question

The problem:

X is a binomial random variable, find $E[\frac{1}{\left(X+1\right)^2}]$

Using the law of unconscious statistician we get

$$ E\left[\frac{1}{\left(X+1\right)^2}\right]=\sum_{k=0}^n\frac{1}{\left(1+k\right)^2}\binom{n}{k}p^k(1-p)^{n-k} $$

If I write

$$ \frac{1}{\left(k+1\right)^2}\binom{n}{k}=\binom{n+1}{k+1}\frac{1}{n+1}\frac{1}{k+1}=\binom{n+2}{k+2}\frac{1}{n+1}\frac{1}{n+2}\frac{k+2}{k+1} $$

But then I don't know how to proceed. Maybe there is an ellegant solution for this?

Answer 1

A 'closed-form' solution exists, although I am not sure it is as elegant as you would like. To illustrate, if $X \sim \text{Binomial}(n,p)$ with pmf $f(x)$:

Then:

... with the help of the Expect function from mathStatica/Mathematica.

Here is a plot of the solution, as $n$ increases from 1 to 20, when:

• $p = \frac14$ (green)

• $p = \frac12$ (blue)

• $p = \frac34$ (red)

Notes

1. HypergeometricPFQ denotes the generalised HypergeometricPFQ function

2. As disclosure, I should add that I am one of the authors of the Expect function used above.