Exponential Variables - $P(X < x | X < Y)$

Question

Find the following probability: $$P(X < x | X < Y)$$, where X and Y are independent exponential random variables with parameters $a,b$ respectively.

I'm not sure how to approach this problem. Would I need to use Bayes rule and then integrate the pdfs? I'm also unsure as to how the integrating will actually work.

After using Bayes rule, we get: $$\frac{P(X < x, X < Y)}{P(X<Y)}$$ From general results, we know that $P(X<Y) = \frac{a}{a+b}$, so all that's left is to find the numerator. Any hints on how to do that?

Answer 1

One small aside: you didn't use Bayes' Rule to start, you used the definition of conditional probability, which states that $\mathbb P(A \mid B) = \mathbb P (A \cap B) / \mathbb P(B)$.

Some hints:

• Remember that to find any probability involving two continuous random variables, your job is to integrate the joint density function over the region describing the desired event.
• The joint density function of two independent random variables is just the product of the individual density functions.
• The desired region in this case is the subset of the $ST$-plane (using $s, t$ instead of $x, y$ because we need to have $x$ in the answer): $\{(s, t): s > 0, t > 0, s < x, s < t\}$

I'd suggest drawing that region of the plane and setting up a double integral over it.

Answer 2

I believe this is the approach. $$\begin{align}\mathrm P(X<x,X<Y)&=\mathrm P(X\lt\min(x,Y))\\[1ex]&=\int_0^\infty f_Y(y)\,\mathrm P(X\lt\min(x,y)\mid Y=y)~\mathrm dy\\[1ex]&=\int_x^\infty f_Y(y)\int_0^x f_X(u)\,\mathrm du\,\mathrm dy+\int_0^x f_Y(y)\int_0^yf_X(u)\,\mathrm du\,\mathrm dy\\[1ex]&=\frac{a}{a+b}(1-e^{-(a+b)x})\end{align}$$

Net result - answer for problem =$(1-e^{-(a+b)x}).$