variance of $1/(X+1)$ where $X$ is Poisson-distributed with parameter $\lambda$ - MathOverflow

variance of $1/(X+1)$ where $X$ is Poisson-distributed with parameter $\lambda$

2010-01-27T20:31:38Z

What is the variance of $1/(X+1)$ where $X$ is Poisson-distributed with parameter $\lambda$! The series for the second moment is horrible!

$E({1\over (X+1)^2})=\sum_{k=1}^{\infty}\frac{1}{k^{2}}\frac{\lambda^{k}e^{-\lambda}}{k!}$

Is there an easy way to do it?

Answer by David Speyer for variance of $1/(X+1)$ where $X$ is Poisson-distributed with parameter $\lambda$

2010-01-27T21:21:54Z

Sorry, I gave a moronic answer before. Let me try to give a better one.

There should be no expression for $f(\lambda) := \sum_{k \geq 1} \lambda^k/(k^2 k!)$ in elementary functions. If there were, then $g(\lambda) = \lambda f'(\lambda) = \sum_{k \geq 1} \lambda^{k}/(k \cdot k!)$ would also be elementary. But $g(\lambda)=\int_0^{\lambda} \frac{e^t-1}{t} dt$ and $e^t/t$ is a standard example of a function without an elementary antiderivative.

Answer by vilvarin for variance of $1/(X+1)$ where $X$ is Poisson-distributed with parameter $\lambda$

2010-01-27T21:22:25Z

the previous answer has vanished. To tell the truth I haven't understood it completly :(

Answer by vilvarin for variance of $1/(X+1)$ where $X$ is Poisson-distributed with parameter $\lambda$

2010-01-27T21:27:52Z

thanks, David. The original problem I had is to compute the variance Y/(X+1) where Y bernoulli distributed with parameter p , and X is poisson distributed. But I don't think it will change anything. So I will just leave the sum (Vilvarin)