driver_writer wrote:
>
> In my readings, I've come across unbounded discrete probability
> functions, e.g.
>
> f(x) = (1/2)^x, x = 1,2,3...
> = 0 elsewhere
>
> How does one take the Expected value or a r.v. X, for an unbounded (or
> in this case bounded on the left side) p.f?
>
> My guess is you have to see if the series converges or diverges. If it
> converges, you take the sum and then ?
Not just the series, but SUM n f(n) , must converge. (SUM f(n) should
converge to 1.) Then
E(X) = SUM n f(n)
If this diverges, there is no expected value.
An example is f(n) = (pi/2)/n^2; this converges but n f(n) diverges.
There may be a "Cauchy principle value" based on cancelling negative
and positive values. However, this is NOT a mean and cannot be expected
to behave well. In particular, sample means may fail to converge to it.
-Robert Dawson
.
.
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