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By the bias-variance decomposition of mean squared error,For $\theta \ne 0$, comparison with the unbiased estimator reduces toIts roots are $(n-1)/(n+1)$ and $1$, soEquality at either endpoint gives the same risk as $\bar{X}$. At $\theta =0$, every risk is zero, so no strict improvement is possible. This illustrates shrinking a sample mean with variance proportional to mean squared: allowing a small bias can reduce variance enough to improve total risk. The best constant is $k=n/(n+1)$, whose mean squared error is ${\theta }^2/(n+1)$.