Probability theory question: bivariate normal and conditional probability

Factual Questions
1

First of all, this is from a homework assignment, but I didn’t see any rule forbidding such questions so I figured it was okay to ask.

Suppose a random variable X has a normal distribution, and for every x the conditional distribution of another random variable Y given that X=x is normal with mean a*x+b and variance t^2 where a,b, and t^2 are constants. Prove that the joint distribution of X and Y is a bivariate normal.

I have tried several approaches to this problem, but so far I haven’t been successful. My first reaction is to multiply the conditional pdf of Y by the marginal on X to get the joint pdf, and then show that this is equivalent to the joint pdf of a bivariate normal. However, for some reason this doesn’t work. Why is this? (algebra below)

Let f[sub]1/sub be the marginal on x, and g(y|x) be the conditional pdf for Y given that X=x. Then the joint pdf f(x,y) is:

f(x,y)=f[sub]1/sub*g(y|x)

(assume X has mean m and variance s[sub]1[/sub]^2) Substitution gives:

f(x,y)=1/(sqrt(2pi)*s[sub]1[/sub])e^(-0.5((x-m[sub]1[/sub])/s[sub]1[/sub])^2)*1/(sqrt(2pi)*t)e^(-0.5((y-ax-b)/t)^2)

Since the above expression does not contain rho, I’m not sure how to get it into the standard form for a bivariate normal, which does include rho. I’m not looking for a step by step explanation, but some general suggestions would be helpful.

2

The basic rule around these parts is:

If you want us to DO your homework, that ain’t allowed.

But you are looking for suggestions. That I’ll allow.

samclem GQ moderator

3

Hint: rho describes the strength of the linear relationship between X and Y. What quantity that you mentioned above has a similar purpose?

4

Rho depends mostly on t. If t is 0 and a is not 0, then rho must be 1. As t gets large rho gets small, but I’m not sure how to use this to determine exactly what rho is in terms of the other variables.