Is the logarithmic transformation monotonic?

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2 total messages Started by mikef Thu, 12 Nov 1992 04:58

#575

Author: mikef
Date: Thu, 12 Nov 1992 04:58

22 lines
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I have needed to make comparisons between variances of some
random variable X, var(X). Specifically, I needed to know if 
var(X=a)>var(X=b). However, I was only able to compute var(lnX), 
and have made the comparison between var(ln{X=a}) and var(ln{X=b}).

I thought that this comparison would be valid since: 

(1) I believed the natural logarithm was a monotonic tranformation, and

(2) because, if (1) is true I thought that it then follows that
var(X=a)>var(X=b) implies that var(ln{X=a})>var(ln{X=b}).

Am I correct? If not, any good suggestions?

Thanks in advance.  


-- 
__
Michael Fahey         | Internet: mikef@pub.health.su.oz.au
Dept of Public Health | UUCP: uunet!munnari.oz.au!pub.health.su!mikef
University of Sydney  | Bitnet: mikef%pub.health.su.oz.au@cunyvm.bitnet

Re: Is the logarithmic transformation monotonic?

#577

Author: Bill Venables
Date: Thu, 12 Nov 1992 16:36

35 lines
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>>>>> "Michael" == mikef <mikef@dph1.health.su.OZ.AU> writes: 

Michael> I have needed to make comparisons between variances of some random
Michael> variable X, var(X). Specifically, I needed to know if var(X=a) >
Michael> var(X=b).  However, I was only able to compute var(lnX), and have
Michael> made the comparison between var(ln{X=a}) and var(ln{X=b}).

As it stands Michael this does not really make any sense.  However if your
question amounts to the following:

"Is it true that var[log(X)] > var[log(Y)]  implies  var[X] > var[Y]  where
 X and Y are positive random variables?"

then the answer is "No, it doesn't", (even though log(x) is indeed a
monotone increasing function for positive values of x).  

Just consider a simple sample case and put, say ,

		 X = {110, 120, 130} and Y = {10, 11, 12}.

Then var[X] = 100 and var[Y] = 1 so var[X] > var[Y].  

On the other hand var[log(X)] = .0069808 and var[log(Y)] = .008316 so
var[log(X)] < var[log(Y)].

It comes about because on the one hand var[c + X] is the same for all
constants c, whereas var[log(c*X)] is the same for all positive constants
c.  By trading these off one against the other you can make the direction
of the inequalities go either way.

Bill
--
___________________________________________________________________________
Bill Venables, Dept. of Statistics, | Email: venables@stats.adelaide.edu.au
Univ. of Adelaide, South Australia. | Tel: +61 8 228 5412  Fax: ...232 5670

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