Kolmogorov's three-series theorem

Condition (i) and Borel–Cantelli give that ${\displaystyle X_{n}=Y_{n}}$ for ${\displaystyle n}$ large, almost surely. Hence ${\displaystyle \textstyle \sum _{n=1}^{\infty }X_{n}}$ converges if and only if ${\displaystyle \textstyle \sum _{n=1}^{\infty }Y_{n}}$ converges. Conditions (ii)-(iii) and Kolmogorov's Two-Series Theorem give the almost sure convergence of ${\displaystyle \textstyle \sum _{n=1}^{\infty }Y_{n}}$.

Suppose that ${\displaystyle \textstyle \sum _{n=1}^{\infty }X_{n}}$ converges almost surely.

Without condition (i), by Borel–Cantelli there would exist some ${\displaystyle A>0}$ such that ${\displaystyle \{|X_{n}|\geq A\}}$ for infinitely many ${\displaystyle n}$, almost surely. 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 ${\displaystyle A=1}$ gives the convergence of ${\displaystyle \textstyle \sum _{n=1}^{\infty }(Y_{n}-\mathbb {E} [Y_{n}])}$. So given the convergence of ${\displaystyle \textstyle \sum _{n=1}^{\infty }Y_{n}}$, we have ${\displaystyle \textstyle \sum _{n=1}^{\infty }\mathbb {E} [Y_{n}]}$ 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 ${\displaystyle \textstyle \sum _{n=1}^{\infty }Z_{n}=\textstyle \sum _{n=1}^{\infty }(Y_{n}-Y'_{n})}$ where for each ${\displaystyle n}$, ${\displaystyle Y_{n}}$ and ${\displaystyle Y'_{n}}$ are IID—that is, to employ the assumption that ${\displaystyle \mathbb {E} [Y_{n}]=0}$, since ${\displaystyle Z_{n}}$ is a sequence of random variables bounded by 2, converging almost surely, and with ${\displaystyle \mathrm {var} (Z_{n})=2\mathrm {var} (Y_{n})}$. So we wish to check that if ${\displaystyle \textstyle \sum _{n=1}^{\infty }Z_{n}}$ converges, then ${\displaystyle \textstyle \sum _{n=1}^{\infty }\mathrm {var} (Z_{n})}$ 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 (${\displaystyle \mathbb {E} [Z_{n}]=0}$; the series of the variances is converging; and the summands are bounded).[2][3][4]