# Distribution of sum of independent exponentials with random number of summands

Let $$\tau_i\sim\exp\left(\lambda\right)$$ be independent and identically distributed exponentials with parameter $$\lambda$$. Then, for given $$n$$, the sum of these values $$T_n := \sum_{i=0}^n \tau_i$$ follows an Erlang-Distribution with probability density function $$\pi(T_n=T| n,\lambda)={\lambda^n T^{n-1} e^{-\lambda T} \over (n-1)!}\quad\mbox{for }T, \lambda \geq 0.$$

I am interested in the distribution of $$T_\tilde n$$ where $$\tilde n$$ is a random variable such that for $$\tau_a \sim \exp(\lambda_a)$$ exponentially distributed, it holds that $$T_\tilde n \leq \tau_a \\T_{\tilde{n}+1} > \tau_a.$$

In other words, $$T_{\tilde n}$$ is truncated by an exponential distribution. I fail in deriving the distribution of $$\tilde n$$ but maybe there is an easier way: $$\pi\left(\tilde n = k\right) = \pi\left(T_n < \tau_a|n=k\right)\\ =1-\int\limits_{\mathbb{R}^+}\sum\limits_{n=0}^{k-1}\frac{1}{n!}\exp\left(-(\lambda+\lambda_a)\tau_a\right)(\tau\lambda_a)^n\lambda_ad\tau_a.$$

However, just sampling and eye-balling looks to me like this density isn't that ugly:

iter <- 20000

lambda_a <- 1
lambda <- 2

df <- data.frame(tau=rep(NA, iter), a=rep(NA, iter))

for(i in 1:iter){
set.seed(i)
a <- rexp(1, rate = lambda_a)
s <- cumsum(rexp(500, rate = lambda))

df[i,] <- c(max(s[1], s[s<a]), a)
}

library(tidyverse)

ggplot(df %>% gather(), aes(x = value, fill = key)) +
geom_density(alpha = .3) + theme_bw()


• I would recommend not using the same notation for both $\tau_i$ and the sum $\tau_n$. Feb 22, 2019 at 11:08
• A more standard name for the Erlang is the Gamma distribution. Feb 22, 2019 at 16:47

## 1 Answer

As detailed in this X validated answer, waiting for a sum of iid exponential $$\mathcal E(\lambda)$$ variates to exceed one produces an Poisson $$\mathcal P(\lambda)$$ variate $$N$$. Hence waiting for a sum of iid exponential $$\mathcal E(\lambda)$$ variates to exceed $$\tau_a$$ produces an Poisson $$\mathcal P(\tau_a\lambda)$$ variate $$N$$, conditional on $$\tau_a$$ (since dividing the sum by $$\tau_a$$ amounts to multiply the exponential parameter by $$\tau_a$$. Therefore \begin{align*} \mathbb P(N=n)&=\int_0^\infty \mathbb P(N=n|\tau_a) \,\lambda_a e^{-\lambda_a\tau_a}\,\text{d}\tau_a\\ &= \int_0^\infty \dfrac{(\lambda\tau_a)^n}{n!}\,e^{-\tau_a\lambda} \,\lambda_a e^{-\lambda_a\tau_a}\,\text{d}\tau_a\\ &=\dfrac{\lambda_a\lambda^n}{n!}\,\int_0^\infty \tau_a^n\,e^{-\tau_a(\lambda+\lambda_a)} \,\text{d}\tau_a\\ &=\dfrac{\lambda_a\lambda^n}{n!}\,\dfrac{\Gamma(n+1)}{(\lambda_a+\lambda)^{n+1}}=\dfrac{\lambda_a\lambda^n}{(\lambda_a+\lambda)^{n+1}} \end{align*} which is a Geometric $$\mathcal G(\lambda_a/\{\lambda_a+\lambda\})$$ random variable. (Here the Geometric variate is a number of failures, meaning its support starts at zero.)

Considering now $$N$$ as a Geometric number of trials, $$N\ge 1$$, the distribution of$$\zeta=\sum_{i=1}^N \tau_i$$the moment generating function of $$\zeta$$ is $$\mathbb E[e^{z\zeta}]=\mathbb E[e^{z\{\tau_1+\cdots+\tau_N\}}]=\mathbb E^N[\mathbb E^{\tau_1}[e^{z\tau_1}]^N]=E^N[\{\lambda/(\lambda-z)\}^N]=E^N[e^{N(\ln \lambda-\ln (\lambda-z))}]$$and the mgf of a Geometric $$\mathcal G(p)$$ variate is $$\varphi_N(z)=\dfrac{pe^z}{1-(1-p)e^z}$$Hence the moment generating function of $$\zeta$$ is $$\dfrac{pe^{\ln \lambda-\ln (\lambda-z)}}{1-(1-(\lambda_a/\{\lambda_a+\lambda\}))e^{\ln \lambda-\ln (\lambda-z)}}=\dfrac{p \lambda}{ \lambda-z-\lambda^2/\{\lambda_a+\lambda\}}$$ where $$p=\lambda_a/\{\lambda_a+\lambda\}$$, which leads to the mfg $$\dfrac{\lambda\lambda_a/\{\lambda_a+\lambda\}}{ \lambda-z-\lambda\lambda_a/\{\lambda_a+\lambda\}^2}=\dfrac{1}{1-z(p\lambda)^{-1}}$$meaning that $$\zeta$$ is an Exponential $$\mathcal{E}(\lambda\lambda_a/\{\lambda_a+\lambda\})$$ variate.

• Thanks a lot @Xi'an! Do I get it right that in your notation $p= \lambda_a/(\lambda_a+\lambda)$ ? Because then the moment generating function in the last line is equivalent to $\frac{1}{1-z\left(p\lambda\right)^{-1}}$ which corresponds to the MGF of an exponential distribution.. Feb 22, 2019 at 12:15