Mean of the exponential distribution

Question

Prove that the mean of the exponential distribution is $1/\lambda$ by computing $E[X] = \int_0^\infty x\,\lambda e^{-\lambda x}\,dx$ by parts.

Solution

We integrate by parts with $u = x$ (so $du = dx$) and $dv = \lambda e^{-\lambda x}\,dx$ (so $v = -e^{-\lambda x}$): $E[X] = \int_0^\infty x\,\lambda e^{-\lambda x}\,dx = \Big[-x\,e^{-\lambda x}\Big]_0^\infty + \int_0^\infty e^{-\lambda x}\,dx.$ The boundary term vanishes: at $x=0$ it equals $0$, and as $x\to\infty$ the exponential dominates the linear growth, so $x\,e^{-\lambda x}\to 0$. There remains $\int_0^\infty e^{-\lambda x}\,dx = \Big[-\frac{1}{\lambda}e^{-\lambda x}\Big]_0^\infty = 0 - \left(-\frac{1}{\lambda}\right) = \boxed{\frac{1}{\lambda}}. \qquad \blacksquare$

Links

Topics: Probability distributions
Concepts: Exponential distribution · Expected value
Skills: Proving · Integrating
Exercise type: Proof