# Calculating the mean from an exponential distribution

could someone please tell me how I can prove/confirm the information on this page? specifically the part about calculating the mean, I have my doubts about the -b.

Many thanks

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Evaluate the integral $\int_b^\infty xf(x)\,dx$ (use integration by parts). –  David Mitra Nov 14 '11 at 18:26

You are right, there is a bad typo. It should be $\dfrac{1}{\lambda}+b$. The easiest way to see it is that the density function is the usual one, shifted to the right by an amount $b$. That shifts the mean of the "ordinary" exponential to the right by $b$.
We could also integrate. There is no real point, since the geometric argument above is enough. But for completeness, the mean is $$\int_b^\infty \lambda x e^{-\lambda (x-b)}dx.$$ To integrate, make the change of variable $u=x-b$. We get $$\int_0^\infty (u+b)\lambda e^{-\lambda u}du.$$ Break this up into the sum $$\int_0^\infty u\lambda e^{-\lambda u}du +\int_0^\infty b\lambda e^{-\lambda u}du.$$ The first integral is the familiar mean $\dfrac{1}{\lambda}$ of the standard exponential. The second is simply $b$.