Jump to content

User:Elizabethevangenline/sandbox

From Wikipedia, the free encyclopedia


Statement of the Theorem.

[edit]

Let (Xn)n∈N be independent random variables. The random series ∑n=1Xn converges almost surely in ℝ if and only if the following conditions hold for some A > 0:

i.n=1(|X|n ≥ A) converges

ii. Let Yn:= Xn1{|X|n ≤ A}, then ∑n=1𝔼(Yn), the series of expected values of Yn , converges

iii.n=1𝕍ar(Yn) converges


Proof.

[edit]
Sufficiency of Conditions ("if")
[edit]

Condition (i) and Borel-Cantelli give that almost surely Xn = Yn for n large, hence ∑n=1Xn converges if and only if ∑n=1Yn converges. Conditions (ii)-(iii) and Kolmogorov's Two-Series Theorem given the almost sure convergence of ∑n=1𝔼(Yn).

Necessity of Conditions ("only if")
[edit]

Suppose that ∑n=1Xn converges almost surely.

Without condition (i), by Borel-Cantelli there would exist some A > 0 such that almost surely {|Xn| ≤ A} for infinitely many values n, but then the series would diverge. Therefore we must have condition (i).

We see that condition (iii) implies condition (ii): Kolmogorov's Two-Series Theorem along with condition (i) applied to the case A =1 gives the convergence of ∑n(Yn - 𝔼(Yn)). So given the convergence of ∑n=1Yn, we have ∑n=1𝔼(Yn) converges, so condition (ii) is implied.

Thus, it only remains to demonstrate the necessity of condition (iii), and we will have obtained the full result. It is equivalent to check condition (iii) for the series ∑n=1Zn = ∑n=1(Yn - Y'n) where for each n, Yn and Y'n are IID-- that is, to employ the assumption that 𝔼(Yn) = 0, since Zn is a sequence of random variables bounded by 2, converging almost surely, and with 𝕍ar(Zn) = 2𝕍ar(Yn). So we wish to check that if ∑n=1Zn converges, ∑n=1𝕍ar(Zn) converges as well. This is a special case of the known, more general result with summands equal to the increments of a martingale sequence and the same conditions (𝔼(Zn) = 0, series of the variances converging, summands bounded).