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Question

Consider a population of known size $N$, from which we sample $n$ individuals without replacement and measure their trait values $x_i$ for all individual $i$. The variable $x$ is bounded [0,1].

Let $\mu = \frac{\sum_i^N x_i}{N}$ be the average trait value in the population and $\sigma^2 = \frac{\sum_i^N (x_i - \mu)^2}{N}$ be the variance in the population.

What is the unbiased estimator for $\frac{\sigma^2}{\mu (1 - \mu)}$?

Attempt

I got estimators for both the numerator ($\sigma^2$) and denominator ($\mu (1-\mu)$).

First few definitions

First let

$$\bar x = \frac{\sum_i^n x_i}{n}$$

and

$$s^2 = \frac{\sum_i^n (x_i - \bar x)^2}{n-1}$$

Numerator

From this post, the unbiased estimator for $\sigma^2$ is

$$\sigma^2 = \operatorname{E}\left[\frac{(N-1)s^2}{N}\right]$$

Denominator

The unbiased estimator of $\mu - \mu^2$ is

$$ \begin{align} \mu - \mu^2 &= \operatorname{E}[\bar x] - \operatorname{E}[\bar x]^2\\ &= \operatorname{E}[\bar x] - \operatorname{E}[\bar x]^2 - \operatorname{var}[\bar X] + \operatorname{var}[\bar x] \\ &= \operatorname{E}[\bar x] - \operatorname{E}[\bar x^2] + \operatorname{var}[\bar x] \end{align}$$

From utexas

$$ \operatorname{var}[\bar X] = \operatorname{E}\left[\frac{s^2}{n}\left(1-\frac{n}{N}\right)\right]$$

Therefore,

$$\mu - \mu^2 = \operatorname{E}\left[\bar x - \bar x^2 + \frac{s^2}{n}\left(1-\frac{n}{N}\right)\right]$$

Putting numerator and denominator together

If the two estimators were independent, then the product of the expectations would be the expectation of the product but the numerator is not independent of the denominator!

I suppose I would now need to compute something like

$$ \begin{align} \frac{\sigma^2}{\mu (1 - \mu)} &= \operatorname{E}\left[\frac{\frac{(N-1)s^2}{N}}{\bar x - \bar x^2 + \frac{s^2}{n}\left(1-\frac{n}{N}\right)} + \operatorname{cov}\left[\operatorname{var}\left[x\right], \frac{1}{\bar x - \bar x^2}\right]\right] \\ &= \operatorname{E}\left[\frac{\frac{(N-1)s^2}{N}}{\bar x - \bar x^2 + \frac{s^2}{n}\left(1-\frac{n}{N}\right)}\right] + \operatorname{E}\left[\frac{\operatorname{var}\left[x\right]}{\bar x - \bar x^2}\right] - \operatorname{E}\left[\operatorname{var}\left[x\right]\right]\operatorname{E}\left[\frac{1}{\bar x - \bar x^2}\right]\\ \end{align} $$

but I am getting lost there!

Remi.b
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    The very appearance of $1/(\mu(1-\mu))$, with its singularities at $\{0,1\}$, suggests a context in which strong assumptions are made about the variable. One would suppose its values are limited to the interval $[0,1]$, for instance. Could you disclose all such assumptions and perhaps give us some context or motivation for looking at this particular combination of parameters? – whuber Aug 17 '17 at 19:31
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    @whuber Yes, applications are in genetics. The variable $x$ are allele frequencies among populations and can take values in the range [0,1] ($\mu - \mu^2$ can therefore take values in the range [0,0.25]). I have added this info to the post! The statistics $\frac{\sigma^2}{\mu-\mu^2}$ is actually quite common (see for example [Nei 1973](http://www.pnas.org/content/70/12/3321.short)). – Remi.b Aug 17 '17 at 20:32
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    Unbiased estimators of nonlinear statistics are extremely rare. You can stumble upon one only by some luck (outside of exponential families with canonical link). I would very seriously doubt that an unbiased estimator for this quantity exists. There won't be any shortage of consistent estimators, and if you simply have the plug-in estimators of $\mu$ and $\sigma^2$, you can obtain the variance of your statistic by the delta method. I don't see where the finite nature of your population comes into play though -- the finite population corrections matter quantitatively when $n/N>0.1$, say. – StasK Aug 17 '17 at 23:47
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    The existence of at least one unbiased estimator cannot assumed; it must be demonstrated. And one certainly cannot assume there is only one. – Michael Hardy Nov 18 '18 at 21:59
  • If $\Pr(x_i\in\{0,1\}) = 1$ then $\mu(1-\mu)$ is the same thing as $\sigma^2.\qquad$ – Michael Hardy Nov 18 '18 at 22:02

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