I want to identify the marginal of a normal distribution subject to a restriction.
Take two normally distributed random variables $x,y$. Their pdfs are denoted by$\phi(x)$ and $\phi(y)$. The moments of the marginal of $y$ and the conditional of $y|x $ are known and given by $y \sim N(\mu_y, \sigma^{2}_y)$ and $y|x \sim N(\mu_{y|x}, \sigma^{2}_{y|x}) $ respectively. How can one calculate the moments of the distribution of $x \sim N(\mu_x, \sigma^{2}_{x})$ such that: \begin{equation} \phi(y) = \int \phi(y|x) \phi(x) dx \end{equation} is satisfied?
Instead of trying to solve the problem brute force, I exploited the following known relationships:
\begin{align} E(Y) &= E_X( E(Y|X) ) \\ Var(Y) &= E_X ( Var(Y|X) ) + Var_X ( E (Y|X)) \end{align}
Employing these two relationships render solving the problem above much easier. Can I indeed use these two relationships?
The constraint you mention isn't really a constraint.
$\int \phi(y|x) \phi(x) dx = \int \phi(x,y) dx = \phi(y)$
where $\phi(x,y)$ is the joint distribution of X and Y, which in this case is bivariate normal. So this doesn't impose any constraints on the moments of X other than the standard ones for a normal distribution.