Posterior computation and error accounting — Examples and Counterexamples

Since $0<\epsilon <1\mathrm{/}2$, each row assigns the nonnegative masses $1-\epsilon$ and $\epsilon$ to the two states, and every subset is measurable; thus $K_{\epsilon }$ is a probability kernel [F1]. Under AC [A1], apply [F6] to initial law ${\delta }_{+1}$ and kernel $K_{\epsilon }$; this supplies the canonical chain used in the Example.

For either singleton $\{x\}$, $\pi K_{\epsilon }(\{x\})=\frac{1}{2}(1-\epsilon )+\frac{1}{2}\epsilon =\frac{1}{2}=\pi (\{x\}),$ so additivity on the two-state space gives $\pi K_{\epsilon }=\pi$ by [F2].

Put ${\lambda }_0:=1-2\epsilon =499\mathrm{/}500$. Summing the two conditional probabilities in [F3] against the values $+1$ and $-1$, conditional on $X_n=x$, and using linearity [F7], $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=(1-\epsilon )x+\epsilon (-x)={\lambda }_{0}x\mathrm{.}$ Taking expectations by [F7] and using $X_0=1$ gives $\mathbb{E}{X}_j={\lambda }_0^j$ for every $j\ge 1$ by induction.

By [F8], $\mathbb{E}[{100}^{-1}{\sum }_{j=1}^{100}{X}_j]={100}^{-1}{\sum }_{j=1}^{100}{\lambda }_0^j=\frac{499}{100}(1-(499\mathrm{/}500{)}^{100})>0$. Since $\pi (f)=0$ and $\pi$ is invariant by step 2.1, the kernel discrepancy vanishes, while [F4] identifies this positive difference as the burn-in component.

Since $0<\epsilon <1\mathrm{/}2$, each row assigns the nonnegative masses $1-\epsilon$ and $\epsilon$ to the two states, and every subset is measurable; thus $K_{\epsilon }$ is a probability kernel [F1]. Under AC [A1], apply [F8] to initial law $\pi$ and kernel $K_{\epsilon }$; this supplies the canonical chain used in the Example.

For either singleton $\{x\}$, $\pi K_{\epsilon }(\{x\})=\frac{1}{2}(1-\epsilon )+\frac{1}{2}\epsilon =\frac{1}{2}=\pi (\{x\})$; additivity gives $\pi K_{\epsilon }=\pi$ by [F2].

Put $\lambda =1-2\epsilon =49\mathrm{/}50$. On the two-point state space, summing the conditional probabilities in [F3] against the values $+1$ and $-1$ and using linearity [F9] gives $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=\lambda X_n$. Since $X_0\sim \pi$, $\mathbb{E}{X}_0=0$; expectation preservation [F9] gives $\mathbb{E}{X}_n=0$ for all $n$. As $X_0$ is bounded and ${\mathcal{F}}_k$-measurable, [F10] and expectation preservation [F9] give $\mathbb{E}[X_0{X}_{k+1}]=\lambda \mathbb{E}[X_0{X}_k]$, and $\mathbb{E}[X_0^2]=1$. Induction yields $\mathrm{Cov}(X_0,X_k)={\lambda }^k$ and, since $\mathrm{Var}(X_0)=1$, ${\rho }_f(k)={\lambda }^k$.

Since $X_0\sim \pi$ and $\pi$ is invariant by step 2.1, [A1] permits the invariant-initial-law result [F4], which makes the chain strictly stationary. Its one-time law is symmetric, so $\mathbb{E}{X}_j=0$ and $\mathrm{Var}(X_j)=1$.

Since the chain is strictly stationary by step 3.1 and has variance one, [F5] applies. Step 2.2 gives ${\rho }_f(k)={\lambda }^k$; these autocorrelations are absolutely summable because $0<\lambda <1$, with ${\sum }_{k\ge 1}{\lambda }^k=\lambda \mathrm{/}(1-\lambda )=49$. Thus ${\tau }_f=1+2\cdot 49=99>0$ and $[F5]$ gives ${\mathrm{ESS}}_f(\mathrm{10,000})=\mathrm{10,000}\mathrm{/}99$. Since ${\gamma }_k={\rho }_f(k)$ here, [F6] applies, but neither it nor the ESS definition supplies a CLT.

With the likelihoods stated in the Example, the exact prior-times-likelihood product is $(2\pi {)}^{-1\mathrm{/}2}{e}^{-u^2}$. The substitution $z=\sqrt{2}u$ and the normalized standard-normal integral in [F2] give ${\int }_{\mathbb{R}}{e}^{-u^2}du=\sqrt{\pi }$, so this product has normalizing integral $(2\pi {)}^{-1\mathrm{/}2}\sqrt{\pi }=1\mathrm{/}\sqrt{2}$. Its normalized density is therefore ${\pi }^{-1\mathrm{/}2}{e}^{-u^2}$, which is $N(0,1\mathrm{/}2)$ by [F2]. For the solver output, the prior-times-likelihood product is $(2\pi {)}^{-1\mathrm{/}2}{e}^{-u^2\mathrm{/}2-(u+2{)}^2\mathrm{/}2}=(2\pi {)}^{-1\mathrm{/}2}{e}^{-(u+1{)}^2-1};$ its normalizing integral is $(2\pi {)}^{-1\mathrm{/}2}{e}^{-1}\sqrt{\pi }=e^{-1}\mathrm{/}\sqrt{2}$, and its normalized density is ${\pi }^{-1\mathrm{/}2}{e}^{-(u+1{)}^2}=N(-1,1\mathrm{/}2)$ by [F2].

The exact density is symmetric about zero, so its mass on $(-\mathrm{\infty },0]$ is $1\mathrm{/}2$. Under the solver density, $Z=\sqrt{2}(U+1)$ is standard normal, and $U\le 0$ is equivalent to $Z\le \sqrt{2}$; hence $\tilde{\mathrm{\Pi }}(f)={\mathrm{\Phi }}_0(\sqrt{2})$. The standard normal density is positive and symmetric, so ${\mathrm{\Phi }}_0(\sqrt{2})>{\mathrm{\Phi }}_0(0)=1\mathrm{/}2$.

In the error ledger set ${\mu }_{\star }={\mu }_{\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{e}\mathrm{l}}=\mathrm{\Pi }$ and ${\mu }_{\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{r}}={\mu }_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{c}}=\tilde{\mathrm{\Pi }}$; then the solver-stage term is ${\delta }_{\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{r}}={\mathrm{\Phi }}_0(\sqrt{2})-1\mathrm{/}2$ and the model and discretization terms vanish. Since $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=1$, [F3] also gives ${\delta }_{\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{r}}\le 2\mathrm{\parallel }\mathrm{\Pi }-\tilde{\mathrm{\Pi }}{\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}$.