Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 4 / a / ii / Solution

The probability density function is normalized because ${\int }_0^{\infty }\alpha e^{-\alpha t}dt=1$. Integration by parts gives $\mathbb{E}[T]={\int }_0^{\infty }t\alpha e^{-\alpha t}dt=1/\alpha$ and $\mathbb{E}[T^2]=2/{\alpha }^2$. ThereforeThese are the mean and variance of the exponential distribution with rate $\alpha$.