Relative spread of a sum of four exponentials

Question

Four independent exponential waiting times, each with mean 1, are added to form a total $T$ . The coefficient of variation (CV) is the standard deviation divided by the mean, a unitless measure of relative spread.

What is the coefficient of variation of $T$ ?

Accepted Answer

Each exponential with mean 1 has rate \lambda = 1, so variance 1/\lambda^2 = 1. For the Erlang sum of n = 4 of them, means and variances add: \mathbb{E}[T] = 4 	imes 1 = 4, \qquad \operatorname{Var}(T) = 4 	imes 1 = 4, \qquad 	ext{sd} = \sqrt{4} = 2. The coefficient of variation is 	ext{CV} = \frac{	ext{sd}}{	ext{mean}} = \frac{2}{4} = 0.5. A single exponential always has CV = 1 (its sd equals its mean). Summing n independent copies shrinks the CV to 1/\sqrt{n}; here 1/\sqrt{4} = 0.5. The total is proportionally tighter than any one wait. Takeaway. A sum of n independent exponentials ha…

Relative spread of a sum of four exponentials

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