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$X$ 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?

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up vote 7 down vote accepted

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 $(x,y)$.

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thank you for your clue – perry zhu Jul 6 '12 at 15:02
You are welcome. Why did you unaccept this answer? – Did Jul 8 '12 at 19:45
Haha! +1 $ $ $ $ – Tim Jul 8 '12 at 20:18

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