🚀 go-pugleaf

>>>>> "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