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 $v-u$?
 A: 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
$$ f(x) = \frac{\lambda \mu}{\lambda+\mu}
\cases{e^{-\mu x} & if $x > 0$\cr
       e^{\lambda x} & if $x < 0$\cr}$$
A: 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:
$$
f_{Y-X}(t) = \frac{\lambda \mu e^{-\mu t}}{\lambda+\mu} \quad \text{for $t \geq 0$}
$$
A: 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. 
