3.5. Hypoexponential Random Variables Definition 1. A hypoexponential random variable is a sum of independent exponential random variables. If 1, …, n are positive numbers then T1,…,n denotes the hypoexponential random variable T1,…,n = T1 + + Tn where T1, …, Tn are independent and each Tj is an exponential random variable with decay rate j. Proposition 1. Let m = (m1,...,mn) be a vector of positive integers and = (1,...,n) be a vector of positive numbers and let X1, …, Xn be independent gamma random variables with Xj ~ (mj,1/j) and Y = X1 + + Xn be their sum. Then Y is a hypoexponential random variable. Proof. This follows from the fact that each Xj is the sum of mj independent exponential random variables each having decay rate j. Since an exponential random variable is a gamma random variable, a hypoexponential random variable is a sum of independent gamma random variables. Thus the density function of a hypoexponential random variable is given by (4) in section 3.1 and also by Moschopolos' formula (11) in section 3.3. However, it turns out that the density function of a hypoexponential random variable can be expressed in terms of elementary functions which is the purpose of this section. First we look at the simpler case where all the j in Definition 1 are distinct. After this we look at the more complicated case where some of the j are equal. The discussion in this case is based on the formulas of Mathai [7] and Cetnar [9]. 3.5 - 1 Proposition 1. Let 1, …, n be distinct positive numbers and T1,…,n = T1 + + Tn where T1, …, Tn are independent and each Tj is an exponential random variable with decay rate j. Let An(t) = An(t; 1,…,n) be the density function of T1,…,n. Then (1) An(t; 1,2) = 1- 2 e-1t + 1- 2 e-2t 2 1 1 2 (2) An(t;1,...,n) = c1e-1t + + cne-nt with (3) ck = ck(1,...,n) = 1n (1-k)(k-1-k)(k+1-k)(n-k) Proof. An(t; 1,2) = 1e-1t * 2e-2t. Doing the integration gives (1). Then (2) follows from (1) and induction and the fact that convolution is associative and commutative. By Theorem 4 in section 3 the moment generating function of Y is (4) 1m12m2 nmn M(s) = (1 - s)m1(2 - s)m2 (n - s)mn Using partial fractions one can write iik+1i,mi-1-k M(s) = (i - s)k+1 i=1 k=0 n (5) mi-1 where (6) iik+1i,mi-1-k = s (7) imj i = (j - i)mj j=1 n ji So 3.5 - 2 n (8) mi-1 A(t;m,) = i=1 k=0 iik+1i,mi-1-ktke-jt k! 3.5 - 3