Sum of independent exponential variables

In a homogeneous Poisson process, the time until the \(k\)'th event is

\[ X \;\; = \; \; \sum_{i=1}^k {Y_i} \]

where \(Y_1\) is the time to the first event, \(Y_2\) is the time from the first event to the second, and so on. From the memoryless property of a homogeneous Poisson process, the \(\{Y_i\}\) are independent and they all have \(\ExponDistn(\lambda)\) distributions.

Since \(X\) is the sum of a random sample of size \(k\) from an exponential distribution, we can use general results about the sum of a random sample to find its mean and variance.

Mean and variance of Erlang distribution

If a random variable, \(X\), has an Erlang distribution with probability density function

\[ f(x) \;\;=\;\; \begin{cases} \dfrac{\lambda^k}{(k-1)!} x^{k-1} e^{-\lambda x} & x \gt 0 \\[0.3em] 0 & \text{otherwise} \end{cases} \]

its mean and variance are

\[ E[X] \;=\; \frac k{\lambda}\spaced{and} \Var(X) \;=\; \frac k{\lambda^2} \]

(Proved in full version)

A final useful property of Erlang distributions that adding together two independent Erlang variables (with the same \(\lambda\)) results in a variable that also has an Erlang distribution.

Additive property of Erlang distributions

If \(X_1 \sim \ErlangDistn(k_1,\; \lambda)\) and \(X_2 \sim \ErlangDistn(k_2,\; \lambda)\) are independent, then

\[ X_1 + X_2 \;\;\sim\;\; \ErlangDistn(k_1 + k_2,\; \lambda) \]

(Proved in full version)