----- Original Message -----
From: schlawalter <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Sunday, December 12, 1999 9:48 AM
Subject: First Moment of Quotient of 2 Random Variables
> Given three continuos random variables X, Y, Z with well-defined means,
> variances, and covariances. Is it possible to represent E(X/Y) and
> E(Z^2/XY) as a function of the means, variances, and covariances of X,
> Y, and Z?
> Thx in advance,
> Christian
No; unless f_X,Y(a,0) = 0 for all a (this is necessary but not
sufficient),
E(X/Y) is undefined. Otherwise there would be a rectangular region
[a-b,a+b] x (0,y] on which f_X,Y(x,y) was greater than some positive K, and
the integral of
f_X,Y(x,y) x/y would diverge on that rectangle, as y approached 0.
Moreover, it is easy to see that such a counterexample can occur with
more or less any combination of means, variances, covariances, etc; this
suggests that even using such formalisms as the Cauchy principal value to
handle the integral we cannot hope to get any such formula.
The special case of the quotient of two normal distributions centered
at
0 being the Cauchy distribution [f_X(x) = 1/(pi(1+x^2))] is well known. The
important thing to realize is that this is not a special feature of the
normal distribution, or anything to do with tail weights, or avoidable by
having E(Y) <> 0. It is simply that if the denominator distribution can be
0 then you are often coming close to dividing by 0, thus getting big
numbers.
It is true that if the probability of Y being close to 0 is very small
the quotient may usually act as if it had a mean. That is, sample means
will
almost always be well behaved. However, this is like the St. Petersburg
paradox, a delusion based on the fact that a finite number too big for most
practical purposes is not really infinite.
An example like the St. Petersburg paradox but less confusing: Suppose
I sell $1 lottery tickets each of which has 1 chance in 1,000 of winning a
$2,000 prize. I will soon be bankrupt, and you can use me as an Awful
Warning of the dangers of not knowing probability theory.
However, if I sell $1 lottery tickets each of which has 1 chance in a
trillion of winning a 2 trillion dollar prize, limiting my sales only to
what I need to finance a multimillionnaire's lifestyle, it is almost
certain
that I will die wealthy - despite the fact that the expected value of the
two types of ticket is the same.
-Robert Dawson
