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

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Richard Ulrich

Posts: 2,862
Registered: 12/13/04
Re: CV of square
Posted: Nov 14, 2012 12:33 PM
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On Wed, 14 Nov 2012 11:47:47 -0000, "David Jones"
<dajhawk@hotmail.co.uk> wrote:

>"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.


Thanks.

I decided a few years ago -- It is perhaps a complication that the
CV is not conventionally defined as the SD of the log-transformed
variable. If we did that, then the ratio of SD to mean becomes the
easy-to-understand approximation of the CV (and, one that you
should be doubtful about when there are any negative numbers
since, you could not take their logs).

This would help to inhibit people from using the CV in cases
where it is not appropriate or particularly useful. That is to say,
I don't think of cases where it is useful where the original does
not deserve the log transform. On the other hand, if it were
defined as the SD of the logs, "coefficient of variation" does not
seem most obvious name for it.

--
Rich Ulrich





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