$Y|X \sim N(\mu X,X^2)$ and $X\sim N(\alpha, \beta)$. How is $Y$ distributed? Dear all,
I have recently been breaking my head over this question. The idea is that a certain variable $Y$ is normally distributed with a parameter $X$ in both mean and variance. 
$Y|X \sim N(\mu X,X^2)$
This parameter $X$ is assumed to be normally distributed as well with parameters $\alpha$ and $\beta$. 
$X\sim N(\alpha, \beta)$
Now I am interested in the distribution of $Y$ (with $X$ marginalized out).  
My current progress:
Simulation shows me that the distribution of Y seems approximately normal as well. This does not proof anything off course. 
The integral $\int_{all x}f_{Y|X}(s;x)f_{X}(s)dx$ seems unsolveable. 
Kind regards
 A: $$N(\mu X,X^2) \sim XN(\mu,1) \sim \mu X + XN(0,1) \sim N(\alpha \mu, \beta \mu^2 + \alpha^2) + \sqrt{\beta}N(0,1)N(0,1)$$
You thus have a sum between a normal distribution and a normal product distribution.
A: The characteristic function is 
$$\eqalign{ E\left[e^{itY}\right] &= E\left[ E\left[ e^{itY}|X \right]\right] \cr
&= E \left[ \exp(it\mu X - t^2 X^2/2 \right] \cr
&= \frac{\exp \left(\left(-(\alpha^2+\beta \mu^2) t^2 + 2 i \mu \alpha t\right)/\left(2 t^2 \beta + 2\right)
\right)}{\sqrt {{t}^{2}
\beta+1}}\cr}$$
It is certainly not a normal distribution, but might be approximated by a normal distribution when $\beta$ is small and $\alpha \ne 0$.  In fact, by expanding this in a series in powers of $\beta$ and taking the inverse Fourier transform, I get a density
$$f(y) =  \frac{\exp(-(y-\mu \alpha)^2/(2 \alpha^2))}{\sqrt{2 \pi |\alpha|}} \left(
1 + 
\frac { \left( 2\;{\alpha}^{4}+4\;x{\alpha}^{3}\mu+{x}^{2} \left( -5+{\mu}^{2} \right) {\alpha}^{2}-2\;{x}^{3}\alpha\mu+{x}^{4
} \right) }{2{\alpha}^{6}} \beta + \ldots\right)
$$
A: Let $\xi_1,\xi_2 \sim \mathcal{N}(0,1)$. Then $X = \alpha + \xi_1 \sqrt{\beta}$ and $Y = \mu X + \xi_2 X$ $ = \mu \alpha + \mu \xi_1  \sqrt{\beta} + \xi_2 \alpha + \xi_1 \xi_2  \sqrt{\beta}$.
Hence $E(Y) = \mu \alpha$, $Var(Y) = \mu^2 \beta  + \alpha^2 + \beta$.
