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I was reading this as a Lemma, however my book doesn't provide proof of it:

Let $X_1,X_2,...$ be a sequence of random variables, then the expression in the title is true.

I'm trying to understand bits of the lemma. Let $S_n$ be defined by:

$$S_n=\sum\limits_{i=1}^{n}X_i.$$ Now let's look at the hypothesis: $$\sum\limits_{n=1}^{\infty}\mathbb E(|X_n|)= \sum\limits_{n=1}^{\infty} \sum\limits_{x_n}^{}|x_n|\mathbb P(X_n=x_n) \leq \sum\limits_{n=1}^{\infty} \sum\limits_{x_n}^{}|x_n|\stackrel{\text{not sure of this}}{<}\infty.$$ I'm not entirely sure if I can say that $(*)$ $\sum\limits_{n=1}^{\infty} \sum\limits_{x_n}^{}|x_n|=\sum\limits_{n=1}^{\infty}X_n $ or $\sum\limits_{n=1}^{\infty} \sum\limits_{x_n}^{}|x_n|<\infty\Rightarrow\sum\limits_{n=1}^{\infty}X_n <\infty$, because in the left-hand side of the equation we are adding numbers and in the right-hand side we are "adding" random variables.

Continuing with my idea of defining $S_n$, I would like to show that $S_n$ converges almost surely, i.e.:

$$\mathbb P\left(\lim_{n\to\infty} S_n=X\right)=1$$

for some random variable $X$.

If I could write some of the two options I pointed out in $(*)$, then I'd be able to prove that $S_n \stackrel{\text{a.s}}{\underset{n\to\infty}{\longrightarrow}} X$, $X=L\in \mathbb R$.

Any help would be appreciated.

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You have $E [ \sum_n |X_n| ] = \sum_n E[|X_n|] < \infty$, hence $\sum_n |X_n| < \infty$ ae. It follows that $\sum_n X_n$ converges absolutely ae.

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  • $\begingroup$ I'm a bit confused with the Lemma. It has 2 parts. The first one says: $\mathbb E \left(\sum_n|X_n|\right)=\sum_n\mathbb E|X_n|.$ The second one is the one I asked, but at the end it says: "... converges almost surely and the first part holds". As if it was necessary to prove the second part to prove the first one. However the property you pointed out is easily proved. $\endgroup$ – Vladimir Vargas Nov 22 '14 at 3:12
  • $\begingroup$ I would guess that its badly worded, Tonelli is the basic result here, vaguely similar to the Borel Cantelli lemma. $\endgroup$ – copper.hat Nov 22 '14 at 4:37
  • $\begingroup$ Alright. Thanks @copper.hat $\endgroup$ – Vladimir Vargas Nov 22 '14 at 4:41

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