Joint pdf of a continuous and a discrete rv Let us consider a manufacturing system. It involves 2 independent components. If one of these components fails then the entire system fails. Let $Y_j$ be distributed $\exp(Q_j)$ where $j=1, 2$.
If component 1 fails first, then $Y_1$ is observed but $Y_2$ is not ($Y_2$ is censored). If component 2 fails first, then $Y_2$ is observed but Y_1 is not ($Y_1$ is censored). Therefore, if the system fails, we can only observe $u = \min(Y_1, Y_2)$ and the binary random variable $V$, which is $1$ if $Y_1 < Y_2$ and $0$ otherwise.
How can I derive the joint pdf of a continuous variable $u = \min(Y_1, Y_2)$ and a discrete variable $V = 1$ if $Y_1 < Y_2$ and $0$ otherwise?
 A: In simplistic terms, there is no such thing as a joint density of a continuous random variable and a discrete random variable because all the probability mass lies
on two straight lines ($v=0$ and $v=1$) and on these lines, the joint 
density, being the probability mass per unit area, is infinite. On the other
hand, the line density of the mass on the two lines
is a (univariate) exponential density (measured in probability mass per unit
length).  More specifically, the line density on the line $v=0$ is the density
of $U_2$ and the line density on the line $v=1$ is the density of $U_1$.
A: Sheldon, Sheldon. How comes that you have to ask a question about math to people like us?
In survival analysis, your setting is called "competing risk". The joint distribution of the earliest failure time and the type of failure is fully described by the so called "cumulative incidence function" (it even allows for censoring, i.e. no failure until end of time horizon). I am quite sure that you will find relevant information in the literature stated in
Assumptions and pitfalls in competing risks model 
A: What you have here is a mixture model, specifically a mixture of exponentials. If I understand your problem setup correctly, I believe what you're looking for looks something like this:
$$
u \sim f(x) =
\begin{cases}
f_{Y_1}(x), & V=1 \\
f_{Y_2}(x), & V=0
\end{cases}
$$
or alternatively
$$u \sim f(x) = \theta f_{Y_1}(x) + (1-\theta)f_{Y_2}(x)$$
Where $\theta$ is the expected proportion of samples generated by $Y_1$ (or using your formulation, $\theta = E[V]$).
You can confirm this experimentally. Here's a mixture model with arbitrarily selected parameters r1, r2 and theta:
n=1e5
theta=.2
v=rbinom(n,1,theta)
r1=5; r2=1
sample=v*rexp(n,r1) + (1-v)*rexp(n,r2)

f=function(x){theta*dexp(x,r1) + (1-theta)*dexp(x,r2)}

plot(density(sample), xlim=c(0,6))
xv=seq(from=0,to=6, length.out=1e4)
lines(xv,f(xv), col='red')


