Choose an action by posterior expected loss

Try

Before observing Bernoulli data, assign p a Beta(2,2) prior. Its density is proportional to p(1−p) on the unit interval. After three successes and one failure, multiply this density by the likelihood p³(1−p). The posterior is Beta(5,3), with mean 5/8. Every exponent has a traceable source: data add counts, while the prior contributes its own parameters. This is a specified model of parameter uncertainty, not a likelihood normalized without declaring a prior.

Predict

Predict whether the posterior mean is always the best number to report. It minimizes posterior expected squared error, but absolute error calls for a median, and asymmetric costs can call for a quantile or a thresholded action. An optimality claim must name the loss. Also predict whether a 95% posterior credible interval automatically has 95% frequentist coverage at every fixed p. It need not. Posterior probability conditions on the observed data under the prior model; frequentist coverage repeats data generation at a fixed parameter.

Experiment

Update the same synthetic record under two proper priors and compare the resulting means. Both calculations are internally valid; their disagreement reveals prior sensitivity at this sample size.

Prior	Observed successes/failures	Posterior	Posterior mean
Beta(1,1)	3 / 1	Beta(4,2)	2/3
Beta(2,2)	3 / 1	Beta(5,3)	5/8

Now imagine thirty successes and ten failures. Derive the two posterior means again and compare the size of their difference. This exercise illustrates diminishing influence in this conjugate setting, not a theorem that all priors become harmless in every model. Nonidentifiability and strongly conflicting priors require their own analysis.

Understand

For a Beta(a,b) prior with a,b>0 and a binomial sample with s successes in n trials, the posterior is Beta(a+s,b+n−s). Its mean is (a+s)/(a+b+n). Integrating the next Bernoulli success probability over the posterior gives the same number as the posterior predictive success probability. A predictive statement concerns a future observation; a parameter estimate and an interval for p concern different objects even when they share this calculation.

Under squared loss, posterior expected loss decomposes into (action−posterior mean)² plus posterior variance, proving the mean’s optimality when the second moment exists. For absolute loss, a median minimizes the expectation: left and right slopes bracket zero exactly when the posterior cumulative probability crosses one half. Atoms can produce nonunique medians, so a derivative argument assuming a smooth density is not the most general proof. These are conditional decisions after observing data. Integrating their risks over a proper prior produces a Bayes-risk comparison under additional integrability conditions.

Apply

Suppose an intervention costs c if the hidden state is zero, while withholding it costs d if the state is one. Let q be the posterior probability of state one. The two posterior expected losses are c(1−q) and dq. Act when q>c/(c+d), with a tie at equality, for positive costs. With c=3 and d=1 the threshold is 3/4. A probability threshold of one half is not universal; it reflects equal error costs in this binary setup.

If an improper prior is used, prove that the posterior normalizing integral is finite and positive. An improper prior is not a probability distribution, so statements about prior-averaged risk cannot be imported without qualification. Sensitivity analysis should compare plausible prior choices and explain which substantive information each represents, rather than presenting conjugacy as evidence that a prior is scientifically appropriate.

Continue

Keep prior, likelihood, posterior, prediction and loss as separate entries in the reasoning. The exact-risk widget in Bias, variance and risk can compare the frequentist risks of beta-prior posterior means at fixed p; that is different from averaging risk over the prior. Continue to Multivariate normal projections for the geometric machinery behind regression inference.