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Let X be a non-negative random variable with finite expected value. Can we use Borel Cantelli Lemma to show that X is finite?

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  • $\begingroup$ do you mean "almost surely" finite? There are many examples of unbounded positive random variables that have a finite mean. $\endgroup$ – user237392 Mar 11 '16 at 5:21
  • $\begingroup$ Is $X$ somehow related to a sequence of events or random variables? If not, there doesn't seem to be much scope for applying Borel-Cantelli. $\endgroup$ – Robert Israel Mar 11 '16 at 5:30
  • $\begingroup$ Yes Bey it is almost surely... $\endgroup$ – user321821 Mar 11 '16 at 5:35
  • $\begingroup$ Robert Israel The question asks to use Borel-Cantelli $\endgroup$ – user321821 Mar 11 '16 at 5:36
  • $\begingroup$ It's asking if you can use Borel-Cantelli...not that you must. $\endgroup$ – user237392 Mar 11 '16 at 15:38
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Let $(a_n)$ be a monotone-increasing sequence of non-negative numbers. By the Markov's inequality,

$$ \Bbb{P}(X \geq a_n) \leq \frac{\Bbb{E}X}{a_n}. $$

Choose $a_n$ such that $\sum_n \frac{1}{a_n} < \infty$. Then $\sum_n \Bbb{P}(X \geq a_n) < \infty$ and hence by the Borel-Cantelli's lemma,

$$ \Bbb{P}(X \geq a_n \text{ i.o.}) = 0. $$

But if $X(\omega) = \infty$, then we must have $X(\omega) \geq a_n$ infinitely often. Therefore we have $\Bbb{P}(X = \infty) = 0$.

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This is simpler than the Borel–Cantelli lemma: if $\Pr(X=\infty)>0$, then $\operatorname{E}(X)=\infty$ by the definition of expectation.

Now suppose $X$ is a discrete random variable equal to the number of events in a sequence $A_1,A_2,A_3,\ldots$ that actually occur. Then one could think of applying Borel–Cantelli. One would have to know that the hypotheses are satisfied.

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  • $\begingroup$ I know the simple method...But the question asks to use the Borel-Cantelli lemma $\endgroup$ – user321821 Mar 11 '16 at 5:38
  • $\begingroup$ @user321821 : As I said in my second paragraph, there is certainly at least one context in which the Borel–Cantelli lemma can be applied. But one would need to know the specifics of the question to say much more than what I wrote. $\qquad$ $\endgroup$ – Michael Hardy Mar 11 '16 at 5:45

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