Shrinking a sample mean with variance proportional to mean squared ($k_{\mathrm{opt}}=n/(n+1)$)

If independent observations have mean $\theta$ and variance ${\theta }^2$, the scaled sample mean hasFor $\theta \ne 0$, 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.