I have a conditional: $p(x|a,d,q) = \mathcal{N}(x|d/(1-a),q/(1-a^2))$, where $\mathcal{N}(t|u,v)$ denotes a normally distributed variable $t$ with mean $u$ and variance $v$.
I have a prior over $a$ and $d$: $p(a,d|\mu_a,\mu_d,\sigma^2_a,\sigma^2_d)$, which is a bivariate normal distribution $\mathcal{N}([a \ d]^\intercal|[\mu_a \ \mu_d]^\intercal,diag(\sigma^2_a,\sigma^2_d))$ that is truncated at 0 and 1 for $a$ but unbounded for $d$.
All variables ($x, a, d, q, \mu_a, \mu_d, \sigma^2_a, \sigma^2_d$) are scalars, $^\intercal$ denotes a transpose and $diag$ denotes a diagonal covariance matrix.
I want to integrate out the unknown $a$ and $d$ from the joint $p(x|a,d,q)p(a,d|\mu_a,\mu_d,\sigma^2_a,\sigma^2_d) = p(x,a,d||q,\mu_a,\mu_d,\sigma^2_a,\sigma^2_d)$ to obtain the marginal $p(x|q,\mu_a,\mu_d,\sigma^2_a,\sigma^2_d)$.
I note that the mean of $p(x|a,d,q)$ is not a linear function of $[a \ d]^\intercal$ and the variance of $p(x|a,d,q)$ depends on $a$. Both these properties violate the standard linear Gaussian model that I am familiar with. Does this marginal have an analytical solution? Is the desired marginal even still a Gaussian? Can anybody tell me what it is, or point me to a derivation for this type of marginal?
Thanks in advance!