On Tue, 13 Nov 2012 13:18:39 -0800 (PST), Paul <firstname.lastname@example.org> wrote:
>Suppose T is a random nonnegative variable with finite mean and variance, and suppose S=Sqrt(T). > >Is there a useful rule of thumb, even approximate, relating the coefficient of variation (CV) of T to the CV of S? For example, if CV(S)=1/10, what does that suggest about CV(T)? > >Reminder: the definition of CV is > CV(T) = Sqrt(V(T)) / E(T) > CV(S) = Sqrt(V(S)) / E(S)
I start out by keeping in mind that the CV is mainly appropriate for describing the error of a distribution that is log-normal. For a log-normal distribution, 1/10 is pretty small. An approximation probably will be a lot worse for large values.
Without getting into theory, I've tried out a few values. It looks like the CV(S) = 1/2 * CV(T) where 1/2 is the derivative of the function. That is a guess, but using a derivative is reasonable.
Try: mean=100, SD=10. CI= (80,120). Transf. by square root: mean=100, CI= (9,11); implies SD= 1/2, CV= 1/20. A smaller original CV works out similarly.
That solves it for CV= 1/10 since every other solution is proportional.
Trying a cube root gave me an answer close to 1/3*CV, so using the derivative is my first guess. I won't be surprised if I've missed something, but this much is all the interest I have in the doing the problem right now.