# Minimum of two random variables with exponential distribution

Let $$X,Y$$ be two random variables with exponential distribution and their rates are $$\gamma, \beta$$. Let $$Z$$ be a random variable such that $$Z = min\{X, Y\}$$.

How do I prove that the density function of $$Z$$ is $$(\gamma+\beta)e^{-(\gamma+\beta)x}$$ for $$x\geq0$$ and $$0$$ for $$x<0$$?

• Hint: you can do this by first finding the CDF, then differentiating it. Try finding an expression for $\mathbb{P}(Z > x)$ (to help do this, ask yourself "when is the minimum of two things greater than $x$?"). – Minus One-Twelfth Feb 20 at 14:50

Following the hint in the comment, note that $$\begin{split} F_Z(z) &= \mathbb{P}[Z \le z] = \mathbb{P}[\min\{X,Y\} \le z] \\ &= \mathbb{P}[X \le z, Y \le z] \quad \text{by independence of X,Y}\\ &= \mathbb{P}[X \le z] \cdot \mathbb{P}[Y \le z] \\ &= F_X(z) \cdot F_Y(z) \end{split}$$ And now you can compute $$f_Z(z) = \frac{d}{dz}\left[F_Z(z)\right].$$
• I get $f_Z(z)=f_X(z)F_Y(z)+F_X(z)f_Y(z)$. I simplify it, knowing that $F_X(x)=1-e^{-x\alpha}$ but it doesn't work out... – Pilpel Feb 20 at 16:29