# How to calculate/approximate expectation of function of a binomial random variable?

Hi, I am stuck at following problem in my research.

Suppose that $M=m$ is a random variable with binomial distribution with parameters $n,p$. The constants $r$ and $\gamma$ are greater than zero. $\mathcal{E} _M$ is the expectation operator with respect to $M$. I need to calculate following expectations: $$\mathcal{E} _M \left[ \frac{m}{m+1} e^{ - \frac{2^{\frac{m+1}{m}\cdot r } - 1}{\gamma}} \right]$$ and $$\mathcal{E} _M \left[ \frac{m}{m+1} \left(e^{ - \frac{2^{\frac{m+1}{m}\cdot r } - 1}{\gamma}}\right)^{m-1} - \frac{m}{m+1} \left(e^{ - \frac{2^{\frac{m+1}{m}\cdot r } - 1}{\gamma}}\right)^{m} \right]$$ The values of $r$ and $\gamma$ are such that the exponential function cannot be approximated.

Can anyone please provide any guidance/reference for how to go about solving/approximating above expectations? Thanks in advance.

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OK, a not very helpful comment but here goes anyway. I don't have any idea about how to solve these exactly. As far as approximation goes: it will depend what sort of values of $n$ and $p$ you are interested in. For example, if $np$ and $n(1-p)$ both become large, then $M$ is well approximated by a normal random variable with mean $np$ and variance $np(1-p)$, and the law of large numbers or the central limit theorem will help you. On the other hand if $n\to\infty$ and $np$ stays bounded, then $M$ is well approximated by a Poisson random variable. – James Martin Nov 29 '11 at 10:53
Can you give some sample values for r and gamma and why the exponential can't be approximated? It seems to me that for the first expectation, the term converges quite quickly to the constant exp(-(2^r-1)/gamma ) for example. – Arthur B Nov 29 '11 at 15:24
James and Arthur...... thanks a lot for your suggestions..... I am working on your answers.... Arthur.... my work is related to Wireless communications... The typical value of r is between 0.3 to 0.8, the value of gamma is between 1 - 10, n is between 5 - 20 and p is between 0.01 - 0.2. Therefore I am not able to approximate them. – Navneet Dec 1 '11 at 4:49