Credible sets and frequentist coverage — Examples and Counterexamples

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 ${\theta }_{\ast }$.

Every squared coordinate is either zero or $M{i}^{-1-2\beta }$, so ${\sup }_i{i}^{1+2\beta }{\theta }_{0,i}^2\le M$ and ${\sum }_i{\theta }_{0,i}^2\le M{\sum }_i{i}^{-1-2\beta }<\mathrm{\infty }$. Also $N_1=2^{1\mathrm{/}s}\in (1,2)$ and ${\rho }_1=2$, hence $N_1\mathrm{/}{\rho }_1<1\le N_1$ and the first zero gap contains $i=1$. Thus ${\theta }_0\in {\mathrm{\Theta }}_{\beta }(M)$ and ${\theta }_{0,1}=0$.

Fix any polished-tail parameters $L_0\ge 1$, integer $N_0\ge 1$, and ${\rho }_{\ast }\ge 2$. The recursion gives $N_j\mathrm{/}{\rho }_j\ge 2{\rho }_j{N}_{j-1}>{\rho }_{j-1}{N}_{j-1}$, so every earlier gap ends below $N_j\mathrm{/}{\rho }_j$. Also $N_j^{\mathrm{\prime }}=\lfloor 2N_j\rfloor +1\le 2N_j+1$; because $N_j\to \mathrm{\infty }$ and ${\rho }_j\to \mathrm{\infty }$, for all sufficiently large $j$ we have $N_j^{\mathrm{\prime }}\ge N_0$ and ${\rho }_{\ast }(2N_j+1)\le {\rho }_j{N}_j$, whence $\lfloor {\rho }_{\ast }{N}_j^{\mathrm{\prime }}\rfloor \le {\rho }_j{N}_j$. Since $N_j^{\mathrm{\prime }}>2N_j$, the block $[N_j^{\mathrm{\prime }},\lfloor {\rho }_{\ast }{N}_j^{\mathrm{\prime }}\rfloor ]$ lies in the current zero gap $(2N_j,{\rho }_j{N}_j]$. For all sufficiently large $j$, $i_j=\lfloor {\rho }_j{N}_j\rfloor +1\le {\rho }_j{N}_j+1<4{\rho }_j{N}_j\le N_{j+1}\mathrm{/}{\rho }_{j+1}$, so $i_j$ is before the next first gap; applying the same separation at later indices puts all still later gaps farther right. Since ${\rho }_j{N}_j>2N_j+1\ge N_j^{\mathrm{\prime }}$ eventually, $i_j>N_j^{\mathrm{\prime }}$; earlier gaps end below $N_j\mathrm{/}{\rho }_j<i_j$. Hence $i_j$ lies outside every zero gap and is in the tail from $N_j^{\mathrm{\prime }}$, with ${\theta }_{0,i_j}^2=M{i}_j^{-1-2\beta }>0$. The block sum is zero while that tail is positive, violating [F2]. Thus ${\theta }_0$ is outside every fixed polished-tail class.

For a fixed $L\ge 1$, take $L_{n_j}=L$ and $K_L=L\mathrm{/}\sqrt{M}$. Since $(1+2p)\mathrm{/}(8+8\beta +8p)>0$ and ${\rho }_j\ge 1$, we have $L_{n_j}=L\le K_L\sqrt{M}{\rho }_j^{(1+2p)\mathrm{/}(8+8\beta +8p)}$ for every $j$. Applying [F3] gives $P_{{\theta }_0}^{(n_j)}\{{\theta }_0\in {\widehat{C}}_{n_j}(L)\}\to 0$. By [F6] this is the frequentist coverage at the fixed truth; because ${\theta }_0\in {\mathrm{\Theta }}_{\beta }(M)$ by step 1.1, the infimum coverage over that hyperrectangle cannot have lower limit at least $1-\gamma$.

The positive coverage statement [F4] has the additional polished-tail hypothesis, which fails for ${\theta }_0$ by step 2.1. Thus the sparse-gap undercoverage does not contradict the polished-tail theorem; it refutes only the broader fixed-inflation claim over the full hyperrectangle.