Exponential distribution

Remarks

The cumulative distribution function is $F(x) = 1 - e^{-\lambda x}$ for $x \ge 0$, so $P(X > x) = e^{-\lambda x}$. The mean is $1/\lambda$, the variance is $1/\lambda^2$ and the median is $\ln 2 / \lambda$. Some authors use the scale $\beta = 1/\lambda$ as the parameter. The exponential distribution is memoryless: $P(X > s + t \mid X > s) = P(X > t)$ for $s, t \ge 0$. If $X_1, \dots, X_n$ are independent with $X_i \sim \operatorname{Exp}(\lambda_i)$, then $\min_i X_i \sim \operatorname{Exp}(\lambda_1 + \dots + \lambda_n)$.