On Tuesday, November 13, 2012 6:18:18 PM UTC-6, Rich Ulrich wrote: > On Tue, 13 Nov 2012 13:18:39 -0800 (PST), Paul 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=10, 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. > > > > > > -- > > Rich Ulrich
Thanks -- that might well be a good approximation! But doesn't your example assume that the square root of the mean is the mean of the square root?