# Pdf of the difference of two exponentially distributed random variables

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 $v-u$?

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 $$f(x) = \frac{\lambda \mu}{\lambda+\mu} \cases{e^{-\mu x} & if x > 0\cr e^{\lambda x} & if x < 0\cr}$$

• why $P(X>Y)=\frac{\lambda}{\mu+\lambda}$? – user65985 Aug 14 '13 at 16:55
• $X$ ~ $Exp(\mu)$ and $Y$ ~ $Exp(\lambda)$ then: $P(Y < X) = \int_0^\infty\int_0^x\mu e^{-\mu y}\lambda e^{-\lambda x}dydx$ – YellowPillow Nov 5 '15 at 5:50
• @YellowPillow Doesn't that give ${\mu \over \mu + \lambda}$? – Columbo Apr 13 '16 at 14:59
• This can also be found by the convolution between two pdfs – Hunle Apr 28 '16 at 4:42
• Isnt this (kind of) symmetrization? – badatmath Dec 20 '19 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.

• Before the first double integral, should that be "below or on the line $y = x + z$"? – Patrick Mar 1 '12 at 0:10
• @Patrick: Thank you for spotting the typo! Fixed. – André Nicolas Mar 1 '12 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:

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

• @StubbornAtom Ah yes, you are correct. Updated my notation. Thanks. – Alex P. Miller May 5 at 20:08