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I have this for homework:

Let $X_1, X_2,\cdots,X_n, \cdots$ be descending sequence of random variables. Show that $X_n$ converges to 0 in probability if and only if $X_n$ converges to 0 almost surely.

I know that the "only if" part is generally true. Now, if $X_n$ converges to zero in probability, that means that there is subequence $X_{p(n)}$ that converges to zero (a.s.). My problem is: can I conclude that $X_n$ converges to zero almost surely since $X_n$ is descending and has almost surely converging subsequence?

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  • $\begingroup$ What are your thoughts on the problem? Where are you stuck? $\endgroup$
    – saz
    Feb 3, 2015 at 13:39
  • $\begingroup$ The "only if" part is generally true. Now, if X_n converges to zero in probability, that means that there is subequence X_p(n) that converges to zero (a.s.). My problem is: can I conclude that X_n converges to zero almost surely since X_n is descending and has almost surely converging subsequence? $\endgroup$
    – Martin
    Feb 3, 2015 at 15:16
  • $\begingroup$ If you add these lines to your question, I'll give you a vote to reopen the question. $\endgroup$
    – saz
    Feb 3, 2015 at 15:23

1 Answer 1

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Sketch of the proof: Fix $\varepsilon>0$. Since $X_n \to 0$ in probability and $X_n$ is decreasing, we have

$$\lim_{n \to \infty} \mathbb{P}(X_n \notin [0,\varepsilon)) =0.$$

In particular, we can choose a subsequence $(n(k))_{k \in \mathbb{N}}$ such that

$$\sum_{k \in \mathbb{N}} \mathbb{P}(X_{n(k)} \notin [0,\varepsilon))<\infty.$$

It follows from the Borel-Cantelli lemma that for almost all $\omega \in \Omega$ we have

$$0 \leq X_{n(k)}(\omega) \leq \varepsilon$$

for all $k \geq K(\omega)$ sufficiently large. Conclude from the monotonicity that

$$0 \leq X_{n}(\omega) \leq \varepsilon$$

for all $n \geq n(K(\omega))$.

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