Conditional density function of a max functiion

Question

Let $X,Y$ be exponential random variables with means $1/\lambda$ and $1/\mu$, respectively. Let $Z=\max\{X,Y\}$.

Find the conditional density function of $Z$ given that $Z=X$.

Answer 1

One approach: compute $$P(Z \le z \mid Z=X) = \frac{P(Z \le z, Z=X)}{P(Z=X)},$$ and take the derivative.

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$$P(Z \le z , Z=X) = P(Y \le X \le z) = \int_0^z \int_y^z f_X(x) f_Y(y) \, dx \, dy = \int_0^z f_Y(y) (e^{-\lambda y} - e^{-\lambda z}) \, dy = - e^{-\lambda z}(1 - e^{-\mu z}) + \frac{\mu}{\mu + \lambda}(1 - e^{-(\lambda + \mu)z}) $$

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Thus $$P(Z \le z \mid Z=X) = \frac{P(Z \le z, Z=X)}{P(X \ge Y)} = - \frac{\mu + \lambda}{\mu} e^{-\lambda z}(1 - e^{-\mu z}) + (1 - e^{-(\lambda + \mu)z})$$

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Taking the derivative with respect to $z$ yields $$\frac{\mu + \lambda}{\mu} (\lambda e^{- \lambda z} - (\lambda + \mu) e^{-(\lambda + \mu) z}) + (\lambda + \mu) e^{-(\lambda + \mu) z} = \frac{(\mu + \lambda)\lambda}{\mu} (e^{-\lambda z} - e^{-(\lambda + \mu) z}).$$