Silly statistics question

[Daniel McLaury]

I've been thinking this over in the back of my mind for some time now
and finally I thought I'd just ask someone because I don't really have
the background to figure it out for myself.

Let X_1, X_2, ... X_N be i.i.d. random variables, and let Y_N = [ X_1
+ X_2 + ... + X_N ] / N. Put f(N) = Median(Y_N).

If Var(X_i) is finite, then as N approaches infinity f(N) approaches E
(X_i). I can prove that much. I do not even know what happens when E
(X_i) exists and Var(X_i) does not.

What interests me is that, if this holds even for infinite variance,
then this is potentially a generalization of the expectation, since it
could presumably be defined in cases where the expectation is not. Of
course, more likely it's either exactly equivalent to expectation or
it's not even defined even for some distributions with well-defined
expectations. Anyway, I just wondered if anyone know how to work with
problems like this.

[Robert Israel]

The weak and strong laws of large numbers are valid for iid random variables
with an expected value. Variance is not needed. The Weak Law says for any
epsilon > 0, P{|Y_N - E[X_i]| < epsilon) -> 1 as N -> infty. As soon as
that probability > 1/2, |f(N) - E[X_i]| < epsilon. So f(N) -> E[X_i]
as N -> infty.

There are cases where Median(Y_N) does have a limit but the expected value
doesn't exist, especially when there is symmetry involved. Take for
example a Cauchy distribution.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

[daniel....@gmail.com]

Awesome, thanks for making me feel like an idiot :)

As a followup, is this limit something people actually think about, or
are the cases where it's defined -- and the expectation isn't -- too
limited to be interesting?

On Mar 31, 3:45 pm, Robert Israel