I am currently studying the textbook Modeling and Analysis of Stochastic Systems, third edition, by Kulkarni. Chapter 5.1 Exponential Distributions says the following:

The probability density function (pdf) $f_X$ of an $\exp(\lambda)$ random variable is called the exponential density and is given by

$$f_X(x) = \dfrac{d}{dx}F_X(x) = \begin{cases} 0 & \text{if}\, x\leq 0\\ \lambda e^{-\lambda x} & \text{if} \, x \ge 0 \end{cases}$$

The density function is plotted in Figure 5.2. The Laplace Stieltjes transform (LST) of $X \sim \exp(\lambda)$ is given by $$\begin{align}\tilde{F}_X(s) &= E\left( e^{-sX} \right) \\&= \int_0^\infty e^{-sx} f_X(x) \ dx \\&= \dfrac{\lambda}{\lambda + s} \,, \ \text{Re}(s) > - \lambda, \tag{5.2}\end{align}$$ where the $\text{Re}(s)$ denotes the real part of the complex number $s$. enter image description here Taking the derivatives of $\tilde{F}_X(s)$ we can compute the $r$th moments of $X$ for all positive integer values of $r$ as follows: $$E\left( X^r \right) = (-1)^r \dfrac{d^r}{ds^r} \tilde{F}_X(s) \Big\vert_{s = 0} = \dfrac{r!}{\lambda^r}.$$ In particular we have $$E(X) = \dfrac{1}{\lambda}, \ \ \ \ \ \text{Var}(X) = \dfrac{1}{\lambda^2}.$$ Thus the coefficient of variation of $X$, $\text{Var}(X)/E(X)^2$, is $1$.

I am confused about this part:

Thus the coefficient of variation of $X$, $\text{Var}(X)/E(X)^2$, is $1$.

The Wikipedia article on 'coefficient of variation' says the following:

The coefficient of variation (CV) is defined as the ratio of the standard deviation $\sigma$ to the mean $\mu$.

And it then says the following:

The standard deviation of an exponential distribution is equal to its mean, so its coefficient of variation is equal to 1.

But $\text{Var}(X)/E(X)^2$ is not the ratio of the standard deviation to the mean, right? So what's going on here?


2 Answers 2


The coefficient of variation is $\sqrt{\mathrm{Var}(X)/E(X)^2}$, so in this case it's $\sqrt{1}$, which is 1.

  • $\begingroup$ So I guess the author isn't strictly wrong, but I'd say that it's misleading (or not precisely enough phrased). $\endgroup$ Apr 30, 2021 at 7:18

You could do all this calculation easier with the moment generating function. But apart from that, you are right...the coefficient of variation is the ratio between the standard deviation (i.e. the square root of variance) and the absolute value of the mean.

So actually the ratio between the variance and the squared mean is the squared coefficient of variation.


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