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Consider $(X_i)$ independent events such that $\sum_{i = 1}^\infty P(X_i) = \infty$

Let $S_n = \sum_{i = 1}^n \mathbb{1}_{X_i}$, then $\frac{S_n}{\mathbb{E}S_n} \to 1$ almost surely.

Do you have any ideas how to prove this? I've been thinkint about applying law of large numbers or usual Borel-Cantelli lemma, but I could not come up with anything.

Turtle5
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  • $M_n= \sum\limits_{i=1}^{n} (1_{X_i}-P(X_i))$ is a martingale. If you can show it converges a.s. you are done. At the moment I don't know why it converges. – geetha290krm Dec 09 '22 at 23:33
  • Perhaps you could try to prove that random variable $T_n = S_n / E S_n$ converges to 1 in probability. Then it also would converge in distribution, and for sequence converging in distribution to any Dirac delta you have convergence almost surely. – Esgeriath Dec 09 '22 at 23:46
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    @Esgeriath That is not true. Convergence in distribution to a degenerate r.v. only gives convergence in probability, not a.s. convergence. – geetha290krm Dec 09 '22 at 23:51
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    This is theorem 2.3.9 of Durrett PTE 5e. I don’t think his proof is very well written, but it’s correct. – Andrew Dec 10 '22 at 05:15
  • Thanks @geetha290krm, I messed it up. Probably I shouldn't try to be clever late at night... – Esgeriath Dec 10 '22 at 08:27

1 Answers1

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Note that $1_{X_i} \sim \text{Bernoulli}(p_i)$ where $p_i = P(X_i)$.

If we consider a sequence of independent Bernoulli random variables $B_i$ where $P(B_i = 1)=p_i \in (0,1)$ then:

$$E[S_n] = \sum_{i=1}^n p_i,\;\; Var[S_n]= \sum_{i=1}^n p_i(1-p_i) $$

For independent but not necessarily identically distributed random variables, the following version of the strong law of large numbers holds:

$$\sum_{i=1}^{\infty} \frac{1}{k^2}Var[B_i] < \infty \implies \bar{B}_n-E[\bar{B}_n] \xrightarrow{a.s.} 0$$

Where $\bar{B}_n := \frac1n \sum_{i=1}^n B_i$ (sample average).

Since $0<Var[B_i] < 1$ we know that:

$$0<\sum_{i=1}^{\infty} \frac{1}{k^2}Var[B_i] < \sum_{i=1}^{\infty} \frac{1}{k^2} < \infty$$

Therefore,

$$\frac1n \sum_{i=1}^n B_i - \frac1n \sum_{i=1}^n p_i \xrightarrow{a.s.} 0$$

If $\frac{\sum_{i=1}^n p_i}{n} \to p$ (Cesaro convergence) and $\sum_{i=1}^n p_i > 0\;\forall n$ then :

$$\frac{\sum_{i=1}^n B_i}{\sum_{i=1}^n p_i} - 1 \xrightarrow{a.s.} 0 \implies \frac{\sum_{i=1}^n B_i}{\sum_{i=1}^n p_i} \xrightarrow{a.s.} 1$$

Stated more suggestively,

$$\frac{S_n}{E[S_n]} \xrightarrow{a.s.} 1\;\;\square$$

Annika
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  • Thank you, I really like this idea, that we can think about indicators as Bernoulli distribution. – Turtle5 Dec 10 '22 at 12:35