# Almost Sure Convergence of a Series of Random Variables

Let $$\psi(x) = x^2$$ when $$|x| \leq 1$$ and $$\psi(x) = |x|$$ when $$|x| \geq 1$$. Show that if $$X_1, X_2, \dots$$ are independent with $$\mathbb{E} X_n = 0$$ and $$\sum_{n=1}^\infty \mathbb{E} \psi(X_n) < \infty$$, then $$\sum_{n=1}^\infty X_n$$ converges a.s.

This is Durrett exercise 2.5.6.

My thoughts:

We know by Kolmogorov's two-series test that if $$\sum_{n=1}^\infty \text{Var}X_n < \infty$$ that the series converges almost surely, and here $$\psi$$ is in some sense a "pseudo variance", but I'm not sure where to get started. Any help would be much appreciated.

Let $$Y_n=X_n I_{|X_n| \leq 1}$$. First observe that $$\sum EY_n^{2} <\infty$$ and $$\sum E|X_n| I_{|X_n| >1} <\infty$$. Using this second property we see that $$\sum P(|X_n| >1) <\infty$$. By Borel Cantelli Lemma $$|X_n| \leq 1$$ for all $$n$$ sufficiently large with probability $$1$$. Now Note that $$\sum var(Y_n) <\infty$$ so $$\sum Y_n$$ converges almost surely. Combining these two facts we see that $$\sum X_n$$ converges almost surely.

• Sorry to nitpick, but how is that Chebyshev? There is no variance here... Isn't it just $$\mathbb{P}\{|X_n| > 1\} = \mathbb{E}[ \mathbf{1}_{\{|X_n| > 1\}}] \leq \mathbb{E}[ |X_n|\mathbf{1}_{\{|X_n| > 1\}}]$$? (Besides that, +1) – Clement C. Dec 5 '19 at 5:49
• Yes, that's what I understood (cf. my comment). Nice argument... – Clement C. Dec 5 '19 at 5:53

The conditions given in the question does not necessarily imply that $$\sum_{n=1}^\infty \text{Var} X_n <\infty$$. Here is an example: Let $$X_n$$ take the values $$2^n$$ and $$-2^n$$, each with probability $$p_n = 1/3^n$$ and take the value $$0$$ with probability $$1-2p_n.$$ Then, $$\psi(X_n) = \min(X_n^2, |X_n|) = |X_n|,$$ hence $$\sum_{n=1}^\infty \text{E}\psi(X_n) = \sum_{n=1}^\infty \text{E}|X_n| = \sum_{n=1}^\infty 2^n \frac{2}{3^n} <\infty.$$ But, $$\sum_{n=1}^\infty \text{Var}X_n =\sum_{n=1}^\infty \text{E}(X_n^2) = \sum_{n=1}^\infty 4^n \frac{2}{3^n} =\infty.$$

The Kolmogorov three series theorem still applies. Define $$Y_n = X_n I_{|X_n|\le 1}.$$ Then,

(i) We have $$P(|X_n|\ge 1) = \text{E}(I_{|X_n|\ge 1})\le \text{E}(|X_n| I_{|X_n|\ge 1}) \le \text{E}\psi(X_n).$$ Hence $$\sum_{n=1}^\infty P(X_n\neq Y_n) = \sum_{n=1}^\infty P(|X_n|\ge 1) \le \sum_{n=1}^\infty \text{E}\psi(X_n) <\infty.$$

(ii) To show that $$\sum_{n=1}^\infty \text{E}Y_n$$ converges, observe that $$\text{E}Y_n=\text{E}(X_n I_{|X_n|\le 1}) = \text{E}(X_n - X_n I_{|X_n|\ge 1}) = - \text{E}(X_n I_{|X_n|\ge 1}),$$ and we already showed above that $$\sum_{n=1}^{\infty} \text{E}(X_n I_{|X_n|\ge 1}) \le \sum_{n=1}^{\infty} \text{E}\psi(X_n)<\infty.$$

(iii) We have $$Y_n^2 \le \psi(X_n),$$ hence $$\sum_{n=1}^\infty \text{E}(Y_n^2) \le \sum_{n=1}^\infty \text{E}\psi(X_n) < \infty.$$ So all the conditions are met and the conclusion follows from Kolmogorov three series theorem.