Paper 2, Section II, K

Principles of Statistics
Part II, 2013

Describe the Weak Sufficiency Principle (WSP) and the Strong Sufficiency Principle (SSP). Show that Bayesian inference with a fixed prior distribution respects WSP.

A parameter Φ\Phi has a prior distribution which is normal with mean 0 and precision (inverse variance) hΦ.h_{\Phi} . Given Φ=ϕ\Phi=\phi, further parameters Θ:=(Θi:i=1,,I)\Theta:=\left(\Theta{ }_{i}: i=1, \ldots, I\right) have independent normal distributions with mean ϕ\phi and precision hΘh_{\Theta}. Finally, given both Φ=ϕ\Phi=\phi and Θ=θ:=(θ1,,θI)\boldsymbol{\Theta}=\boldsymbol{\theta}:=\left(\theta_{1}, \ldots, \theta_{I}\right), observables X:=(Xij:i=1,,I;j=1,,J)\boldsymbol{X}:=\left(X_{i j}: i=1, \ldots, I ; j=1, \ldots, J\right) are independent, XijX_{i j} being normal with mean θi\theta_{i}, and precision hXh_{X}. The precision parameters (hΦ,hΘ,hX)\left(h_{\Phi}, h_{\Theta}, h_{X}\right) are all fixed and known. Let X:=(Xˉ1,,XˉI)\overline{\boldsymbol{X}}:=\left(\bar{X}_{1}, \ldots, \bar{X}_{I}\right), where Xˉi:=j=1JXij/J\bar{X}_{i}:=\sum_{j=1}^{J} X_{i j} / J. Show, directly from the definition of sufficiency, that X\overline{\boldsymbol{X}} is sufficient for (Φ,Θ)(\Phi, \Theta). [You may assume without proof that, if Y1,,YnY_{1}, \ldots, Y_{n} have independent normal distributions with the same variance, and Yˉ:=n1i=1nYi\bar{Y}:=n^{-1} \sum_{i=1}^{n} Y_{i}, then the vector (Y1Yˉ,,YnYˉ)\left(Y_{1}-\bar{Y}, \ldots, Y_{n}-\bar{Y}\right) is independent of Yˉ\bar{Y}.]

For data-values x:=(xij:i=1,,I;j=1,,J)\boldsymbol{x}:=\left(x_{i j}: i=1, \ldots, I ; j=1, \ldots, J\right), determine the joint distribution, Πϕ\Pi_{\phi} say, of Θ\Theta, given X=x\boldsymbol{X}=\boldsymbol{x} and Φ=ϕ.\Phi=\phi . What is the distribution of Φ\Phi, given Θ=θ\boldsymbol{\Theta}=\boldsymbol{\theta} and X=x?\boldsymbol{X}=\boldsymbol{x} ?

Using these results, describe clearly how Gibbs sampling combined with RaoBlackwellisation could be applied to estimate the posterior joint distribution of Θ\Theta, given X=x\boldsymbol{X}=\boldsymbol{x}.