A unit-variance normal working model can miscalibrate credible intervals

The model and truth are dominated by Lebesgue measure. The supplied Gaussian density is normalized by [F1]. Its log likelihood is ${\mathrm{\ell }}_{\theta }(x)=-\frac{1}{2}(x-\theta {)}^2-\frac{1}{2}\log (2\pi )$, so $E_0[{\mathrm{\ell }}_0(X)-{\mathrm{\ell }}_{\theta }(X)]=\frac{1}{2}{\theta }^2$ because $E_{0}X=0$. Thus ${\theta }^{\ast }=0$ is the unique expected-log-likelihood maximizer and $V=1$. The score at zero is $s(X)=X$, hence $I=E_0{X}^2={\sigma }^2$. For $\mathrm{\mid }{\theta }_1\mathrm{\mid },\mathrm{\mid }{\theta }_2\mathrm{\mid }\le r$, $\mathrm{\mid }{\mathrm{\ell }}_{{\theta }_1}(X)-{\mathrm{\ell }}_{{\theta }_2}(X)\mathrm{\mid }\le \mathrm{\mid }{\theta }_1-{\theta }_2\mathrm{\mid }(\mathrm{\mid }X\mathrm{\mid }+r)$, a square-integrable local envelope. The local log likelihood ratio at $h\mathrm{/}\sqrt{n}$ is exactly $h\sqrt{n}\bar{X}-h^2\mathrm{/}2$. Also $E_0(\sqrt{n}\bar{X}{)}^2={\sigma }^2$, and $R^2{\mathbf{1}}_{\{\mathrm{\mid }\sqrt{n}\bar{X}\mathrm{\mid }>R\}}\le n{\bar{X}}^2$ gives $P_0(\mathrm{\mid }\sqrt{n}\bar{X}\mathrm{\mid }>R)\le {\sigma }^2\mathrm{/}{R}^2$, so $\sqrt{n}\bar{X}=O_{P_0}(1)$.

Completing the square in ${\prod }_i\varphi (X_i-\theta )\pi (\theta )$ gives posterior mean $m_n=n\bar{X}\mathrm{/}(n+{\tau }^{-2})$ and variance $v_n=(n+{\tau }^{-2}{)}^{-1}$. The empirical log likelihood is maximized at $\bar{X}$, since its $\theta$-dependent part is $n\bar{X}\theta -n{\theta }^2\mathrm{/}2$; for each $\epsilon >0$, ${\epsilon }^2{\mathbf{1}}_{\{\mathrm{\mid }\bar{X}\mathrm{\mid }>\epsilon \}}\le {\bar{X}}^2$, so $P_0(\mathrm{\mid }\bar{X}\mathrm{\mid }>\epsilon )\le {\sigma }^2\mathrm{/}(n{\epsilon }^2)$ and the estimator is consistent. The exact posterior variance satisfies $n{v}_n\to 1$, while the sample-mean variance is ${\sigma }^2\mathrm{/}n$.

The scaled posterior mean is $a_n=\sqrt{n}{m}_n=\frac{n}{n+{\tau }^{-2}}\sqrt{n}\bar{X}=O_{P_0}(1)$ and its variance parameter is $b_n=n\mathrm{/}(n+{\tau }^{-2})\le 1$. Given any deterministic $M_n\to \mathrm{\infty }$, on $\mathrm{\mid }{a}_n\mathrm{\mid }\le R$ its tail outside $[-M_n,M_n]$ is at most $P(\mathrm{\mid }Z\mathrm{\mid }>M_n-R)$ for $Z\sim N(0,1)$. First take $R$ large using boundedness in probability of $a_n$, then let $n\to \mathrm{\infty }$; this proves the every-diverging-radius posterior contraction premise.

Let $z=z_{1-\alpha \mathrm{/}2}={\mathrm{\Phi }}^{-1}(1-\alpha \mathrm{/}2)$. The exact central posterior interval is $[m_n-z\sqrt{v_n},m_n+z\sqrt{v_n}]$. Its coverage event for ${\theta }^{\ast }=0$, after multiplication by $\sqrt{n}$, is $\mathrm{\mid }\sqrt{n}\bar{X}\mathrm{\mid }\le z\sqrt{(n+{\tau }^{-2})\mathrm{/}n}$ because $m_n=n\bar{X}\mathrm{/}(n+{\tau }^{-2})$. By [F3], $\sqrt{n}\bar{X}\Rightarrow N(0,{\sigma }^2)$; the limiting interval endpoints are continuity points of this nondegenerate normal law, so [F4] gives limiting coverage $\mathrm{\Phi }(z\mathrm{/}\sigma )-\mathrm{\Phi }(-z\mathrm{/}\sigma )=2\mathrm{\Phi }(z\mathrm{/}\sigma )-1$.

Since $z>0$, the coverage in step 2.2 equals $1-\alpha$ when $\sigma =1$, is below $1-\alpha$ when $\sigma >1$, and is above $1-\alpha$ when $0<\sigma <1$. Indeed, $\mathrm{\Phi }$ is strictly increasing and $\mathrm{\Phi }(z)=1-\alpha \mathrm{/}2$. Thus the exact posterior width and sampling width disagree away from $\sigma =1$, giving nonnominal coverage.