Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 1 / 18H (Statistics)

Independence gives

Expanding the exponent givesThe Fisher-Neyman factorization theorem therefore shows that $T=X_1+X_2$ is sufficient.

It is also minimal sufficient. For two sample points $x,y\in {\mathbb{R}}^2$,where $C$ is independent of $\mu$. This ratio is independent of $\mu$ exactly when $T(x)=T(y)$. The likelihood-ratio criterion for minimal sufficiency applies, proving the claim and the general normal sample sum with known variance result in this case.

Writing $X_1=\mu +Z$ with $Z\sim N(0,1)$ givesso $S$ is unbiased. Its mean square error is therefore its variance. Sincewe obtain

The pair $(X_1,T)$ is jointly normal, withThe conditional distribution of a bivariate normal variable is thereforeEquivalently, its conditional density is

The Rao-Blackwell theorem suggests conditioning $S$ on $T$. Part (d) givesIt remains unbiased. Since $T\sim N(2\mu ,2)$ and a normal variable with mean $m$ and variance $v$ satisfies $\mathrm{var}(T^2)=2v^2+4m^{2}v$,This is strictly below $2+4{\mu }^2$ for every $\mu$. The calculation is recorded as the Rao-Blackwell estimator of a squared normal mean.