Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 27I / Solution

A sufficient statistic $T$ has a conditional distribution of the data given $T$ that does not depend on the parameter. In a dominated family, the factorization criterion for sufficiency states that this is equivalent to $p_{\mu }(x)=h(x)g_{\mu }(T(x))$ for a nonnegative parameter-free $h$. Bayes' formula then gives a posterior distribution proportional to $\pi (\mu )g_{\mu }(T(x))$, because $h(x)$ cancels from numerator and normalizing denominator. Thus the posterior depends on the data only through $T$.

For the normal sample,The first term is parameter-free, so the factorization proves sufficiency of $\bar{X}$. Combining its likelihood with the normal prior and completing the square givesThe prior predictive distribution of the sample mean is obtained from $\bar{X}=M+\epsilon$, $\epsilon \sim N(0,1/n)$ independent of $M$:

Define the Bayes factor $B_{01}$ as the marginal likelihood under $H_0$ divided by that under $H_1$. The common residual factor cancels, so it is the ratio of the densities of $\bar{X}$ under these two hypotheses:The reverse Bayes factor is its reciprocal. At $|\bar{x}|=1.96/\sqrt{n}$,for a fixed positive $\tau$. Thus the frequentist test is at its rejection threshold while fixed prior odds give increasingly strong posterior support for the point null. This is the Jeffreys-Lindley paradox for a Gaussian point null: the standardized statistic stays fixed, but the observed effect tends to zero and the fixed-scale alternative spreads predictive mass much more widely. There is no logical contradiction, since the procedures compare hypotheses using different calibrations. For the degenerate $\tau =0$ the hypotheses coincide and $B_{01}=1$.