Bennett acceptance ratio

In probability theory, Kolmogorov's Three-Series Theorem, named after Andrey Kolmogorov, gives a criterion for the almost sure convergence of an infinite series of random variables in terms of the convergence of three different series involving properties of their probability distributions. Kolmogorov's three-series theorem, combined with Kronecker's lemma, can be used to give a relatively easy proof of the Strong Law of Large Numbers.

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 a more general result from martingale theory 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).

Example

As an illustration of the theorem, consider the example of the harmonic series with random signs:

$\sum _{n=1}^{\infty }\pm {\frac {1}{n}}.$

Here, "$\pm$" means that each term $1/n$ is taken with a random sign that is either $1$ or $-1$ with respective probabilities $1/2,\ 1/2$, and all random signs are chosen independently. Letting $X_{n}$ in the theorem denote a random variable that takes the values $1/n$ and $-1/n$ with equal probabilities, one can check easily that the conditions of the theorem are satisfied, so it follows that the harmonic series with random signs converges almost surely. On the other hand, the analogous series of (for example) square root reciprocals with random signs, namely

$\sum _{n=1}^{\infty }\pm {\frac {1}{\sqrt {n}}},$

diverges almost surely, since condition (3) in the theorem is not satisfied. Note that this is different from the behavior of the analogous series with alternating signs, $\sum _{n=1}^{\infty }(-1)^{n}/{\sqrt {n}}$, which does converge. In fact one can simply check that

$\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}}=({\sqrt {2}}-1)\zeta ({\frac {1}{2}})\approx -0.604899.$

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