= Shrinking a sample mean with variance proportional to mean squared
{title2=$k_{\rm opt}=n/(n+1)$}
If independent observations have mean $\theta$ and variance $\theta^2$, the scaled <sample mean> has
$$
\operatorname{MSE}(k\bar X)=\theta^2[(k-1)^2+k^2/n].
$$
For $\theta\ne0$, it improves on the unbiased sample mean precisely when $(n-1)/(n+1)<k<1$. The optimal scale is $k=n/(n+1)$ and its risk is $\theta^2/(n+1)$. At $\theta=0$ all risks vanish and strict improvement is impossible.
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