Can the coefficient of variation for two samples be added? I have a quick question that should hopefully require a quick answer (my apologies if this is a rather simple question).  I have two estimates of relative abundance from two sub-populations and want to sum them to obtain one overall number.  Being that they're just estimates, I also have their corresponding measures of uncertainty (i.e. coefficients of variation, CVs).  I know that if the output is a linear combination of the inputs with coefficients equal to 1, I can come up with an estimate of the variance for the output by adding together the variances of the inputs.  For example:
\begin{align} y &= x_{1} + x_{2} \\ var(y) &= var(x_{1}) + var(x_{2}) \end{align}
However, I'm obtaining the estimates from published reports and they express uncertainty in CVs and not variances.  Can I add the CVs together like variances?
Thanks in advance for your help!!
 A: 
I would still like to have some measure of uncertainty for the pooled values. Any ideas of how to pool CVs? Or is this just not possible, at least conceptually?

You almost did it in your question; you have rules for expectations of sums (covered in your question) and for variances of sums (the one you used is correct if you have independence, but if you don't there's still a rule for it).
What's the definition of CV?
You have already specified how to get the variance of the sum and the mean of the sum.
You need the standard deviation and the mean. How will you get that from what you have?
A: Well, you can add them, but what would you get? :-).
Let's do a toy example:
x1 <- c(10,20,30)
x2 <- c(5, 15, 20)
x <- c(x1, x2)
sd(x1)/mean(x1)
sd(x2)/mean(x2)
sd(x)/mean(x)

If you add .5 to .57 you get something which seems pretty meaningless; I think you probably want to take the CV of the whole sample.
But what are you trying to get?
Edit to respond to comment: I don't think so.  The cv of the combined vectors can be greater (or less) than either, e.g. 
x1 <- c(10,20,30,40, 40)
x2 <- c(5, 15, 20)
x <- c(x1, x2)
(cv1 <- sd(x1)/mean(x1))
(cv2 <- sd(x2)/mean(x2))
sd(x)/mean(x)

