# prove a bivariate normal distribution

$X$ has a normal distribution. The conditional distribution of another random variable $Y$ given that $X=x$ is a normal distribution with mean $ax+b$ and variance $t^2$, where $a$, $b$, and $t^2$ are constants. How can I prove that the joint distribution of $X$ and $Y$ is bivariate normal distribution?

-

By definition, $$f_{X,Y}(x,y)=f_X(x)\cdot f_{Y\mid X}(y\mid x)=\frac1{\sqrt{2\pi\sigma^2}}\mathrm e^{-(x-\mu)^2/(2\sigma^2)}\cdot\frac1{\sqrt{2\pi t^2}}\mathrm e^{-(y-ax-b)^2/(2t^2)}$$ Hence the task is to solve for $M$ in $\mathbb R^2$ and $C$ a definite positive $2\times2$ matrix, the fact that for every $z=(x,y)^t$, one has the identity $$f_{X,Y}(z)=\frac1{2\pi\sqrt{\det C}}\mathrm e^{-\frac12(z-M)^tC^{-1}(z-M)}.$$ Hints: note that the diagonal of $C^{-1}$ is $(1/\sigma^2+a^2/t^2,1/t^2)$ and that $M=(\mu,a\mu+b)^t$. The rest of the proof relies on some simple algebraic manipulations of a degree $2$ polynomial in $(x,y)$.
Haha! +1   – Tim Jul 8 '12 at 20:18