# How to estimate bias-corrected variance of a half-normal distribution?

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

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.