Note that only the results for $\alpha\in\{0.5,1\}$ are presented for $d_{\mathrm{MMD}}(\tilde{\pi}_t,\pi_t)$. []{data-label="fig:toy_mmd"}](figs/toy_example_mmdBar.pdf "fig:")

It is easily verified that, for this example, the posterior distribution $\pi_t$ converges ${\mathbb{P}}$-a.s. to $\delta_{{\mathbb{E}}[Y_1]}$, and it is of interest to find conditions on $(h_t)_{t\geq 0}$ that preserve this mode of convergence. If ${\mathbb{E}}[Y_1^2]<\infty$, so that $\theta_\star$ is well defined, Parts \[part2b\] and \[part2c\] of Proposition \[prop:gaussian\_toy\] imply that if $h_t\sim c t^{-\alpha}$ then the almost sure convergence of $\tilde{\pi}_t$ towards $\delta_{{\mathbb{E}}[Y_1]}$ may not hold for $\alpha\leq 0.5$. Notice that if $\alpha=0.5$ then $\ h^2_t/\sigma_{t}^2={\mathcal{O}}(1)$, where $\sigma_t^2={\mathcal{O}}(1/t)$ is the posterior variance of $\theta$. If we relax the assumption ${\mathbb{E}}[Y^2_1]<\infty$ then $\tilde{\pi}_t$ may not converge almost surely to $\delta_{{\mathbb{E}}[Y_1]}$ when $\alpha<1$, by Parts \[part2a\] and \[part2c\] of the proposition. Proposition \[prop:gaussian\_toy\] therefore shows that $h_t$ must converge very quickly to zero to guarantee that $\tilde{\pi}_t$ preserves the almost sure convergence property of $\pi_t$.

To illustrate the result of Proposition \[prop:continuity\], namely that $\tilde{\pi}_t$ gets closer to $\pi_t$ as $\max\{h_s,\, s<t\}$ decreases, Figure \[fig:toy\_mmd\] (left plot) shows the evolution, as $t$ increases, of the maximum mean discrepancy (MMD) distance $d_{\mathrm{MMD}}(\tilde{\pi}_t,\pi_t)$ between $\tilde{\pi}_t$ and $\pi_t$, when $h_t=t^{-\alpha}$ for all $t\geq 1$ and $\alpha\in\{0.5,1\}$. The MMD distance is computed using 