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As the title states, is there a closed form formula for the expectation of the square root of a hypergeometric variable.

Edit:

Closed form approximate solutions based on related distributions or expansions are also welcome.

hyper
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    There is every reason not to expect one. After all, there must be at least one term involving $\sqrt{p}$ for every prime $p$ less than the maximum value of the variable; in the expectation it will be multiplied by a rational number (the probability) and (as a consequence of Galois theory) those terms will not simplify in the summation. – whuber Dec 20 '11 at 18:31
  • Ah, well my hopes are dashed. Is there something approximate that I could use, for instance using a binomial or normal approximation to the distribution function? – hyper Dec 20 '11 at 20:16

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You can use the delta method to get an approximate answer, although it's not guaranteed to be a good answer. The delta method approximation with two terms for $E(\sqrt{x})$ in terms of the mean $\mu$ and variance $\sigma^2$ of $x$ is:

$E(\sqrt{x}) \approx \mu^{1/2}\left(1 - \frac{1}{8}\frac{\sigma^2}{\mu^2}\right)$

For a Gamma(5,1) variate, which has mean 5 and standard deviation 2.24, the delta method estimates $E(\sqrt{x}) = 2.180$; a simulation of 100,000 Gamma(5,1) variates resulted in a sample mean of 2.181, with a std. error less than 0.001. That worked well; on the other hand, sampling from a Poisson(1) results in a delta-method estimate of 0.875 and a sample mean of 0.774, with a standard error of 0.002. So your mileage may vary.

jbowman
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