Distribution of a Gaussian variable with a normally distributed mean Let $X\sim N(0,1)$ and $Y\sim(X,1)$, where $Y-X$ is independent of $X$. Then what is the PDF of $Y$? Specifically, I am interested in computing $P(Y<0\vert X>0)$. For those interested in mathematical finance, this is equivalent to pricing a Slalom option (a derivative that pays out 1 if some index is above it's initial value at $T_{1}$ and below it's initial value at $T_{2}$ and 0 otherwise) in a Normal model without drift.

Gordon's answer of $1/4$ is nice enough to make one wonder if there's an easier way to calculate it.

There is. Specifically, let $Z = Y - X$. Then $X$ and $Z$ are independent standard normals and you are looking for $P((X+Z) < 0 | X > 0)$. By the definition of conditional probability, this is

$$ P(X+Z < 0, X > 0) \over P(X > 0) $$

and the denominator is $1/2$. To evaluate the numerator, note that the joint distribution of $X$ and $Z$ is rotationally symmetric around the origin in the $XZ$-plane. The region of the plane with $X + Z < 0$ and $X > 0$ simultaneously is a 45-degree sector, so the probability of falling in it is $1/8$; thus the conditional probability you're looking for is $(1/8)/(1/2) = 1/4$.