Newsgroup sci.stat.math 12189

Directory

DENRPN@Dames.com (Rock Neveau) in <58s4r1$8ia@news.cerf.net> writes:
> Why would someone want to see the geometric mean
> and compare it to the arithmetic mean (and same for
> standard deviation). . . . Oh, and I am having a devil of
> a time finding the equation used to calculated the
> geometric Standard Deviation.
I have never heard of geometric standard deviation but I assume you would
define it as e to the power of the standard deviation of the natural
logarithms of the data (the geometric mean can be defined as e to the
power of the mean of the natural logarithms of the data).
Geometric means make sense when your data (or data errors) are likely to
be multiplicative rather than additive. In other words when the logarithm
of the data is really a more natural variable. The other common
justification of geometric means is that they emphasize the smaller
observations more than the larger ones compared to an arithmetic mean.
For example in the United States, Bill Gates adds about $30 to the
arithmetic mean family income. To reduce the arithmetic mean family income
by the same $30 we would have to add about 100,000 poor families. So the
arithmetic mean is strongly influenced by Bill Gates and virtually ignores
the poor people.
On the other hand Bill Gates only adds about 0.4 cents to the geometric
mean family income. Adding one family with an income of $3 reduces the
geometric mean by about the same amount. So the geometric mean ignores
Bill Gates and emphasizes poor people compared to the arithmetic mean.
Aaron C. Brown
New York, NY

Theodore Sternberg  in <58srg9$lun@samba.rahul.net>
writes:
> Suppose we know the three marginal probabilities, as
> well as all the two-way joint probabilities p(A,B), p(B,C)
> and p(A,C). But we don't know the three-way joint
> probabilities.  What is the most "natural" guess at the
> three-way probabilities?
First of all you can set limits:
p(A,B) + p(B,C) + p(A,C) - 1 <= p(A,B,C) <= Min[p(A,B),p(B,C),p(A,C)]
One natural guess that meets these limits is:
p(A,B,C) = p1*(p2*p3)^.5
where p1 is the minimum of p(A,B),p(B,C),p(A,C) and p2 and p3 are the
other two two-way probabilities. There's no complex logic here.
Aaron C. Brown
New York, NY