Let X, Y, Z be r.v.s such that X ⇠ N (0, 1) and conditional on X = x, Y and Z are i.i.d. N (x, 1).
(a) Find the joint PDF of X, Y, Z.
(b) Find the joint PDF of Y and Z. You can leave your answer as an integral, though the integral can be done with some algebra (such as completing the square) and facts about the Normal distribution.
Is my reasoning correct?
I did the following:
a) $f(X=x,Y=y,Z=z)=f(X=x)f(Y=y|X=x)f(Z=z|X=x,Y=y)=f(X=x)f(Y=y|X=x)f(Z=z|X=x)=$
$$\frac{1}{2\sqrt \pi} e^{-x^2/2} \, \frac{1}{2\sqrt \pi} e^{-(y-x)^2/2} \, \frac{1}{2\sqrt \pi} e^{-(z-x)^2/2} \,$$
b) $$f(Y,Z)=\int_{-\infty}^{\infty}( \frac{1}{2\sqrt \pi} e^{-x^2/2} \, \frac{1}{2\sqrt \pi} e^{-(y-x)^2/2} \, \frac{1}{2\sqrt \pi} e^{-(z-x)^2/2} \, dx)$$
For part (a), your solution is almost correct. I am unsure if you made a typographical error, but the PDF of $f(X=x)$ for $X\sim N(0,1)$ is: $$\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$ In other words, all the constants in the denominator should have 2's inside the square root, not outside.
Your answer for part (b) as an integral is correct, but you might be able to simplify some algebra here by combining constants and also recognizing something about the fact that Y and Z are iid.
Motivating question: We have that $Y$ and $Z$ are conditionally independent depending on $X$. However, do we know whether $Y$ and $Z$ are also unconditionally independent?