Learned reconstruction targets and risk — Examples and Counterexamples

Since $Y=0$ almost surely, the constant kernel $K(0,\cdot )={\mathrm{\Pi }}_0$ satisfies the conditional-law identity both for the empty event and for the whole sample space; hence it is a posterior version for the stated joint law.

Direct integration gives $\mu (0)=\frac{1}{2}(-1)+\frac{1}{2}(1)=0$. Since $S=\{-1,1\}$, this posterior mean is not an admissible signal, and $S$ is nonconvex because it omits the midpoint of its two elements.

For every $a\in \mathbb{R}$, $\int (t-a{)}^2{\mathrm{\Pi }}_0(dt)=\frac{1}{2}(-1-a{)}^2+\frac{1}{2}(1-a{)}^2=1+a^2$. The unique minimizer over $\mathbb{R}$ is $0$; on the restricted set $S$, both values $a=\pm 1$ have risk $2$. This also agrees with the full-space proposition [F2], whose AC hypothesis is stated in the Example, while the restricted-action calculation is separate.

Under either prior, the deterministic observation is $Y=F(T)=0$, so the training and test data marginals are both ${\delta }_0$. The target equals $0$ almost surely under $P_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{i}\mathrm{n}}$ and equals $1$ almost surely under $P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}}$.

The rule $g(0)=0$ is measurable on the one-point data space and takes values in $Q$, so it is a point reconstruction rule.

By [F3], $R_{P_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{i}\mathrm{n}}}(g)=(0-0{)}^2=0$ and $R_{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}}}(g)=(1-0{)}^2=1$. Therefore the risk changes from zero to one while the input marginal remains fixed; no risk comparison follows from that marginal agreement alone.

Because $Y=0$ with probability one, the constant kernel ${\mathrm{\Pi }}_0=\frac{1}{2}{\delta }_{-1}+\frac{1}{2}{\delta }_1$ satisfies the conditional-law identity for the empty event and the whole sample space, so it is a posterior version.

The sampler kernel is $S_0={\mathrm{\Pi }}_0$, hence $\mathcal{L}(\tilde{T}\mid Y=0)={\mathrm{\Pi }}_0$ and exact posterior calibration holds by [F2]. Its mean is $\frac{1}{2}(-1)+\frac{1}{2}(1)=0$.

Under posterior risk, the action $0$ has loss expectation $\frac{1}{2}(-1-0{)}^2+\frac{1}{2}(1-0{)}^2=1$. By [F3], whose AC hypothesis is stated in the Example, it is the unique Bayes point action.

Conditional independence gives probability $1\mathrm{/}2$ to equal draws, with squared difference $0$, and probability $1\mathrm{/}2$ to unequal draws, with squared difference $4$. Therefore $E[(T-\tilde{T}{)}^2\mid Y=0]=\frac{1}{2}(0)+\frac{1}{2}(4)=2$, which is the randomized rule risk under [F4].

The exact posterior sample has conditional law ${\mathrm{\Pi }}_0$, but its random squared-error risk $2$ exceeds the posterior mean action's risk $1$; therefore posterior calibration of a sample does not make the sample the point Bayes action.