# Divergent series of independent RV

I'm trying to prove that if $\{X_n\}_{n=1}^{\infty}$ is a sequence of independent random variables with the same distribution and $P(X_1 \neq 0)>0$, then the series $\sum_{n=1}^{\infty} X_n$ is divergent almost surely.

$\sum_{n=1}^{\infty} X_n$ is convergent if, by Cauchy criterion, $$\forall \varepsilon >0 \ \ \exists N \ge 1 : \ \forall m, n \ge N : \ |S_m - S_n|< \varepsilon.$$

So, if the series is divergent, we have: $$\exists \varepsilon >0 \ \ \forall N \ge 1 : \ \exists m, n \ge N : \ |S_m - S_n| \ge \varepsilon.$$

Equivalently, $$\exists k \in \mathbb{N_+} \ \ \forall N \ge 1 : \ \exists m, n \ge N : \ |S_m - S_n| \ge \frac{1}{k}.$$

So we need to check that: $$P(\omega \in \Omega \ | \ \bigcup_{k \in \mathbb{N_+}} \bigcap_{N \in \mathbb{N_+}} \bigcup_{m, n \ge N} \{ |S_m(\omega) - S_n(\omega)| \ge \frac{1}{k} \})=1.$$

$$|S_m(\omega) - S_n(\omega)| = X_{n+1}(\omega) + ... + X_{m}(\omega)$$

I guess I could set $A_{N, \varepsilon} := \bigcup_{m, n \ge N} \{ |S_m(\omega) - S_n(\omega)| \ge \frac{1}{k} \}$. The sequence $\{A_{N, \varepsilon} \}_{N \in \mathbb{N_+}}$ is decreasing. But I doubt that's helpful. I suppose I should prove that the intersection $\bigcap_{N \in \mathbb{N_+}} A_{N, \varepsilon}$ has measure $1$, because then the union over $k \in \mathbb{N}$ would also have measure one. But I'm not sure whether this is true or how to show this.

Could you give me some hints? What could I do to prove that the series is divergent?

## 1 Answer

Use the Borel-Cantelli lemma.

Essentially, for the series to converge almost surely, the probability X is bigger than 0 necessarily has to converge to zero, which means that P({X=0 i.o.}) = 1. Clearly that's not the case since P({X≠0}) > 0 and constant for all X. Therefore, the sum of the probabilities diverges. And, because the sum of the probabilities diverges, the expectation of the series diverges (Markov's inequality). Hence, The series itself diverges.

reference:

http://www.columbia.edu/~ks20/stochastic-I/stochastic-I-BC.pdf

• Isn't $P(X_n >0) \to 0$ a necessary condition for the series to converge, not diverge? Also, could you explain to me why $P(X_1 \neq 0) >0$ implies that the series: $\sum_{n=1}^{\infty}P(X_n \neq 0)$ diverges? We want to use contraposition of lemma 1, is that correct? – Hagrid Apr 18 '15 at 10:42
• By iid assumption, P(X≠0) = c (a constant) for all X. Therefore, the partial sums are exactly cn. Thus, the sum of the probabilities is O(n) which diverges. Yes, contrapositive seems the right way to go. And, I'll fix the first part, that was a mistake. – Ragnar Apr 18 '15 at 10:45
• I see, you mean Proposition 0.1 in the article. From that we have that $P(A(i.o.)) = 1$, is that right? This gives us divergence of the series of $P(X_n)$, what needs to be done to prove that $\sum X_n$ is divergent? – Hagrid Apr 18 '15 at 10:48
• Expectation gives us $L^1$ convergence/ divergence. How can I get almost sure divergence from that? – Hagrid Apr 18 '15 at 10:59
• I've found the following on wikipedia: If $S_n$ is a sum of $n$ real independent random variables: $S_n = X_1+\cdots+X_n$, then $S_n$ converges almost surely if and only if $S_n$ converges in probability. Could you tell me how to prove that? – Hagrid Apr 18 '15 at 11:03