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 $x<0$?

  • 2
    $\begingroup$ 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 $x$?"). $\endgroup$ – 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 $$ f_Z(z) = \frac{d}{dz}\left[F_Z(z)\right]. $$

  • $\begingroup$ 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... $\endgroup$ – Pilpel Feb 20 at 16:29

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