total expectation. The second equation exploits the fact that once we condition on , it must follow that . The third equation relies on the memoryless property to replace with . Sum of two independent exponential random variables The probability distribution function (PDF) of a sum of two independent random variables is the convolution of their individual PDFs. If and are independent exponential random variables with respective rate parameters and then the probability density of is given by The entropy of this distribution is available in closed form: assuming (without loss of generality), then where is the Euler-Mascheroni constant, and is the digamma function. In the case of equal rate parameters, the result is an Erlang distribution with shape 2 and parameter which in turn is a special case of gamma distribution. Related distributions If X ~ Laplace(μ, β−1), then |X − μ| ~ Exp(β). If X ~ Pareto(1, λ), then log(X) ~ Exp(λ). If X ~ SkewLogistic(θ), then . If Xi ~ U(0, 1) then The exponential distribution is a limit of a scaled beta distribution: Exponential distribution is a special case of type 3 Pearson distribution. If X ~ Exp(λ) and X ~ Exp(λ) then: , closure under scaling by a positive factor. 1 + X ~ BenktanderWeibull(λ, 1), which reduces to a truncated exponential distribution. keX ~ Pareto(k, λ). e−X ~ Beta(λ, 1). e ~ PowerLaw(k, λ) , the Rayleigh distribution , the Weibull distribution . , a geometric distribution on 0,1,2,3,... , a geometric distribution on 1,2,3,4,... If also Y ~ Erlang(n, λ) or then If also λ ~ Gamma(k, θ) (shape, scale parametrisation) then the marginal distribution of X is Lomax(k, 1/θ), the gamma mixture λX − λY ~ Laplace(0, 1). min{X1, ..., Xn} ~ Exp(λ1 + ... + λn). If also λ = λ then: Erlang(k, λ) = Gamma(k, λ−1) = Gamma(k, λ) (in (k, θ) and (α, β) parametrization, respectively) with an integer shape parameter k. X − X ~ Laplace(0, λ−1). If also X are independent, then: ~ U(0, 1) has probability density function . This can be used to obtain a confidence interval for . If also λ = 1: , the logistic distribution μ − σ log(X) ~ GEV(μ, σ, 0). Further if then (K-distribution) If also λ = 1/2 then ; i.e., X has a chi-squared distribution with 2 degrees of freedom. Hence: If and ~ Poisson(X) then (geometric