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Topic: CV of square
Replies: 5   Last Post: Nov 14, 2012 2:56 PM

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David Jones

Posts: 61
Registered: 2/9/12
Re: CV of square
Posted: Nov 14, 2012 6:47 AM
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"Paul" wrote in message
news:40b71613-a0cd-440b-98c6-88f4487153ad@googlegroups.com...

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?

---------------------------------------------------

If the CV is small, then the CV is approximately the standard deviation of
the logarithm of the original variable. The results found by Rich correspond
to this (logarithm of the power, etc.). The OP might consider using the
standard deviation of the logarithm as the basic measure of
variability-adjusted-for-scale, in which case there an exact result for the
effects of taking powers.




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