TITLE: simplifying the coefficient of variation
QUESTION [1 upvotes]: The coefficient of variation $c$ is defined as the ratio of the standard deviation to the mean, and so,
$c^2=\frac{(\mathbb{E}[(x-\mu)^2])}{(\mathbb{E}[\mu])^2}$
I am given the result that
$c^2=\frac{\mathbb{E}[x^2]}{(\mathbb{E}[\mu])^2}-1$
But what has happened to the term $\mathbb{E}[-2x\mu]$ which would (?) come from $(x-\mu)^2$?

REPLY [1 votes]: Since $\mu = \mathbb{E}[x]$ is a constant and using the linearity of expectation, you get
$$\begin{equation}\begin{aligned}
\mathbb{E}[(x-\mu)^2] & = \mathbb{E}[x^2 - 2\mu x + \mu^2] \\
& = \mathbb{E}[x^2] - \mathbb{E}[2\mu x] + \mathbb{E}[\mu^2] \\
& = \mathbb{E}[x^2] - 2\mu\mathbb{E}[x] + \mu^2 \\
& = \mathbb{E}[x^2] - 2\mu(\mu) + \mu^2 \\
& = \mathbb{E}[x^2] - \mu^2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus, using that $\mathbb{E}[\mu] = \mu$, the original equation becomes
$$\begin{equation}\begin{aligned}
c^2 & = \frac{\mathbb{E}[(x-\mu)^2]}{(\mathbb{E}[\mu])^2} \\
& = \frac{\mathbb{E}[x^2] - \mu^2}{(\mathbb{E}[\mu])^2} \\
& = \frac{\mathbb{E}[x^2]}{(\mathbb{E}[\mu])^2} - \frac{\mu^2}{(\mathbb{E}[\mu])^2} \\
& = \frac{\mathbb{E}[x^2]}{(\mathbb{E}[\mu])^2} - 1
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
{"set_name": "stack_exchange", "score": 1, "question_id": 3414475}