Let $S=\sigma^2+\tau^2$. For one posterior draw, the Gaussian quadratic-exponential moment is finite precisely when $\tau<\sigma$, and then
$$
\mathbb E_{\theta\mid y}[p(y\mid\theta)^{-2}]
=2\pi\sigma^2\sqrt{\frac{S}{\sigma^2-\tau^2}}
\exp\!\left[
\frac{\sigma^2y^2}{S(\sigma^2-\tau^2)}
\right].
$$
Because the $m$ posterior draws are independent,
$$
\operatorname{Var}_{\theta\mid y}(\widehat I)
=\frac{2\pi}{m}\left[
\sigma^2\sqrt{\frac{S}{\sigma^2-\tau^2}}
\exp\!\left\{\frac{\sigma^2y^2}{S(\sigma^2-\tau^2)}\right\}
-S\exp\!\left\{\frac{y^2}{S}\right\}
\right].
$$
For $\tau\geq\sigma$ the second moment, and hence the variance, is infinite. The <Harmonic mean estimator of Bayesian model evidence> is therefore unstable in the usual diffuse-prior regime: posterior sampling does not adequately control the reciprocal likelihood in the posterior tails.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 219 / 4 / a / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-219/4/a/iv/solution