on the behaviour of the sequence $(\tilde{\pi}_t)_{t\geq 0}$ is studied in more details in the following proposition.

\[prop:gaussian\_toy\] Assume that $h_t\rightarrow 0$. Then,

1.  \[part1\] $\tilde{\pi}_t\Rightarrow \delta_{{\mathbb{E}}[Y_1]}$ in ${\mathbb{P}}$-probability.

2.  \[part2\] If in addition we have $h_t\sim c t^{-\alpha}$ for some constants $(c,\alpha)\in(0,\infty)^2$ then

    1.  \[part2a\] for $p\in[1,2)$ we have ${\mathbb{P}}(\tilde{\pi}_t\Rightarrow \delta_{{\mathbb{E}}[Y_1]})=1$ for all $(Y_t)_{t\geq 1}$ such that ${\mathbb{E}}[|Y_1|^p]<\infty$ if and only if $t^{1/p} \tilde{\sigma}_t^2={\mathcal{O}}(1)$,

    2.  \[part2b\] the condition $t^{1/2} \tilde{\sigma}_t^2={\mathcal{O}}(1)$ is necessary but not sufficient to have ${\mathbb{P}}(\tilde{\pi}_t\Rightarrow \delta_{{\mathbb{E}}[Y_1]})=1$ for all $(Y_t)_{t\geq 1}$ such that ${\mathbb{E}}[ Y_1^2]<\infty$,

    3.  \[part2c\] for all $p\geq 1$, we have $t^{1/p} \tilde{\sigma}_t^2={\mathcal{O}}(1)$ only if $\alpha\geq  1/p$,

    4.  \[part2d\] $ h_t^2/\tilde{\sigma}_t^2\rightarrow 0$.

Part \[part2d\] of the proposition shows that $\tilde \sigma_t^2$, the variance of $\theta$ under $\tilde{\pi}_t$, converges to $0$ at a strictly slower rate than that of $h^2_t$ whenever $h_t\sim c t^{-\alpha}$. More precisely, the numerical results given in Table \[table:ex\] shows that, for this example, $\tilde{\sigma}_t^2\approx t^{-\alpha}$ when $h_t=t^{-\alpha}$ for all $t\geq 1$. As argued in the previous subsection, we observe in Table \[table:ex\] that $\bar{\sigma}_t^2$, the variance of $\theta$ under $\bar{\pi}_t$, does not converge to zero at a faster rate than $\tilde{\sigma}_t^2$. Indeed, in this example, both $\tilde{\sigma}_t^2$ and $\bar{\sigma}_t^2$ converge at 