Unbiased estimator of $2\left(\frac{E[x^2]}{E[x]^2}-1\right)^{-1}$ For a random variable $x$, how would I go about creating an unbiased estimator of the following quantity?
$$R=2\left(\frac{E[x^2]}{E[x]^2}-1\right)^{-1}$$
For instance, when $x$ comes from from chi-squared distribution with 100 degrees of freedom, R=100 and sample estimate is an over-estimate. Any tips for practical bias correction when $x\sim$ generalized chi-squared?

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Motivation: this represents "degrees of freedom" in the underlying distribution. When $x=\|y\|^2$ and $y$ is centered/normally distributed with covariance $\Sigma$, this quantity is equal to the "effective rank" $R=\frac{(\text{Tr}\Sigma)^2}{\text{Tr}\Sigma^2}$ defined in Eq 3. of https://arxiv.org/abs/2106.09276
 A: The quantity is
$$2\frac{\mathbb E[X]^2}{\text{var}(X)}$$
Given an iid sample $X_1,\ldots,X_n$, the estimator$$X_1X_2$$is an unbiased estimator of the numerator. In the case $\text{var}(X)^{1/2}=\sigma$ is a scale parameter, if $X_3-X_4$ has a first order negative moment, i.e.$$\mathbb E[|X_3-X_4|^{-1}]<\infty$$then$$\mathbb E[(X_3-X_4)^{-1}]\propto\sigma^{-1}$$
and, assuming further that the missing constant does not depend on an unknown parameter, $$\frac{1}{(X_3-X_4)(X_5-X_6)}$$leads to an unbiased estimator of $\sigma^{-2}$. If a lower order negative moment exists, as e.g.
$$\mathbb E[|X_3-X_4|^{-1/4}]<\infty$$a similar reasoning leads to an unbiased estimator proportional to
$$\frac{1}{(X_3-X_4)^{1/4}\cdots(X_9-X_{10})^{1/4}}$$
Note however that this solution is distribution-dependent since (i) it depends on the existence of some negative moments and (ii) the proportionality constant depends on the distribution. I do not think there exists a universal unbiased estimator of $\sigma^{-2}$.
For instance, if the $X_i$'s are distributed from an Exponential $\mathcal E(\theta)$ distribution,
$$\mathbb E[X_i]=\theta^{-1}\qquad\text{var}(X_i)=\theta^{-2}
\qquad\mathbb E[(X_i+X_j)^{-1}]=\theta$$
Therefore
$$\frac{1}{(X_1+X_2)(X_3+X_4)}$$
is an unbiased estimator of $\theta^2$. (For the exponential, the ratio is always equal to $2$ and the estimation problem is non-existing.)
For the Normal $X_1,X_2,X_3\sim\mathcal N(0,\sigma^2)$ distribution,
$$\mathbb E[(X_1^2+X_2^2+X_3^2)^{-1}]=\sigma^{-2}$$
Hence for a Normal $X_1,\ldots,X_8\sim\mathcal N(\mu,\sigma^2)$ distribution,
$$2\mathbb E[X_1X_2([X_3-X_4]^2+[X_5-X_6]^2+[X_7-X_8]^2)^{-1}]=\mu^2\sigma^{-2}$$
