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I would like some help with the following problem. Thanks for any help in advance.

Demonstrate by way of a suitable example that if $(X_n)_{j=1}^\infty$ is a sequence of nonnegative random variables such that

$P({X_n} = 0i.o.(n)) =1$ and $\lim_{n\to\infty} EX_n = 0$

then it is not necessarily true that $\lim_{n\to\infty} X_n = 0 a.c.$

I don't see how the above cannot be true. Wouldn't Borel-Cantelli imply that $\lim_{n\to\infty} X_n = 0 a.c.$ ?

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shmiggens
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1 Answers1

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The Borel Cantelli lemma plays no role here. Note that $P(X_n \ne 0 \text { i.o.}) = 0$ would imply $X_n \to 0$ almost surely. But $\{X_n \ne 0 \text{ i.o.}\}$ is not the complementary event to $\{X_n = 0 \text{ i.o.}\}$.

Counterexamples can be found e.g. here.

To expand on the sliding humps example: Let $A_n$ be the sequence of intervals $[0, 1]$, $[0, \tfrac{1}{2}]$, $[\tfrac{1}{2}, 1]$, $[0, \tfrac{1}{3}]$, $[\tfrac{1}{3}, \tfrac{2}{3}]$, $[\tfrac{2}{3}, 1]$, $[0, \tfrac{1}{4}]$ etc.. We consider the random variables $X_n(\omega) = I\{ \omega \in A_n\}$.

Now every real numbers $\omega \in [0, 1]$ lies in infinitely many of the intervals $A_i$, so $X_n(\omega)$ assumes the value $1$ infinitely often. But at the same time, $\omega$ is also in the complement of infinitely many $A_i$, so $X_n(\omega)$ attains the value $0$ also infinitely often.

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Dominik
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  • Would the first counterexample work here? I don't see how $P({X_n} = 0i.o.(n)) =1$ is satisfied. – shmiggens Dec 08 '16 at 13:56
  • The example with the sliding bumps works. Note that for every $\omega \in [0, 1]$ you have that $X_n(\omega)$ attains the values $0$ and $1$ both infinitely often. – Dominik Dec 08 '16 at 14:16
  • Even for the third example, I don't see how both my conditions are satisfied. Could you explain this a bit more? I will give you the check mark if you just provide an explanation of how they are satisfied for the sliding bumps example. – shmiggens Dec 08 '16 at 14:23
  • I have added some further explanations. – Dominik Dec 08 '16 at 14:36
  • Thanks for your help. If I may impose upon you again, there is one more problem that I have which is based upon me finding a suitable example of something. – shmiggens Dec 08 '16 at 15:21
  • To keep things tidy, you should open a new question with your new problem. – Dominik Dec 08 '16 at 15:23
  • Here is the link to my other problem. – shmiggens Dec 08 '16 at 16:45