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I have the following set of equations: $$ y = \alpha x + \epsilon $$ $$ z=\beta y + \mu $$ and $\epsilon$ is N(0,$\sigma_{1}^2$), $\mu$ is N(0,$\sigma_{2}^2$) and $x$ is N(0,1) and they are all independent from each other. I can easily find the distribution of $z|y$, $y$ and $z$. However, I could not find the distribution of $y|z$ and $y|x,z$. Do you have any ideas?

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up vote 1 down vote accepted

Introduce the shorthand $g(t,u)=\exp\left(-t/(2u)\right)$.

The density of the distribution of $y$ is proportional to $g(y^2,\alpha^2+\sigma_1^2)$. The density of $(y,\mu)$ is proportional to $g(y^2,\alpha^2+\sigma_1^2)\cdot g(\mu^2,\sigma_2^2)$. Since $\mu=z-\beta y$, the density of $(y,z)$ is proportional to $$ g(y^2,\alpha^2+\sigma_1^2)\cdot g((z-\beta y)^2,\sigma_2^2)=h(z)\cdot g((y-\nu z)^2,\tau^2), $$ for some constants $\nu$ and $\tau^2$ to be determined and some irrelevant function $h$. Hence $y$ conditionally on $z$ is normal with mean $\nu z$ and variance $\tau^2$.

Algebraically, one must solve $$ \frac{y^2}{\alpha^2+\sigma_1^2}+\frac{(z-\beta y)^2}{\sigma_2^2}=\frac{(y-\nu z)^2}{\tau^2}+k(z), $$ for some function $k$ independent on $y$. This yields $$ \frac1{\tau^2}=\frac1{\alpha^2+\sigma_1^2}+\frac{\beta^2}{\sigma_2^2}, \qquad \frac{\nu}{\tau^2}=\frac{\beta}{\sigma_2^2}, $$ which determine uniquely $\nu$ and $\tau^2$.

The distribution of $y$ conditionally on $(x,z)$ can be computed by the same method.

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