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Re: CV of square
Posted:
Nov 13, 2012 7:18 PM
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On Tue, 13 Nov 2012 13:18:39 -0800 (PST), Paul
<paulvonhippel@yahoo.com> 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.
--
Rich Ulrich
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