Gaussian distribution with poisson variance Let's $r_t\sim \mathcal{N}(0,V_t)$ where $V_t = (1+m)^{J_t}$, $m\geq0$ and $J_t\sim \mathcal{P}oi(\lambda)$. How can we compute the distribution of $r_t$ when the parameters are $m$ and $\lambda$. In other words I would like to compute $$\sum_{j\geq0}\frac{e^\lambda}{j!}\lambda^j\frac{1}{\sqrt{2\pi(1+m)^{j}}}\exp\left\{-\frac{r_t^2}{2(1+m)^{j}}\right\}\,.$$
I know this sum is convergent as for all $j$, $\exp\left\{-\frac{r_t^2}{2(1+m)^{j}}\right\}\leq 1$ then $$\sum_{j\geq0}\frac{e^\lambda}{j!}\lambda^j\frac{1}{\sqrt{2\pi(1+m)^{j}}}\exp\left\{-\frac{r_t^2}{2(1+m)^{j}}\right\} \leq \sum_{j\geq0}\frac{e^\lambda}{j!}\lambda^j\frac{1}{\sqrt{2\pi(1+m)^{j}}} = \frac{\exp\left\{\lambda+\lambda(1+m)^{-1/2}\right\}}{\sqrt{2\pi}}\,.$$
My objective at the end will be to estimate this model by MCMC. So, I would like to sample parameters $m$ and $\lambda$ after computing their posterior. An accepted answer should then help me to compute those posteriors.
Thanks
 A: Since $J_t$ has support on the non-negative integers, you have a Gaussian mixture model, with marginalised sampling density:
$$p_{R_t}(r | \lambda,m) 
= \frac{1}{\sqrt{2 \pi}} \sum_{j=0}^\infty \frac{1}{j!} \Big( \frac{\lambda}{\sqrt{1+m}} \Big)^j \exp \bigg( - \frac{r^2}{2(1+m)^j} - \lambda \bigg).$$
This distribution is non-trivial, but we know it is symmetric around zero (and therefore has zero mean and zero skewness).  Using the law of iterated variance its variance is:
$$\begin{align}
\mathbb{V}(R_t | \lambda, m)
&= \mathbb{E}[\mathbb{V}( R_t |J_t)] + \mathbb{V}[\mathbb{E}( R_t |J_t)] \\[6pt]
&= \mathbb{E}[(1+m)^{J_t}] + \mathbb{V}[0] \\[6pt]
&= G_{J_t}(1+m) \\[6pt]
&= \exp(\lambda m). \\[6pt]
\end{align}$$
(In this working $G_{J_t}$ denotes the probability generating function of $J_t$.)  It is also simple to establish the marginal kurtosis.  As a preliminary step we have:
$$\begin{align}
\mathbb{E}(R_t^4 | \lambda, m)
&= \sum_{j=0}^\infty \frac{\lambda^j e^{-\lambda}}{j!} \cdot \mathbb{E}(R_t^4 | J_t) \\[6pt]
&= \sum_{j=0}^\infty \frac{\lambda^j e^{-\lambda}}{j!} \cdot \frac{3}{(1+m)^{2j}} \\[6pt]
&= 3 e^{-\lambda} \sum_{j=0}^\infty \frac{1}{j!} \cdot \Big( \frac{\lambda}{(1+m)^2} \Big)^j \\[6pt]
&= 3 \exp({-\lambda}) \exp({\lambda/(1+m)^2}) \\[6pt]
&= 3 \exp \Bigg( - \lambda \cdot \frac{2m + m^2}{(1+m)^2} \Bigg). \\[6pt]
\end{align}$$
We therefore have the marginal kurtosis
$$\begin{align}
\mathbb{Kurt}(R_t | \lambda, m)
&= \frac{\mathbb{E}(R_t^4 | \lambda, m)}{\mathbb{V}(R_t^4 | \lambda, m)^2}. \\[6pt]
&= 3 \exp \Bigg( - \lambda \cdot \frac{2m + m^2}{(1+m)^2} \Bigg) \exp(-2 \lambda m) \\[6pt]
&= 3 \exp \Bigg( - \lambda \cdot \frac{4m + 5m^2 + 2 m^3}{(1+m)^2} \Bigg). \\[6pt]
\end{align}$$
Given that the this Gaussian mixture is symmetric with a known variance and kurtosis, you could approximate it well by a generalised error distribution with appropriate choice of parameters using the method of moments using the above marginal variance and kurtosis.  This would give you a substantially simplified approximation to your problem.  In any case, your MCMC simulations of the posterior will depend on the prior distribution you use for $\lambda$ and $m$, which you have not specified here.
