# Coefficient of variation for exponential distribution: $\text{Var}(X)/E(X)^2$?

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 $$s$$. 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 $$1$$.

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?

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