Editor Anirban DasGupta writes:

Well done to Lu Mao (pictured below), of the department of biostatistics at the University of North Carolina at Chapel Hill, who sent a carefully written solution.

Lu Mao

The problem was to find a minimax estimator of $m=|\mu |$ under squared error loss when there are $n$ iid observations from $N(\mu ,1),\mu \in \mathcal{R}$. The claim is that the plug-in estimator $|\overline{X}|$ is unique minimax.

First, note that by the triangular inequality,

$su{p}_{\mu }{E}_{\mu }[(|\overline{X}|-m{)}^2]\le su{p}_{\mu }{E}_{\mu }[(\overline{X}-\mu {)}^2]=\frac{1}{n}.$

Consider now any other estimator $\delta (X_1,···,X_n)$ of $m$; we may assume $\delta$ to be a function of $\overline{X}$ by Rao–Blackwell. Then,

$su{p}_{\mu }{E}_{\mu }[(\delta -m{)}^2]\ge su{p}_{\mu \ge 0}{E}_{\mu }[(\delta -\mu {)}^2]\ge \frac{1}{n},$

the last inequality following from a standard application of the Cramér–Rao inequality. Thus |\bar{X}| is minimax; unique minimaxity follows because the last inequality above is strict by completeness. The exact risk function of |\bar{X}| equals

$R(\theta ,|\overline{X}|)=\frac{1}{n}[1+4{\theta }^2(1-\Phi (\theta ))-4\theta \phi (\theta )]$, where $\theta =|\sqrt{n}\mu |$.

By elementary calculus, the function $g(x)=x^2(1-\Phi (x))-x\phi (x)$, $x\ge 0$ has a unique minimum at $x_0$ where $x_0$ is the unique root of $\frac{x(1-\Phi (x))}{\phi (x)}=\frac{1}{2}$; $x_0\approx .61200$.

Plugging this back into the exact risk formula, one gets that the minimum risk of $|\overline{X}|$ is $\frac{1-2\times 0\phi (x0)}{n}=\frac{.59509}{n}$.