# Variance of $\frac{1}{X+1}$ when $X$ has Poisson distribution

I'm trying to find the variance of $$\frac{1}{X+1}$$ where $$X \sim \mbox{Poisson}(\lambda)$$. I found the previous answers here on how to get the expected value of $$\frac{1}{X+1}$$, and they were helpful and easy to follow. (original: Expectation of $\frac{1}{x+1}$ of Poisson distribution Duplicate: Expected value $\frac{1}{x+1}$ of Poisson distribution ) However, I was unable to generalize this to finding the variance of the same random variable. I tried using the identity $$Var(X) = E[X^2] - E[X]^2$$ but was unable to derive a neat closed form of $$E[\frac{1}{(1+X)^2}]$$. While at times it felt like I was getting close, I think I may have hit a dead end. Any help is greatly appreciated.

Let the random variable (r.v.) $$X$$ on a probability space $$(\Omega,\mathcal F,\Bbb P)$$ correspond to a Poisson r.v. with parameter $$\lambda$$, so $$\Bbb P(X=n) =\exp(-\lambda)\lambda^n /n!$$. Then: \begin{aligned} \Bbb E\left[\frac 1{(X+1)^2}\right] &= e^{-\lambda} \sum_{k\ge 0}\frac1{(k+1)^2}\cdot\frac{\lambda^k}{k!} \\ &\qquad\text{ we use now n=k+1} \\ &= e^{-\lambda} \cdot\frac1\lambda \sum_{n\ge 1}\frac1{n}\cdot\frac{\lambda^n}{n!} \\ &= e^{-\lambda} \cdot\frac1\lambda \int_0^\lambda \frac{e^s-1}s\; ds\ . \end{aligned} and this is a complicated (hypergeometric) function ($${}_2F_2(1,1;2,2;\lambda)$$ ).

sage: var('a,k');
sage: sum( 1/(k+1)^2/factorial(k)*a^k, k, 0, oo )
hypergeometric((1, 1), (2, 2), a)

• Wolfram seems to suggest an expression in terms of the incomplete Gamma function. wolframalpha.com/input/… Aug 15, 2019 at 16:07
• Thank you! That's a huge help.
– Elan
Aug 15, 2019 at 17:20

Thank you @dan_fulea. It took me a while to parse your answer, so I wanted to expand on it to ensure I'm not mistaken and to clarify for any others who have this question. As you said:
$$E\left[\frac 1{(X+1)^2}\right] = \frac{e^{-\lambda}}{\lambda} \int_0^\lambda \frac{e^s-1}{s} ds$$ After some trial and error I realized what you meant was $${}_2F_2(1,1;2,2;\lambda) = \frac{1}{\lambda} \int_0^\lambda \frac{e^s-1}{s} ds$$ So this means that $$Var\left(\frac{1}{1+X}\right) = {}_2F_2(1,1;2,2;\lambda)*e^{-\lambda} - E\left[\frac{1}{1+X}\right]^2$$ Which combined with the answer to earlier user's question (linked in original question) gives us: $$Var\left(\frac{1}{1+X}\right) = {}_2F_2(1,1;2,2;\lambda)*e^{-\lambda} - \frac{(1-e^{-\lambda})^2}{\lambda^2}$$ Just to be safe, I confirmed this answer by simulating around 20,000 Poisson draws for a few different lambdas and in each case it closely matched the measured variance. Thanks again! For those interested here's link to associated Wolfram page with y being used as the $$\lambda$$ in this case https://www.wolframalpha.com/input/?i=HypergeometricPFQ%5B%7B1,1%7D,+%7B2,2%7D,+y%5D*e%5E-y+++-+(1-e%5E-y)%5E2%2F(y%5E2)