How to Find the Probability of a Distribution based on another Distribution.

Question

The question is : Suppose we first sample $X \sim N(0, 1)$ and then sample $Y \sim N(X, 1)$. What is the probability that $P(Y \leq 2)$?

I know how to sample the $X$ distribution, but I'm not sure how to get the $Y$ distribution based on $X$.

Answer 1

$X\sim N(0,1)$ and $Y|(X=x)\sim N(x,1)$ as Gabriel Romon pointed out.

We are in the continuous case so we have that $f_{XY}(x,y)=f_{Y|X}(y|x)f_X(x)$.

$f_X(x)=\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$

$f_{Y|X}(y|x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{(y-x)^2}{2}}$

$\implies f_{XY}(x,y)=\frac{1}{2\pi}e^{-\frac{2x^2+y^2-2xy}{2}}$

and $f_Y(y)=\int f_{XY}(x,y)dx$.

Finally, $P(Y\leq 2)=\int_{-\infty}^2f_Y(y)dy$. You'll have to do the last integrals yourself.