Exponential distribution

If events arrive according to a Poisson process with rate \(\lambda\), the time until the first one is exponential with parameter \(\lambda\). The density \(\lambda e^{-\lambda x}\) decreases from its maximum at 0, and the distribution function is \(F(x)=1-e^{-\lambda x}\) for \(x\ge0\); hence \(P(X>t)=e^{-\lambda t}\). We write \(X\sim\operatorname{Exp}(\lambda)\).

The expectation and variance are \(E[X]=1/\lambda\) and \(\operatorname{Var}(X)=1/\lambda^2\), so the standard deviation equals the mean. Units matter: if \(\lambda\) is failures per hour, \(1/\lambda\) is hours per failure. The median, \(\ln 2/\lambda\), lies below the mean because the right tail is long.

Its distinctive property is memorylessness: \(P(X>s+t\mid X>s)=P(X>t)\). If a part has already worked for \(s\) hours, its remaining lifetime has the same distribution as that of a new part. Among continuous distributions on \([0,\infty)\), only the exponential has this property.

A typical mistake is to use \(\lambda\) instead of \(1/\lambda\) as the mean, or to apply the model to components that wear out, where the failure risk grows with time and memorylessness is unrealistic. The exponential is the continuous counterpart of the geometric and is linked to the Poisson, which counts the events of the same process.