Let $X_n$ be a sequence of random variables with $X_n<\infty$ almost sure for all $n\in \mathbb N$. Show that there are constants $c_n\rightarrow \infty$ such that $\frac{X_n}{c_n}\rightarrow 0$
I guess I have to use the Borel Cantelli Lemma, right?
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Hints:
- Show that for $n \in \mathbb{N}$ and $\varepsilon>0$ there exists $R>0$ such that $$\mathbb{P}(|X_n|>R) < \varepsilon.$$ To this end, use that $X_n< \infty$ almost surely.
- By step 1, we can choose $R_n$ such that $$\mathbb{P}(|X_n| >R_n) \leq 2^{-n}.$$ Show for $c_n := n \cdot R_n$ that $$\sum_{n \geq 1} \mathbb{P} \left(|X_n| \geq \frac{c_n}{n} \right) < \infty.$$ Using the Borel-Cantelli theotem, conclude that $$ \frac{|X_n|}{c_n} \leq \frac{1}{n}$$ for $n$ sufficiently large.
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$\begingroup$ I just want to know if understand everything correctly: 1) The first inequality holds because $X_n<\infty$ a.s. that means that the values of $X_n$ are bounded on all non-zero sets.(in terms of measure), hence if R is large enough the probability can be smaller than any given $\epsilon$ (Because P doesn't consider zero-sets). 2): Just follows from 1). 3): Is just a usage of the geometric series, right? 4): The (first?) Borel-Cantelli lemma yields: If the sum is finite then the measure of all $\omega$ which infinitely often fulfills the inequality is zero. Did not understand how you use it here $\endgroup$– MarcCommented Oct 26, 2014 at 22:58
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3$\begingroup$ @Thingsgonnachange 1) I wouldn't put it like that. In general, $X_n< \infty$ almost surely does not imply that $X_n$ is bounded on sets of measure $>0$. Instead, $\mathbb{P}(|X_n|=\infty)=0$ and the continuity of the measure $\mathbb{P}$ implies $$\mathbb{P}(|X_n|>R) \stackrel{R \to \infty}{\to} \mathbb{P}(|X_n|=\infty)=0.$$ 2) Yes. 3) Yes. $\endgroup$– sazCommented Oct 27, 2014 at 6:13
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3$\begingroup$ 4) Borel Cantelli states that $\sum_n \mathbb{P}(A_n)<\infty$ implies that for almost all $\omega \in \Omega$ we can find $N$ such that for all $n \geq N$, we have $\omega \notin A_n$. (Just think about the complement: If the measure of $\omega$ which satisfy $$|X_n(\omega)| \geq \frac{c_n}{n}$$ infinitely often is zero, then ... ) $\endgroup$– sazCommented Oct 27, 2014 at 6:15