Discussion
Let \(\{X_{n}\}\) be a sequence of independent random variables such that \(E[X_{n}] = 0\) and \(\mathrm{Var}(X_{n}) = \dfrac{1}{n^{2}}\), \(n = 1, 2, \ldots\). Let \(E = \left\{\omega \mid \sum_{n} X_{n}(w)\ \text{converges}\right\}\).
Then which of the following is always correct ?
Answer: D
Always \(P(E) = 1\).
By Kolmogorov's one-series (two-series) theorem, a series of independent random variables with mean 0 converges almost surely when the sum of the variances is finite:
\[ \begin{aligned} \sum_{n} \mathrm{Var}(X_{n}) &= \sum_{n} \frac{1}{n^{2}} \\ &= \frac{\pi^{2}}{6} \lt \infty \end{aligned} \]
So \(\sum X_{n}\) converges with probability 1. (Kolmogorov's 0-1 law already says that P(E) is 0 or 1, since convergence of the series is a tail event; this rules out 1/2 and anything strictly between 0 and 1.)
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