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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.

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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.

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  • $\begingroup$ 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) $\endgroup$ – Clement C. Dec 5 '19 at 5:49
  • $\begingroup$ Yes, that's what I understood (cf. my comment). Nice argument... $\endgroup$ – Clement C. Dec 5 '19 at 5:53
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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.

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