Reciprocal moments of an estimated variance

Reciprocal moments of an estimated variance determine whether a precision estimate such as $1/S^2$ has a finite expected value or variance. Small positive values of a sample variance become large values after inversion. An unbiased estimator of a variance therefore need not give an unbiased or even integrable estimate of its reciprocal.

For $n\ge 2$, let $X_1,\dots ,X_n$ be independent normal random variables with common mean $\mu$ and variance ${\sigma }^2>0$. For the unbiased sample variance $S^2=(n-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$, put $\nu =n-1$. Then $Z=\nu S^2/{\sigma }^2$ has a chi-squared distribution. Its probability density function is $z^{\nu /2-1}{e}^{-z/2}/[2^{\nu /2}\Gamma (\nu /2)]$ for $z>0$.^[ref]

For any real $p>0$, direct integration and the Gamma function giveThe convergence restriction is essential: near zero the integrand has exponent $\nu /2-p-1$, so the integral diverges when $\nu \le 2p$. Substitution of a negative Gamma function argument into the displayed quotient does not define a nonexistent moment.

Rescaling gives the useful general resultIn particular, the mean reciprocal exists for $n>3$, while its variance requires $n>5$:For $n=6$ the mean is $5/(3{\sigma }^2)$, the variance is $50/(9{\sigma }^4)$ and the coefficient of variation is $\sqrt{2}$. For $n=4$ the mean is finite but the variance is infinite. For $n=3$ even the mean diverges. Increasing the number of groups does not repair a nonexistent moment of each raw reciprocal estimate.

Inverse-variance weighting commonly normalizes these reciprocals. Those normalized weights lie between zero and one, so their moments can exist even when the raw reciprocal moments fail. This distinction matters in uncertainty in estimated inverse-variance weights: neither the moment calculation nor a plug-in precision formula alone establishes the accuracy of the final weighted estimator. The results here use the exact normal-sample law; a nonnormal sample requires a different analysis.