Discussion
Let X be a single observation with unknown mean \(E(X) = \mu \in (-\infty, \infty)\) and variance \(Var(X) = \mu^{2} + 1\). Consider the problem of estimating \(\mu\) based on X under the squared error loss \(L(\mu, a) = (a - \mu)^{2}\). Let \(T_{1}\) and \(T_{2}\) be defined as follows :
\(T_{1} = X\), \(T_{2} = X/2 + 1/2\).
Which of the following statements is true ?
Answer: B
The true statement is Risk of \(T_{2} = (1 - \mu + \mu^{2})/2\).
Risk under squared error is variance plus squared bias. \(T_{1} = X\) is unbiased, so its risk is \(\mu^{2} + 1\), not \(\mu^{2}\). For \(T_{2} = X/2 + 1/2\): bias \((\mu + 1)/2 - \mu = (1 - \mu)/2\), variance \((\mu^{2} + 1)/4\), so
\[ \begin{aligned} R(T_{2}) &= \frac{(1 - \mu)^{2} + \mu^{2} + 1}{4} \\ &= \frac{1 - \mu + \mu^{2}}{2} \end{aligned} \]
Then \(R(T_{1}) - R(T_{2}) = (\mu^{2} + \mu + 1)/2 \gt 0\) for every \(\mu\), so \(T_{2}\) is uniformly better. That makes C false and shows \(T_{1}\) is inadmissible, so D is false too.
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