How to compute $P(X>Y\mid Y<1)$ given pdf of $(X,Y)$?

Question

I have the following function

$$f(x,y)=\begin{cases} e^{-(x+y)} &, x,y > 0 \\ 0 &, \text{otherwise} \end{cases}$$

I want to compute the following conditional:

$$P(X>Y\mid Y<1)$$

I'm trying to solve this using the following :

$$f(x\mid y) = \frac{f(x,y)}{f_2(y)}$$

$$f_2(y) = \int_0^{\infty} f(x,y) \,dx = e^{-y}\int_0^{\infty}e^{-x}\,dx = e^{-y}$$

So the conditional probability should be like:

$$\int_y^{\infty}\int_0^1\frac{e^{-(x+y)}}{e^{-y}}dy\,dx$$

I would like to know if this is a good approach.

Answer 1

$X>Y, Y<1$ is equivalent to $Y <X\wedge 1$. Hence the required probability is $$\frac {\int_0^{1} \int_0^{x} e^{-(x+y)}dydx +\int_1^{\infty} \int_0^{1} e^{-(x+y)} } { \int_0^{\infty} \int_0^{1} e^{-(x+y)} dydx}.$$ I will let you carry out the integrations.

Answer 2

First note that $f(x,y)=e^{-x}\cdot e^{-y}$ so the two marginals are $Exp(1)$ independent.

Then start with doing a draw of your $(x,y)$ domain and then observe that your probability is

$$\mathbb{P}[X>Y|Y<1])= \frac{1}{1-e^{-1}}\int_0^1 e^{-y}dy\int_y^{+\infty}e^{-x}dx$$

This because the probability

$\mathbb{P}[X>Y \cap Y<1]$ is obtained integrating the joint density in the purple area below

While $\mathbb{P}[Y<1]=1-e^{-1}$ using CDF of Y.