A proper Gaussian prior yields a credible interval with undercoverage

Under [A1], the standard normal probability law of [F2] is supplied. Its density $\varphi$ is positive and symmetric and integrates to one, so $\mathrm{\Phi }$ is continuous and strictly increasing, with $\mathrm{\Phi }(0)=1\mathrm{/}2$ and $\mathrm{\Phi }(y)\to 1$ as $y\to \mathrm{\infty }$. Choose $B>0$ with $\mathrm{\Phi }(B)>0.975$. By [F3] on $[0,B]$, there is a unique $z>0$ with $\mathrm{\Phi }(z)=0.975$.

The prior density times likelihood is proportional to $\exp \{-\frac{1}{2}[{\theta }^2+(Y-\theta {)}^2]\}=\exp \{-(\theta -Y\mathrm{/}2{)}^2\}$ times a factor independent of $\theta$. Normalization gives posterior law $N(Y\mathrm{/}2,1\mathrm{/}2)$.

Under the fixed truth ${\theta }_{\ast }=2r+1$, write $Y={\theta }_{\ast }+Z$. The event ${\theta }_{\ast }\in I(Y)$ is equivalent to $\mathrm{\mid }Z-{\theta }_{\ast }\mathrm{\mid }\le 2r$, hence to $1\le Z\le 4r+1$.

The interval membership graph is $\{(Y,\theta ):\mathrm{\mid }\theta -Y\mathrm{/}2\mathrm{\mid }\le r\}$, the inverse image of the closed interval $[-r,r]$ under a continuous map. It is closed, hence Borel, so the sampling membership event is measurable.

Under that posterior, $(\theta -Y\mathrm{/}2)\mathrm{/}\sqrt{1\mathrm{/}2}$ is standard normal. The interval radius $r=z\mathrm{/}\sqrt{2}$ therefore gives posterior mass $\mathrm{\Phi }(z)-\mathrm{\Phi }(-z)=0.975-0.025=0.95$ for every observed $Y$.

Since the standard normal density is positive and symmetric, $P(1\le Z\le 4r+1)<P(Z\ge 0)=1\mathrm{/}2<0.95$. Thus the interval has posterior mass $0.95$ but frequentist coverage below $0.95$ at ${\theta }_{\ast }$.