Conditional distribution of minimum of two Independent Exponential Random variables We have $X \sim Exp(\alpha)$ and $Y \sim Exp(\beta)$. We also have $Z = \min(X,Y)$. We want to find the probability that $P(Z \leq z \mid Z > \tau)$. 

I started as: 
$P(Z \leq z) = \min(X,Y)$
$= P(X > z ) P(Y > z )$
Then, $f_{Z \mid Z>\tau} = \frac{f_Z(z)}{P(Z > \tau)}$
and $P(Z \leq z \mid Z > \tau) = \frac{d}{dz} f_{Z \mid Z > \tau}(z)$
Is this the correct way to derive the probability? Thank you.
 A: Recall that $Z\sim\mathrm{Exp}(\alpha+\beta)$. So we have (assuming $z>\tau$)
\begin{align}
\mathbb P(Z\leqslant z\mid Z>\tau) &= \frac{\mathbb P(Z\leqslant z, z>\tau)}{\mathbb P (Z>\tau)}\\
&= \frac{\mathbb P(\tau< Z\leqslant z)}{\mathbb P (Z>\tau)}\\
&= \frac{e^{-(\alpha+\beta)\tau}- e^{-(\alpha+\beta)z}}{e^{-(\alpha+\beta)\tau}}\\
&= 1 - e^{-(\alpha+\beta)z},
\end{align}
by definition of conditional probability. Note that $\tau$ does not appear in the expression, due to the memoryless property of the exponential distribution.
A: $$\begin{aligned}
P(Z \le z | Z > \tau) = \dfrac{P(\min(X,Y) \le z, \min(X,Y) > \tau)}{P(\min(X,Y) > \tau)}
\end{aligned}$$
If $\tau \ge z$, the numerator is $0$.  So suppose $\tau < z$.
The denominator is just
$$\begin{aligned}
P(\min(X,Y) > \tau) &= P(X > \tau, Y > \tau)\\
&=(1-F_X(\tau))(1-F_Y(\tau))
\end{aligned}$$
for the numerator we have
$$\begin{aligned}
&P(\min(X,Y) \le z, \min(X,Y) > \tau)\\
 &= P( \tau < X \le z, \tau < Y \le z) + P(\tau < X \le z, Y > z) + P( \tau < Y \le z, X > z)
\end{aligned}$$
And you can use independance and the given distributions to find the answer.
edit: It may be easier to find 1 - probability of the complement of the event in the numerator.
