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 $1$.
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?
