# Calculating the distribution of the minimum of two exponential functions

Suppose X and Y are two independent exponential random variables with rates $\alpha$ and $\beta$ respectively. I know the following equality to be true but I don't know why it's true: $\mathbb{P}(Y \ge X, X > x) = \int_x^\infty \alpha e^{-\alpha t}\mathbb{P}(Y\ge t)dt$. So if someone could explain this to me step-by-step, it would be much appreciated. I understand that the first part of the integrand comes from the pdf of $X$ and the second part comes from wanting $Y\ge X$ but I don't understand why they are multiplied since $\{Y \ge X\} \text{ and } \{X > x\}$ are dependent.

Let $Z = \min\{X,Y\}$. Then $$P(Z\le z) = 1 - P(Z\ge z) = 1- P(X\ge z)P(Y\ge z) = 1 - e^{-\alpha z}e^{-\beta z} = 1 - e^{-(\alpha+\beta) z}.$$ So $Z\sim\text{Exp}(\alpha+\beta)$.
The inequality follows from the law of total probability: $$P(Y>X;X>x) = \int_{0}^{\infty}P(Y>X;X>x|X=t)f_X(t)dt = \int_{x}^{\infty}P(Y>X|X=t)f_X(t)dt = \int_{x}^{\infty}P(Y>t)f_X(t)dt = \int_{x}^{\infty}\alpha e^{-\alpha t}P(Y>X)dt.$$
• Yes, but now it is given that $X=t$. It is true that $P(X=t)=0$ for all $t$, however, now we calculate $P(Y>X;X>x)$ given that we know that $X=t$. – Marc Apr 5 '14 at 10:57