A high ELBO alone does not certify posterior calibration

Fix any $c\in \mathbb{R}$. By the onto clause of [F5], choose $a>0$ with $\log a=c$. Define ${\gamma }_c=a\gamma$. Homogeneity [F4] gives $Z_c=\int {\gamma }_{c}d\mu =aZ$, so ${\pi }_c={\gamma }_c\mathrm{/}{Z}_c=\pi$ and both $Q$ and $\mathrm{\Pi }$ remain unchanged. Since ${\pi }_c=\pi$, the support condition in [F1] is unchanged. On the q-support, $\mathrm{\mid }\log {\gamma }_c\mathrm{\mid }\le \mathrm{\mid }\log \gamma \mathrm{\mid }+\mathrm{\mid }c\mathrm{\mid }$, so the logarithmic integrability condition remains finite.

On the q-support the logarithm product law [F5] gives $\log {\gamma }_c=\log \gamma +c$. Using [F2], ${\mathrm{ELBO}}_c(q)=\int q(\log \gamma +c-\log q)d\mu =\mathrm{ELBO}(q)+c$. Since $c$ is arbitrary, the absolute ELBO can be made arbitrarily large while $Q$, $\mathrm{\Pi }$, and any fixed discrepancy between them stay unchanged.

Under [A1], the decomposition [F3] gives $\log Z-\mathrm{ELBO}(q)=D(Q\mathrm{\parallel }\mathrm{\Pi })$. Without control of $\log Z$, a separate calibration theorem, or another posterior-discrepancy bound, the absolute ELBO alone supplies no calibration guarantee.