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It has been well known that if $X_1$,...,$X_n$ are independent exponential random variables with common parameter $\mu$, then we have a gamma distribution $\sum_{i = 1}^{n}X_i \sim \Gamma(n,\mu)$ with two parameters $(n, \mu)$.

My question is do we have $\sum_{i = 1}^{n}X_i \sim \Gamma(n,\sum_{j}\mu_j)$ if we have $X_i$'s are independent exponential random variables with parameter $\mu_i$?

As I explored the internet and found this post, which says that the answer to my above question is negative (also confirmed by using moment generating function method).

So I wonder if there is a name of the distribution for such sum of exponential random variables with different parameters?

Liäm
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1 Answers1

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Yes. The sub-topic is called the convolution of exponential random variables. And, the type of distribution that you may be looking for is the hypoexponential distribution.

A hypoexponential distribution is the distribution of the sum of exponential random variables with different means.

Hypoexponential Random Variables

Let $X_i, i=1..n$ be Exponential random variables with rates $\lambda_i, i = 1.. n$ such that $\lambda_i \ne \lambda_j$ for $i \ne j$.

$Y := \sum_{i=1}^{n}X_i$

Here, Y is said to be a hypoexponential random variable.

PDF - Simple Case

Consider two Exponential random variables $X_1$ and $X_2$ with rates $\lambda_1$ and $\lambda_2$ respectively.

Then, the density function:

$f(X_1+X_2 = t) = \frac{\lambda_1}{\lambda_1-\lambda_2}\lambda_2e^{-\lambda_2t} + \frac{\lambda_2}{\lambda_2-\lambda_1}\lambda_1e^{-\lambda_1t} $

Generalization

$f_{X_1+X_2+..+X_n}(t) = \sum_{i=1}^{n}C_{i,n}\lambda_ie^{-\lambda_it}$

where

$C_{i,n} = \prod_{j\ne i}\frac{\lambda_j}{\lambda_j-\lambda_i}$

Note

I've used this extensively while trying to calculate the probability of the sum of several Exponential random variables with different rates being greater than some constant. If you ever need to do that, you could use the following:

Let $X_i, i=1..n$ be Exponential random variables with rates $\lambda_i, i = 1.. n$ such that $\lambda_i \ne \lambda_j$ for $i \ne j$.

$Y := \sum_{i=1}^{n}X_i$

Then,

$P(Y>t) = \sum_{i=1}^{n}C_{i,n}e^{-\lambda_it}$

where

$C_{i,n} = \prod_{j\ne i}\frac{\lambda_j}{\lambda_j-\lambda_i}$

I learned a lot about this, primarily from Introductions to Probability Models by Sheldon Ross which covers this topic. I would recommend going through that book if you want this explained in more detail.

Rahul P
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