Log Transformation for a Normal Variance Estimate

Why Take the Logarithm of a Sample Variance?

The sampling variability of a variance estimate depends on the scale of the underlying population. In a normal model, the sample variance has a standard error that grows in proportion to the population variance. Taking a logarithm changes that scale: for large samples, the log sample variance has approximately the same variance regardless of the value of the population variance. This also gives a direct way to form a confidence interval whose endpoints must be positive.

The result depends on normal sampling. Let \(X_1,\ldots,X_n\) be independent and identically distributed with \(X_i\sim N(\mu,\sigma^2)\), where \(\sigma^2>0\), and estimate the population variance with the usual unbiased sample variance \(S^2=(n-1)^{-1}\sum_{i=1}^n(X_i-\bar X)^2\). Write \(\nu=n-1\), the degrees of freedom.

Exact Sampling Distribution and Log-Scale Variance

A standard result for a normal sample is that \(\nu S^2/\sigma^2\) has a chi-square distribution with \(\nu\) degrees of freedom. Thus, if \(Q=\nu S^2/\sigma^2\), then \(Q\sim\chi^2_\nu\), and \(\log S^2-\log\sigma^2=\log(Q/\nu)\). This exact distribution gives both the large-sample limit and a precise way to see why the log-scale variance loses its dependence on \(\sigma^2\).

Proof of the limiting distribution. The chi-square central limit theorem gives \((Q-\nu)/\sqrt{2\nu}\xrightarrow{d}N(0,1)\). In particular, \(U_\nu=(Q-\nu)/\nu\xrightarrow{p}0\), and \(\sqrt{\nu}U_\nu\) converges in distribution to \(N(0,2)\). Since \(Q/\nu=1+U_\nu\), \(\log(Q/\nu)=\log(1+U_\nu)\). On the event \(|U_\nu|\le 1/2\), Taylor’s theorem gives \(|\log(1+U_\nu)-U_\nu|\le C U_\nu^2\) for a finite constant \(C\). The probability of that event tends to one, and \(U_\nu=O_p(\nu^{-1/2})\), so \(\sqrt{\nu}U_\nu^2=O_p(\nu^{-1/2})\xrightarrow{p}0\). Therefore \(\sqrt{\nu}\log(1+U_\nu)-\sqrt{\nu}U_\nu\xrightarrow{p}0\). Slutsky’s theorem yields the stated limit. \(\square\)

The limiting distribution suggests the constant variance \(2/\nu\). The exact variance confirms the same asymptotic behavior. To see this, set \(T=Q/2\), which has a gamma distribution with shape \(a=\nu/2\) and scale \(1\). Its moments satisfy \(E[T^t]=\Gamma(a+t)/\Gamma(a)\) for \(t>-a\). Differentiating the log of this moment-generating expression for \(\log T\) twice at zero gives \(\operatorname{Var}(\log T)=\psi_1(a)\), where \(\psi_1(a)=\frac{d^2}{da^2}\log\Gamma(a)\) is the trigamma function. Since \(\log S^2=\log\sigma^2+\log(2T/\nu)\), adding constants does not change variance, so \(\operatorname{Var}(\log S^2)=\psi_1(\nu/2)\).

For \(a>0\), the standard series representation \(\psi_1(a)=\sum_{k=0}^{\infty}(a+k)^{-2}\) and integral comparison give \(\int_0^\infty(a+x)^{-2}\,dx\le\psi_1(a)\le a^{-2}+\int_0^\infty(a+x)^{-2}\,dx\). Thus \(1/a\le\psi_1(a)\le1/a+1/a^2\), so \(\psi_1(a)\sim1/a\). Substituting \(a=\nu/2\) shows that \(\operatorname{Var}(\log S^2)\sim2/\nu\). This is an asymptotic statement: the exact variance is not constant for finite \(n\), but it does not involve \(\sigma^2\).

A Large-Sample Confidence Interval for the Variance

Dividing the theorem’s limiting statistic by \(\sqrt{2}\) gives \((\log S^2-\log\sigma^2)/\sqrt{2/\nu}\xrightarrow{d}N(0,1)\). For a desired confidence level \(1-\alpha\), let \(z_{1-\alpha/2}\) be the corresponding standard normal quantile. The approximate central coverage statement is \(P\{-z_{1-\alpha/2}\le(\log S^2-\log\sigma^2)/\sqrt{2/\nu}\le z_{1-\alpha/2}\}\approx1-\alpha\). Rearranging for \(\log\sigma^2\) and exponentiating gives the interval below.

The interval is symmetric around \(\log S^2\) on the log scale, not around \(S^2\) on the original scale. Its upper endpoint is the lower endpoint multiplied by \(\exp(2z_{1-\alpha/2}\sqrt{2/\nu})\), and both endpoints are positive. The approximation uses \(\sqrt{2/\nu}\) as the standard error on the log scale; it does not account for every finite-sample feature of the distribution of \(\log S^2\).

Common Pitfalls and Interpretation

• Using the wrong variance convention. The result here uses \(S^2\) with denominator \(n-1\), so the degrees of freedom are \(\nu=n-1\). Replacing it with a variance estimator using denominator \(n\) changes the finite-sample distribution and should not be done without adjusting the derivation.
• Calling the interval exact. The chi-square distribution is exact for \(\nu S^2/\sigma^2\), but the interval above comes from a normal approximation to \(\log S^2\). For finite samples, its coverage is approximate.
• Forgetting which parameter is being estimated. The endpoints estimate \(\sigma^2\). An interval for \(\sigma\) would require taking square roots of the variance endpoints.
• Assuming log transformation fixes every problem. The constant asymptotic variance follows from the normal sampling model and the large-sample limit. It does not make the distribution exactly normal at every sample size, nor does this derivation establish the result for nonnormal populations.