Let, $X,Y$ are independent rv with pdf.s
$f_X(x\mid \lambda)=\dfrac{1}{\lambda}e^{-x/\lambda}$, $x>0$ and $f_Y(y\mid \mu)=\dfrac{1}{\mu}e^{-x/\mu}$, $y>0$.
Let $Z=\min\{X,Y\}$ and
$W=1$ if $Z=X$ and $0$ if $Z=Y$
Find joint distribution of $(Z,W)$?
I found pmf of $W$ which is $f_W(w)=\Big(\dfrac{\mu}{\lambda+\mu}\Big)^w\Big(\dfrac{\lambda}{\lambda+\mu}\Big)^{1-w}$
But have problem while calculating $P(Z\leq z, W=1)$.
$P(Z\leq z, W=1)=P(Z\leq z\mid W=1)P(W=1)=P(X\leq z)P(W=1)$
The last line does not look right to me. What should I do? Thanks for any help!
First, develop joint distribution of $W$ and $Z$ for both $W$ cases:
$$\begin{align}f(Z=z,W=0)&=f(Z=z,Z=Y)\\&=f(Y=z,Y<X)\\&=f(Y=z,X>z)\\&=f(Y=z)~P(X>z)\\[2ex]f(Z=z,W=1)&=f(Z=z,Z=X)\\&=f(X=z,X<Y)\\&=f(X=z,Y>z)\\&=f(X=z)~P(Y>z)\end{align}$$
Last two terms can be written as follow:
$$\begin{align}f(Z=z,W=0) &= f(Y=z)~P(X>z) \\&= \frac{1}{\mu}e^{-z/\mu} e^{-z/\lambda}\\[2ex]f(Z=z,W=1) &= f(X=z)~P(Y>z) \\&= \frac{1}{\lambda}e^{-z/\lambda} e^{-z/\mu}\end{align}$$
Finally, one can write: $$\begin{align}f(Z=z,W=w)&=e^{\raise{1ex}{-\left(\frac{1}{\lambda} + \frac{1}{\mu} \right)z}}~\left(\frac{1}{\lambda}\right)^w\left(\frac{1}{\mu}\right)^{1-w}\end{align}$$