Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 36 / 4

Compute the Bayesian model evidence for each model by integrating its sampling density against its parameter prior:The posterior odds are the prior odds multiplied by the Bayes factor:The parameter priors used for evidence must be normalized; arbitrary normalization constants in model-specific improper priors would make this comparison undefined.

Within the positive component, normal-normal conjugacy givesCompleting the square yieldsHere positive denotes membership in the variable-effect component, not the condition ${\theta }_i>0$; its normal prior permits effects of either sign.

Integrating the positive component's normal distribution prior gives its prior predictive distribution:The negative component gives $Y_i\mid -\sim N(0,1)$. Let $f_1(y)={\varphi }_{1+V}(y)$ and $f_0(y)={\varphi }_1(y)$ denote those centred normal densities. The point-null mixture prior assigns component odds $q/(1-q)$, so the posterior odds of the positive component areIts posterior probability is $O_i/(1+O_i)$. Evidence depends on $y_i^2$, so large effects of either sign favour the positive component. This is the marginal distribution before observing $y_i$, not the posterior predictive law for a second observation conditional on it.

Under the independent-gene version of the model, integrating both component labels and effects givesThis product uses independence of the gene effects and observation errors conditional on the shared parameters. Maximize its logarithm over $0\le q\le 1$, for example with a one-dimensional bounded optimizer or a grid followed by refinement. Its derivatives areThus the log likelihood function is concave. An interior maximum-likelihood estimate solves the score equation; otherwise it is at zero or one, determined by the endpoint score signs. An alternative expectation-maximization algorithm iteratesFor $V=0$, the components coincide and $q$ is not identifiable; for $V>0$ the full observation values carry information about the mixing fraction.

Use prior elicitation to make the qualitative judgments quantitative: ask whether $10\%$ represents a mean, median or mode, and what probability “unlikely above $15\%$” should mean, for example a $5\%$ upper-tail probability. A beta distribution on $(0,1)$ is a convenient family. If $0.1$ is taken as the mean, write its parameters as $a_q=0.1\kappa$, $b_q=0.9\kappa$, then choose $\kappa$ so that the desired tail probability above $0.15$ is obtained. Check its quantiles and prior predictive distribution for plausibility, and assess sensitivity to reasonable alternative elicited values. The information suggests an informative prior near $0.1$, but does not uniquely determine its parameters.

Introduce independent component indicators $z_i\mid q\sim \mathrm{Bernoulli}(q)$, and auxiliary slab effects $u_i\sim N(0,V)$, independent of the indicators. Set ${\theta }_i=z_i{u}_i$. Then inactive genes have exactly zero effect, while active genes have the stipulated normal distribution. With $V>0$ supplied as known data and elicited positive beta shapes, one suitable WinBUGS model isThe normal arguments are precisions, not variances. Supply all observed y values, monitor q, theta and optionally z, and assess Markov chain Monte Carlo convergence diagnostics. The proper auxiliary prior remains defined for inactive genes, where $u_i$ is not informed by the likelihood. This represents the exact point-null mixture prior, rather than replacing its point mass by a narrow continuous spike. Conditional on the indicators, $q\mid z$ has beta distribution $\mathrm{Beta}(a_q+{\sum }_i{z}_i,b_q+N-{\sum }_i{z}_i)$, providing a useful check on the update.

Let $p_0=\overline{\Phi }(2)$ and $p_1=\overline{\Phi }(2/\sqrt{1+V})$, where $\overline{\Phi }$ is the standard normal distribution upper-tail probability. The mixture model givesFor $V>0$, threshold inference for a mixture proportion therefore givesThis is a method-of-moments estimate when it lies in $[0,1]$. More formally, if $K$ of the $N$ independent genes exceed the threshold, then $K\sim \mathrm{Binomial}(N,(1-q)p_0+q{p}_1)$, and its constrained maximum-likelihood estimate clips the displayed value to $[0,1]$. No numerical estimate is determined without the given value of $V$.

Here $p_0\approx 0.02275$. Matching $15\%$ exactly requires $p_1\ge 0.15$, equivalently $V\ge [2/{\Phi }^{-1}(0.85){]}^2-1\approx 2.724$. For smaller positive $V$, the maximum is at $q=1$, and a large-sample exceedance fraction this high suggests a mismatch with the prescribed component variances. Sampling fluctuation is still possible; an observed fraction need not equal its expectation. If $V=0$, the denominator vanishes and this summary gives no information about $q$. Because $p_1<1/2$ for finite $V$, matching $15\%$ in expectation also forces $q>(0.15-p_0)/(0.5-p_0)\approx 0.267$. Thus the summary suggests tension with the earlier prior concentrated near $0.1$, under this mean-zero alternative model. The event is the one-sided exceedance $Y_i>2$, not $|Y_i|>2$.