Suppose we have $v$ and $u$, both are independent and exponentially distributed random variables with parameters $\mu$ and $\lambda$, respectively.

How can we calculate the pdf of $v-u$?


3 Answers 3


I too prefer to call the random variables $X$ and $Y$. You can think of $X$ and $Y$ as waiting times for two independent things (say $A$ and $B$ respectively) to happen. Suppose we wait until the first of these happens. If it is $A$, then (by the lack-of-memory property of the exponential distribution) the further waiting time until $B$ happens still has the same exponential distribution as $Y$; if it is $B$, the further waiting time until $A$ happens still has the same exponential distribution as $X$. That says that the conditional distribution of $X-Y$ given $X > Y$ is the distribution of $X$, and the conditional distribution of $X-Y$ given $X < Y$ is the distribution of $-Y$. Since $P(X>Y) = \frac{\lambda}{\mu+\lambda}$, that says the PDF for $X-Y$ is $$ f(x) = \frac{\lambda \mu}{\lambda+\mu} \cases{e^{-\mu x} & if $x > 0$\cr e^{\lambda x} & if $x < 0$\cr}$$

  • 12
    $\begingroup$ why $P(X>Y)=\frac{\lambda}{\mu+\lambda}$? $\endgroup$
    – user65985
    Aug 14, 2013 at 16:55
  • 5
    $\begingroup$ $X$ ~ $Exp(\mu)$ and $Y$ ~ $Exp(\lambda)$ then: $P(Y < X) = \int_0^\infty\int_0^x\mu e^{-\mu y}\lambda e^{-\lambda x}dydx$ $\endgroup$ Nov 5, 2015 at 5:50
  • 3
    $\begingroup$ @YellowPillow Doesn't that give ${\mu \over \mu + \lambda}$? $\endgroup$
    – Columbo
    Apr 13, 2016 at 14:59
  • $\begingroup$ This can also be found by the convolution between two pdfs $\endgroup$
    – makansij
    Apr 28, 2016 at 4:42
  • $\begingroup$ Isnt this (kind of) symmetrization? $\endgroup$
    – user515599
    Dec 20, 2019 at 10:35

The right answer depends very much on what your mathematical background is. I will assume that you have seen some calculus of several variables, and not much beyond that. Instead of using your $u$ and $v$, I will use $X$ and $Y$.

The density function of $X$ is $\lambda e^{-\lambda x}$ (for $x \ge 0$), and $0$ elsewhere. There is a similar expression for the density function of $Y$. By independence, the joint density function of $X$ and $Y$ is $$\lambda\mu e^{-\lambda x}e^{-\mu y}$$ in the first quadrant, and $0$ elsewhere.

Let $Z=Y-X$. We want to find the density function of $Z$. First we will find the cumulative distribution function $F_Z(z)$ of $Z$, that is, the probability that $Z\le z$.

So we want the probability that $Y-X \le z$. The geometry is a little different when $z$ is positive than when $z$ is negative. I will do $z$ positive, and you can take care of negative $z$.

Consider $z$ fixed and positive, and draw the line $y-x=z$. We want to find the probability that the ordered pair $(X,Y)$ ends up below that line or on it. The only relevant region is in the first quadrant. So let $D$ be the part of the first quadrant that lies below or on the line $y=x+z$. Then $$P(Z \le z)=\iint_D \lambda\mu e^{-\lambda x}e^{-\mu y}\,dx\,dy.$$

We will evaluate this integral, by using an iterated integral. First we will integrate with respect to $y$, and then with respect to $x$. Note that $y$ travels from $0$ to $x+z$, and then $x$ travels from $0$ to infinity. Thus $$P(Z\le x)=\int_0^\infty \lambda e^{-\lambda x}\left(\int_{y=0}^{x+z} \mu e^{-\mu y}\,dy\right)dx.$$

The inner integral turns out to be $1-e^{-\mu(x+z)}$. So now we need to find $$\int_0^\infty \left(\lambda e^{-\lambda x}-\lambda e^{-\mu z} e^{-(\lambda+\mu)x}\right)dx.$$ The first part is easy, it is $1$. The second part is fairly routine. We end up with $$P(Z \le z)=1-\frac{\lambda}{\lambda+\mu}e^{-\mu z}.$$ For the density function $f_Z(z)$ of $Z$, differentiate the cumulative distribution function. We get $$f_Z(z)=\frac{\lambda\mu}{\lambda+\mu} e^{-\mu z} \quad\text{for $z \ge 0$.}$$ Please note that we only dealt with positive $z$. A very similar argument will get you $f_Z(z)$ at negative values of $z$. The main difference is that the final integration is from $x=-z$ on.

  • $\begingroup$ Before the first double integral, should that be "below or on the line $y = x + z$"? $\endgroup$
    – Patrick
    Mar 1, 2012 at 0:10
  • $\begingroup$ @Patrick: Thank you for spotting the typo! Fixed. $\endgroup$ Mar 1, 2012 at 0:23

There is an alternative way to get the result by applying the the Law of Total Probability:

$$ P[W] = \int_Z P[W \mid Z = z]f_Z(z)dz $$

As others have done, let $X \sim \exp(\lambda)$ and $Y \sim \exp(\mu)$. What follows is the only slightly unintuitive step: instead of directly calculating the PDF of $Y-X$, first calculate the CDF: $ P[X-Y \leq t]$ (we can then differentiate at the end).

$$ P[Y - X \leq t] = P[Y \leq t+X] $$

This is where we'll apply total probability to get

$$ = \int_0^\infty P[Y \leq t+X \mid X=x]f_X(x) dx $$ $$ = \int_0^\infty P[Y \leq t+x]f_X(x) dx = \int_0^\infty F_Y(t+x) f_X(x) dx $$ Note substituting the CDF here is only valid if $t \geq 0$, $$ = \int_0^\infty (1- e^{-\mu(t+x)}) \lambda e^{-\lambda x} dx = \lambda \int_0^\infty e^{-\lambda x} dx - \lambda e^{-\mu t} \int_0^\infty e^{-(\lambda+\mu)x} dx $$ $$ = \lambda \left[ \frac{e^{-\lambda x}}{-\lambda} \right]^\infty_0 - \lambda e^{-\mu t} \left[ \frac{e^{-(\lambda+\mu)x}}{-(\lambda+\mu)} \right]^\infty_0 =1 - \frac{\lambda e^{-\mu t}}{\lambda+\mu} $$

Differentiating this last expression gives us the PDF:

$$ f_{Y-X}(t) = \frac{\lambda \mu e^{-\mu t}}{\lambda+\mu} \quad \text{for $t \geq 0$} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.