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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2025

Paper 3

Given $X=x$, the <posterior expected loss> from reporting $d$ is
$$
\begin{aligned}
\mathbb E\!\left[\theta^{-1}(\theta-d)^2\mid X=x\right]
&=\mathbb E(\theta\mid X=x)-2d
+d^2\mathbb E(\theta^{-1}\mid X=x).
\end{aligned}
$$
This is a strictly convex quadratic in $d$ whenever the conditional inverse moment is finite and positive. Differentiating with respect to $d$ gives
$$
-2+2d\,\mathbb E(\theta^{-1}\mid X=x)=0,
$$
so the <Bayes estimator under reciprocal weighted quadratic loss> is
$$
\boxed{\delta_\pi(x)
=\big(\mathbb E(\theta^{-1}\mid X=x)\big)^{-1}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 28L / b / iii / Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 28L / b / iii / Solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 28L / b / iii / Solution

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