Posterior Prediction, Credible Sets, and Bayes Actions — Examples

By [F1], $Q^{\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}}(\{1\})=\frac{1}{2}{Q}_0(\{1\})+\frac{1}{2}{Q}_1(\{1\})=\frac{1}{2}(0)+\frac{1}{2}(1)=\frac{1}{2}$. Since its support is $\{0,1\}$, this is $\mathrm{Bernoulli}(1\mathrm{/}2)$, with mean $1\mathrm{/}2$ and variance $1\mathrm{/}4$.

The conditional means are $\mu (0)=0$ and $\mu (1)=1$, and both conditional variances are zero. Thus ${\mathbb{E}}_{\mathrm{\Pi }}[\mathrm{Var}(Z\mid \theta )]=0$ and ${\mathrm{Var}}_{\mathrm{\Pi }}(\mu )=1\mathrm{/}4$.

The predictive variance in step 1.1 equals the sum of those two terms in step 1.2, as asserted by [F2]. Here the entire predictive variance comes from posterior uncertainty about $\theta$, while future sampling conditional on $\theta$ contributes zero.

Under zero–one loss, the risks are $1-\mathrm{\Pi }(\{0\})=0.7$ for action $0$ and $1-\mathrm{\Pi }(\{1\})=0.3$ for action $1$. Thus action $1$ is the unique zero–one Bayes action, in agreement with [F2].

Under the asymmetric loss, the risk of action $1$ is $3\mathrm{\Pi }(\{0\})=0.9$, while the risk of action $0$ is $\mathrm{\Pi }(\{1\})=0.7$. Both risks are finite and the lower one is attained uniquely by action $0$.

The two loss functions therefore select different actions under the same posterior: zero–one loss selects $1$, while the stated asymmetric loss selects $0$. This demonstrates that posterior modes alone do not determine a Bayes action without a specified loss.

For $m\in [0,1]$, $\mathrm{\Pi }((-\mathrm{\infty },m])\ge 1\mathrm{/}2$ and $\mathrm{\Pi }([m,\mathrm{\infty }))\ge 1\mathrm{/}2$, so [F1] makes every such $m$ a median.

For $m\in [0,1]$, $\rho (m)=\frac{1}{2}\mathrm{\mid }m\mathrm{\mid }+\frac{1}{2}\mathrm{\mid }1-m\mathrm{\mid }=\frac{1}{2}$. For $a<0$, $\rho (a)=\frac{1}{2}(-a)+\frac{1}{2}(1-a)=\frac{1}{2}-a>\frac{1}{2}$; for $a>1$, $\rho (a)=\frac{1}{2}a+\frac{1}{2}(a-1)=a-\frac{1}{2}>\frac{1}{2}$. Thus precisely the interval $[0,1]$ minimizes risk, in agreement with [F2].

The density is normalized because ${\int }_0^{1}6\theta (1-\theta )d\theta =6(1\mathrm{/}2-1\mathrm{/}3)=1$. Also $\theta (1-\theta )=1\mathrm{/}4-(\theta -1\mathrm{/}2{)}^2$, so the density has its unique maximum at $1\mathrm{/}2$.

Absolute continuity gives $\mathrm{\Pi }(\{a\})=0$ for every point action $a$. Hence $\rho (a)=\int {\mathbf{1}}_{\{\theta \mathrm{\ne }a\}}\mathrm{\Pi }(d\theta )=1-\mathrm{\Pi }(\{a\})=1$ for every $a$. All actions tie under this loss, even though the density has a unique mode.

The joint density/mass is $1\cdot \theta$ at $Y=1$ and $1\cdot (1-\theta )$ at $Y=0$. Its marginal masses are ${\int }_0^1\theta d\theta =1\mathrm{/}2$ and ${\int }_0^1(1-\theta )d\theta =1\mathrm{/}2$. Therefore the posterior densities are $p(\theta \mid Y=1)=2\theta$ and $p(\theta \mid Y=0)=2(1-\theta )$; each conditional law integrates to one and satisfies the defining conditioning identity.

Their CDFs on $[0,1]$ are $F_1(x)=x^2$ and $F_0(x)=1-(1-x{)}^2$. Hence the central 95% intervals are $C(1)=[\sqrt{0.025},\sqrt{0.975}]$ and $C(0)=[1-\sqrt{0.975},1-\sqrt{0.025}]$. Each has posterior mass $0.975-0.025=0.95$, also in accordance with [F1, F2].

For fixed true ${\theta }_0=0.005$, $\sqrt{0.025}>0.005$ because $0.025>{0.005}^2$. Also $1-\sqrt{0.975}>0.005$ because $0.975<{0.995}^2$. Thus ${\theta }_0$ lies below the lower endpoint of both possible intervals, so $P_{{\theta }_0}\{{\theta }_0\in C(Y)\}=0$, not $0.95$. This verifies that posterior credibility is not a frequentist coverage guarantee.