User:Elizabethevangenline/sandbox

Statement of the Theorem.

Let (X_n)_n∈N be independent random variables. The random series ∑^∞_n=1X_n 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 Y_n:= X_n1_{{|X|n ≤ A}}, then ∑^∞_n=1𝔼(Y_n), the series of expected values of Y_n , converges

iii. ∑^∞_n=1𝕍ar(Y_n) converges

Proof.

Sufficiency of Conditions ("if")

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

Necessity of Conditions ("only if")

Suppose that ∑^∞_n=1X_n converges almost surely.

Without condition (i), by Borel-Cantelli there would exist some A > 0 such that almost surely {|X_n| ≤ 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(Y_n - 𝔼(Y_n)). So given the convergence of ∑^∞_n=1Y_n, we have ∑^∞_n=1𝔼(Y_n) 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=1Z_n = ∑^∞_n=1(Y_n - Y'_n) where for each n, Y_n and Y'_n are IID-- that is, to employ the assumption that 𝔼(Y_n) = 0, since Z_n is a sequence of random variables bounded by 2, converging almost surely, and with 𝕍ar(Z_n) = 2𝕍ar(Y_n). So we wish to check that if ∑^∞_n=1Z_n converges, ∑^∞_n=1𝕍ar(Z_n) 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(Z_n) = 0, series of the variances converging, summands bounded).