Vtyjkyuk

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 $P(X>Y) = fraclambdamu+lambda$, that says the PDF for $X-Y$ is
$$ f(x) = fraclambda mulambda+mu
casese^-mu x & if $x > 0$cr
e^lambda x & if $x < 0$cr$$

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
$$lambdamu e^-lambda xe^-mu y$$
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 $Zle z$.

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
$$P(Z le z)=iint_D lambdamu e^-lambda xe^-mu y,dx,dy.$$

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
$$P(Zle x)=int_0^infty lambda e^-lambda xleft(int_y=0^x+z mu e^-mu y,dyright)dx.$$

The inner integral turns out to be $1-e^-mu(x+z)$. So now we need to find
$$int_0^infty left(lambda e^-lambda x-lambda e^-mu z e^-(lambda+mu)xright)dx.$$
The first part is easy, it is $1$. The second part is fairly routine. We end up with
$$P(Z le z)=1-fraclambdalambda+mue^-mu z.$$
For the density function $f_Z(z)$ of $Z$, differentiate the cumulative distribution function. We get
$$f_Z(z)=fraclambdamulambda+mu e^-mu z quadtextfor $z ge 0$.$$
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.

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

$$
P[W] = int_Z P[W mid Z = z]P[Z=z]dz
$$

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

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

Differentiating this last expression will gives us the PDF:

$$
f_Y-X(t) = fraclambda mu e^-mu tlambda+mu quad textfor $t geq 0$
$$