How can you prove that a conditional bivariate Gaussian is a univariate Gaussian? Is there a way to prove that a bivariate Gaussian becomes a univariate Gaussian when conditioned on one of the two variables?
 A: Consider the density of a jointly Gaussian vector $(X,Y)$
$$f_{XY}(x,y)=\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2)}}\exp\left\{ -\frac{1}{2(1-\rho^2)}\left[ \frac{(x-\mu_X)^2}{\sigma_X^2}-2\rho\frac{(x-\mu_X)(y-\mu_Y)}{\sigma_X\sigma_Y}   +\frac{(y-\mu_Y)^2}{\sigma_Y^2} \right]  \right\}=$$
$$=\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2)}}\exp\left\{- \frac{1}{2(1-\rho^2)} \frac{(x-\mu_X)^2}{\sigma_X^2}  \right\}\cdot\exp\left\{ -\frac{1}{2(1-\rho^2)}\left[ \frac{(y-\mu_Y)^2}{\sigma_Y^2}-2\rho\frac{(x-\mu_X)(y-\mu_Y)}{\sigma_X\sigma_Y}    \right]  \right\}=$$
Now sum and subtract the following quantity in the argument of the second exp
$$\frac{\rho^2(x-\mu_X)^2}{\sigma_X^2}$$
...and after some (easy) algebraic manipulations you get
$$f_{XY}(x,y)=f_X(x)\cdot f_{Y|X}(y|x)$$
Where
$$X\sim N(\mu_X;\sigma_X^2)$$
$$(Y|X=x)\sim N\left[\mu_Y+\rho\frac{\sigma_y}{\sigma_X}(x-\mu_X);\sigma_Y^2(1-\rho^2)   \right]$$
Observe that if you set $\rho=0$ (uncorrelation) the joint density becomes
$$f_{XY}(x,y)=f_X(x)\cdot f_Y(y)$$
meaning that, in this model, uncorrelation and independence are equivalent.
