Wikipedia says that for a given numbers $\{x_i\}_{i=1}^{n}$ drawn from a half-normal distribution, the variance of that distribution can be estimated by sample variance $\hat\sigma^2 = \frac{1}{n} \sum_{i=1}^{n}{x_{i}^{2}}$.

The bias-corrected estimator is written as ${\hat {\sigma \,}}_{\text{mle}}^{*}={\hat {\sigma \,}}_{\text{mle}}-{\hat {b\,}},$ where $b\equiv \operatorname {E} {\bigg [}\;({\hat {\sigma }}_{\mathrm {mle} }-\sigma )\;{\bigg ]}=-{\frac {\sigma }{4n}}$.

How can I derive the expression for bias correction, and how it can be calculated for a given numbers $x_i$? Is it simply $\hat\sigma^{\ast} = \hat\sigma \left(1+\frac{1}{4n}\right)$?


1 Answer 1


If $Y_1, \ldots, Y_n \sim \mathcal{N}(0, \sigma^2)$ and $X_i=|Y_i|$ for $i=1,\ldots,n$, we say that $X_1,\ldots,X_n$ is a random sample from a half-normal distribution with scale parameter $\sigma$. Note that $\sigma^2$ is not the variance of the half-normal, but rather that of the underlying normal.

As you point out, the MLE of $\sigma^2$ is $\hat{\sigma}^2=\frac1n \sum_{i=1}Y_i^2=\frac1n \sum_{i=1}X_i^2$. By the invariance property of MLEs, it follows that $\hat{\sigma}= \sqrt{\frac1n \sum_{i=1}X_i^2}$.

Notice that $\frac{n\hat\sigma^2}{\sigma^2} \sim \chi^2_n$, so the quantity $\frac{\sqrt{n}\hat\sigma}{\sigma}$ follows a chi distribution on $n$ degrees of freedom. From the properties of this distribution, we have $$ \mathbb{E}(\hat{\sigma})= \underbrace{\left(\sqrt\frac2n \frac{\Gamma\left(\frac{n+1}2\right)}{\Gamma\left(\frac{n}2\right)} \right)}_{:=k_n}\sigma $$ We deduce that $\hat\sigma^*=\hat\sigma/k_n$ is an unbiased estimator of $\sigma$. It also holds that $$ k_n = 1-\frac1{4(n+1)} + O(n^{-2}) $$ (see here), which gives you a simple approximation for the bias-corrected MLE when $n$ is large, but this is now only asymptotically unbiased.


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