# Jointly distributed exponential random variables

Attempting to solve the following problem I am confused about what to use as the probability density function

## Problem

The time that it takes to service a car is an exponential
random variable with rate 1.

If A.J. brings his car in at time 0 and M.J.
brings her car in at time t, what is the probability
that M. J.'s car is ready before A. J.'s car?
(Assume that service times are independent and service
begins upon arrival of the car.)


## Exponential Random Variable

A continuous random variable whose probability density function is given, for some $\lambda > 0$, by

$f(x) = \left\{ \begin{array}{l l} \lambda e^{-\lambda x} &\text{if } x \ge 0\\ 0 &\text{if } x < 0 \end{array} \right.$

is said to be an exponential random variable (or, more simply, is said to be exponentially distributed) with parameter $\lambda$.

## How I'm approaching the problem

My thought is to let $\lambda = 1$, let $f( x, y) = \lambda^2 e^{-\lambda x} e^{-\lambda y}$, let $X$ denote A. J.'s car, let $Y$ denote M. J.'s car and solve for $P\{ X < Y\} = \int\limits_0^\infty \int\limits_t^y f( x, y) dx dy$ but immediately I am troubled with the though that this is not the proper joint distribution function?

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You need to shift $Y$ by 1. Try writing down the definition of your random variables explicitly (instead of saying they represent the car, they should represent some sort of time interval). – Memming Jul 16 '12 at 0:22
It is said that service times, which correspond to, $x$ and $y$ are independent. In case of independency, joint probability is the multiplication of two marjinals, as you did. I don't see any problem to get the joint density as you did. What was your concern? – Seyhmus Güngören Jul 16 '12 at 0:28

I would approach the problem as follows. Assume that $t \ge 0$. First find the probability that the service time for A.J.'s car is $\le t$. Of course if this happens, then A.J. will beat M.J. If the first car is not serviced by time $t$, then by the memorylessness of the exponential, the probability that A.J. finishes before M.J. is $1/2$. Now you can put the pieces together.

Or else more simply do the same calculation from M.J.'s point of view. For M.J. to "win," we need first of all to have A.J.'s service time be $\ge t$. If that happens, the probability that M.J. wins is $1/2$.

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Just so I know I am on the right page, the probability A.J's car gets serviced when $time \le t$ is $\int\limits_0^t \lambda e^{-\lambda x} dx = 1 - e^{-t}$ (letting $\lambda = 1$) and combining it with the probability when $time > t$ (where it is $\frac{ 1}{2}$ we find that the probability M.J's car is serviced first is $e^{-t} - \frac{ 1}{ 2}$. – rudolph9 Jul 16 '12 at 3:34
Not quite. I wish they had better names. The probability M.J. wins is $e^{-t}(1/2)$. – André Nicolas Jul 16 '12 at 4:07
So your multiplying the two probabilities right? i.e. the probability M.J. wins would be $(1/2)(1 - e^{-t})$? – rudolph9 Jul 16 '12 at 7:57
Probability M.J. wins = (Prob M.J. wins given A.J. waits at least $t$) times (Prob A.J. waits at least $t$). This is $(1/2)(e^{-t})$. – André Nicolas Jul 16 '12 at 12:59
Ohh yeah, since $P\{\text{A.J. waits at least } t\} = 1 - (1 - e^{-t})$, hence $P\{\text{A.J. waits at least } t \cap \text{A.J. Wins after time} t\} = e^{-t} \frac{ 1}{ 2}$. Thank you! – rudolph9 Jul 16 '12 at 13:09

The exponential distribution is memoryless. So by the time MJ arrives, AJs service starts out as a fresh new exponential. So $P=0.5$ since both have the same distribution

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