Confusion related to probability distribution

Question

I was reading a book where it said that if the distribution of a multivariate gaussian with three variables $x,y,z$ given by

$\pi(x,y,z) = f(x,z)g(y,z)$ then $x$ and $y$ are conditionally independent given z. How it be true? Can anyone help me?

Answer 1

Given a value of $Z=z$ the joint density for $X$ and $Y$ factors into a product of two functions $f'(X|Z=z)$ and $g'(Y|Z=z)$ (Note: I am using the symbol $'$ to distinguish between $f(.)$ and $g(.)$ and is not meant to stand for the derivative of the respective functions). You can show this formally by dividing the joint by $\int\int\pi(x,y,z) \,dx \,dy$

We have:

$$h(x,y|Z=z) = \frac{\pi(x,y,z)}{\int\int\pi(x,y,z) \,dx \,dy}$$

But,

$$\int\int\pi(x,y,z) \,dx \,dy =\int\int f(x,z) g(y,z) \,dx \,dy = \int f(x,z)dx \int g(y,z) dy$$

Therefore, we have:

$$h(x,y|Z=z) = \frac{f(x,z)}{\int f(x,z)dx} \frac{g(y,z)}{\int g(y,z)dy}$$

Now, set:

$$f'(x|Z=z) = \frac{f(x,z)}{\int f(x,z)dx}$$

and

$$g'(y|Z=z)=\frac{g(y,z)}{\int g(y,z)dy}$$

Hence, we have the desired result. You could also show that the above conditional $f'(.)$ and $g'(.)$ densities integrate to 1.