# Coefficient of variation of the product of a random variable and a constant

I have a random variable $X$ and I need to calculate the coefficient of variation ($\text{CV}$) of the product of a constant, $c$, and $X$. In other words $\text{CV}(X \cdot c)$. To do this I can apply the following:

\begin{align} \text{CV}(X \cdot c) &= \frac{\text{sd}(X \cdot c)}{E(X \cdot c)}\\[1em] \text{CV}^{2}(X \cdot c) &= \frac{\text{Var}(X \cdot c)}{E^2(X \cdot c)}\\[1em] \text{CV}^{2}(X \cdot c) &= \frac{c^2\text{Var}(X)}{c^2E^2(X)}\\[1em] \text{CV}^{2}(X \cdot c) &= \frac{\text{Var}(X)}{E^2(X)}\\[1em] \text{CV}(X \cdot c) &= \frac{\text{sd}(X)}{E(X)}\\ \end{align}

So basically multiplying a random variable by a constant doesn't change the CV. Is my math correct?

• The CV is constructed to be independent of the units of measurement: when the units change, both the SD and the expectation change in the same proportion, whence their ratio is constant. Multiplication of $X$ by a (positive) constant is just a change of units, QED. Your math is technically incorrect because it yields the wrong answer for negative constants (although that's not really relevant for the CV, which is intended to apply only to random variables with non-negative support).
– whuber
Aug 15, 2013 at 14:27
• @whuber What would the math look like if $c$ were a negative constant? Because $c$ is squared ($c^2$), does it really matter if c is positive or negative? Aug 15, 2013 at 14:33
• The principal error occurs at the last step: the square root of $\mathbb{E}^2(X)$ is $\pm\mathbb{E}(X)$. You need to choose the negative sign when $c$ is negative. A minor error occurs in the penultimate step, whose validity requires $c$ to be nonzero. A simpler derivation exploits the observations that $\text{SD}(cX)=\vert c\vert \text{SD}(X)$ and $\mathbb{E}(cX)=c\mathbb{E}(X)$ whence, for $c\ne 0$, $\text{CV}(cX)=\vert c\vert/c$ (and for $c=0$ the CV is undefined). These mathematical statements express exactly what @Nick Cox says in English in the next comment.
– whuber
Aug 15, 2013 at 14:39
• I don't think you need math here as much as statistical thinking, not that the two are contradictory. With a negative constant the mean is negated but the standard deviation will always remain zero or positive. So the CV is negated.... As @whuber says the CV is for non-negative variables. (Although I have seen published examples of negative CV for Celsius temperatures with negative means, compounded with the comment that using Fahrenheit temperatures would be better! In fact, CVs require ratio scale measurements as well as values $\ge 0$.) Aug 15, 2013 at 14:40
• Thanks @whuber and Nick Cox!! These forums are great, it's always nice to get feedback from experts in the field. Aug 15, 2013 at 15:03

The principal error occurs at the last step: the square root of $\mathbb{E}^2(X)$ is $\pm\mathbb{E}(X)$. You need to choose the negative sign when $c$ is negative. A minor error occurs in the penultimate step, whose validity requires $D(c$ to be nonzero. A simpler derivation exploits the observations that $\text{SD}(cX)=\vert c\vert \text{SD}(X)$ and $\mathbb{E}(cX)=c\mathbb{E}(X)$ whence, for $c\not =0$, $\text{CV}(cX)=\vert c\vert/c$ (and for $c=0$ the CV is undefined). These mathematical statements express exactly what @Nick Cox says in English in the next comment. – whuber
I don't think you need math here as much as statistical thinking, not that the two are contradictory. With a negative constant the mean is negated but the standard deviation will always remain zero or positive. So the CV is negated.... As @whuber says the CV is for non-negative variables. (Although I have seen published examples of negative CV for Celsius temperatures with negative means, compounded with the comment that using Fahrenheit temperatures would be better! In fact, CVs require ratio scale measurements as well as values $\ge 0$.) – Nick Cox