Distribution of a Gaussian variable with a normally distributed mean

Question

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.

Answer 1

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 $(1/8)/(1/2) = 1/4$.

Answer 2

Let $X\sim N(0,1)$ and $Y\sim N(X, 1)$. Note that, for any $t\in\mathbb{R}$, \begin{align*} E\left(e^{it(Y-X)}\right) &= E\left(E\left(e^{it(Y-X)} \mid X\right)\right)\\ &=E\left(e^{-it X} E\left(e^{it Y} \mid X\right)\right)\\ &=E\left(e^{-it X} e^{it X -\frac{t^2}{2}} \right)\\ &=e^{-\frac{t^2}{2}}. \end{align*} That is, $Y-X\sim N(0, 1)$. Moreover, for any $t_1, t_2 \in \mathbb{R}$, \begin{align*} E\left(e^{i\big((t_1X + t_2 (Y-X)\big)} \right) &= E\left(E\left(e^{i\big((t_1X + t_2 (Y-X)\big)}\mid X \right)\right)\\ &=E\left(e^{it_1X-it_2X}E\left(e^{it_2Y}\mid X \right)\right)\\ &=E\left(e^{it_1X-it_2X} e^{it_2X -\frac{t_2^2}{2}}\right)\\ &=e^{-\frac{t_1^2+t_2^2}{2}}\\ &=E\left(e^{it_1X}\right)E\left(e^{it_2(Y-X)}\right). \end{align*} That is, $X$ and $Y-X$ are independent. Then, \begin{align*} P(Y<0 \mid X >0) &= \frac{P(Y<0, X>0)}{P(X>0)}\\ &=2E\Big( P\big(Y<0, X>0 \mid X\big) \Big)\\ &=2E\Big(\pmb{1}_{\{X>0\}}\, P\big(Y-X<-X \mid X\big) \Big)\\ &=2E\big(\pmb{1}_{\{X>0\}}\, \Phi(-X)\big)\\ &=2E\big(\pmb{1}_{\{X>0\}} - \pmb{1}_{\{X>0\}}\,\Phi(X)\big)\\ &=2\bigg[\frac{1}{2} - \int_0^{\infty} \Phi(x) \Phi'(x) dx \bigg]\\ &= \frac{1}{4}, \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable.