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.
> 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.
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
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?
<isr...@math.MyUniversitysInitials.ca> wrote: > > 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.
> 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
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