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

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paulvonhippel at yahoo

Posts: 72
Registered: 7/13/05
Re: CV of square
Posted: Nov 14, 2012 12:56 AM
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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).
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> >
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> >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)?
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> >
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> >Reminder: the definition of CV is
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> > CV(T) = Sqrt(V(T)) / E(T)
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> > CV(S) = Sqrt(V(S)) / E(S)
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>
>
>
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> I start out by keeping in mind that the CV is mainly appropriate for
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> describing the error of a distribution that is log-normal. For a
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> log-normal distribution, 1/10 is pretty small. An approximation
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> probably will be a lot worse for large values.
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>
>
> Without getting into theory, I've tried out a few values.
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> It looks like the CV(S) = 1/2 * CV(T) where 1/2 is the derivative
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> of the function. That is a guess, but using a derivative is
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> reasonable.
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>
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> Try: mean=100, SD=10. CI= (80,120).
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> Transf. by square root: mean=10, CI= (9,11);
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> implies SD= 1/2, CV= 1/20.
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> A smaller original CV works out similarly.
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>
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> That solves it for CV= 1/10 since every other solution is
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> proportional.
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> Trying a cube root gave me an answer close to 1/3*CV,
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> so using the derivative is my first guess. I won't be
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> surprised if I've missed something, but this much is all
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> the interest I have in the doing the problem right now.
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>
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> --
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> 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?



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