Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 29K (Principles of Statistics)

Since ${\overline{X}}_n\sim N(\theta ,1/n)$, the estimator has variance
$a^2/n$ and bias $(a-1)\theta +b$. Its quadratic risk function is

If $a>1$, compare withIts bias is the negative of the original bias, so the squared-bias terms agree, whileIts risk is therefore strictly smaller for every $\theta$. Thus
${\hat{\theta }}_{a,b}$ is not an admissible estimator.

Put ${\theta }_0=-b/(a-1)$ and compare with the constant estimator
$\delta ={\theta }_0$. Its risk is $(\theta -{\theta }_0{)}^2$. For $a<0$,for every $\theta$, since $|a-1|>1$. Hence the original estimator is inadmissible.

The risk of ${\overline{X}}_n$ is constantly $1/n$. Suppose an estimator $\delta$ dominates it, and put
$m(\theta )={\mathbb{E}}_{\theta }\delta$ and $b(\theta )=m(\theta )-\theta$. The biased Cramer-Rao bound, using total information $n$, givesDomination would implyfor every real $\theta$. The standard differential-inequality argument shows that the only globally defined differentiable solution is $b\equiv 0$: a nonzero value forces $b^{\prime }$ to retain a sign and magnitude that makes $b$ leave the permitted bounded interval in one time direction.

Thus $m(\theta )=\theta$. The Cramer--Rao inequality now gives
${\mathrm{Var}}_{\theta }\delta \ge 1/n$, so domination forces equality everywhere. Equality in Cramer--Rao makes the centred estimator proportional to the Gaussian score:and hence $\delta ={\overline{X}}_n$ almost surely. No strict improvement exists, so ${\overline{X}}_n$ is admissible.

No. Every estimator in parts (b) and (c) has $a\ne 1$, so its risk contains the nonconstant quadratic term
$((a-1)\theta +b{)}^2$ and has infinite supremum over $\theta \in \mathbb{R}$. The estimator ${\overline{X}}_n$ has constant risk $1/n$, so those estimators cannot be minimax.