# The expectation of a function of a binomial random variable: $E(\frac{1}{1+N})$

So I need to find: $$\mathbb{E}(\frac{1}{1+N})$$ where N is a binomial random variable with paramaters n,p.

I know that when $$Y = g(N)$$ $$\mathbb{E}(Y) = \sum_{\infty} g(x)p(x)$$

where p(x) is the frequency function of N.

So am I on the right track when I right $$\mathbb{E}(\frac{1}{1+N}) = \sum_{k=0}^n (\frac{1}{1+k}){n \choose k}p^k(1-p)^{n-k}$$

If so how do I proceed from here?

Thanks for any help.

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We have $$\frac 1{k+1}\binom nk=\frac{n!}{(k+1)\cdot k!\cdot (n-k)!}=\frac{n!}{(k+1)!\cdot ((n+1)-(k+1))!}=\frac 1{n+1}\binom{n+1}{k+1},$$ and using this in the expression of $E\frac 1{1+N}$, we get $$E\frac 1{1+N}=\sum_{k=0}^n\frac 1{n+1}\binom{n+1}{k+1}p^k(1-p)^{n-k}=\frac 1{n+1}\sum_{j=1}^{n+1}\binom{n+1}jp^{j-1}(1-p)^{n-j+1}.$$ Now we can conclude using binomial theorem: this gives $$E\frac 1{1+N}=\frac 1{n+1}\frac 1p\left(1-(1-p)^{n+1}\right).$$
So if I am doing this correctly would that be $$\mathbb{E}(\frac{1}{1+N}) = \frac{1}{1+n}1-(1-p)^{n+1}$$ – Lok Oct 6 '12 at 15:18
Maybe a factor $\frac 1p$ is missing, but you got the idea. – Davide Giraudo Oct 6 '12 at 20:07