When estimating population mean, how can one half of the sample mean have lower risk than the sample mean itself?

I read Efron and Morris (1977) Stein's Paradox in Statistics with interest yesterday and stumbled upon the statement that, if and only if the population mean is close to zero, than the risk (mean squared error, MSE) of using half of the sample mean as an estimate for the population mean is lower than that of using the mean.

I tried to model this out (unfortunately in Crystal Ball / Excel, did not use R for this) but could not replicate this result. In my examples, the risk (as per MSE) of the half-of-the-mean estimate became closer to the mean estimate, but never became lower. I might, clearly, have misunderstood the concept, but would be very interested if someone can show / further explain this to me.

• Create a simulation in which the true mean is a tiny fraction of the standard deviation and the sample size is small: that will cause the sample means typically to vary much further from the true mean than $0$ differs from the true mean. – whuber Feb 6 '14 at 15:32

Imagine that the population is described by $\mathcal N(1,10)$, i.e. population mean is $\mu=1$ and standard deviation is $\sigma=10$. Let your sample size be $n=10$. The variance of the sample mean (MSE) will be around $\sigma^2/n=10$, so you will get values of the sample mean that are quite far from the true mean $\mu=1$. Taking one half of the sample mean, will reduce the variance four times, bringing the estimation much closer to zero, and, as a consequence, much closer to $\mu=1$.