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Let $ {a_n} $ be a sequence of numbers in $ (0,1) $ such that $ a_n\to0 $, but $\sum_{i=0}^\infty a_n=\infty$. Suppose $X_1, X_2, \ldots$ be independent random variables with $$P(X_n=1)=P(X_n=-1)=\frac{a_n}{2}\ \qquad \text{and} \qquad P(X_n=0)=1-a_n $$ for all $n\ge1$. Define $Y_1=X_1$ and for all $n \ge 2$, $$Y_n = \begin{cases} X_n, & \text{if } Y_{n-1} =0 \\[0.2cm] nY_{n-1}|X_n|, & \text{if } Y_{n-1} \neq 0 \end{cases}$$

Show that $Y_n$ converges in probability but not almost surely.

I have already shown $Y_n\to0$ in probability. I need help with the second part.

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  • $\begingroup$ Have you tried to apply the second Borel–Cantelli lemma to the event $E_n\stackrel{\rm def}{=} \{X_n=0\}$, since $\mathbb{P}\{Y_n=0\} = \mathbb{P}\{X_n=0\}$? $\endgroup$
    – Clement C.
    Dec 11 '14 at 14:22
  • $\begingroup$ @ClementC. OP is trying to show that $Y_n$ does not converge a.s., which requires the first Borel-Cantelli lemma. Unfortunately, this is not satisfied for $\{Y_n=0\}$ so other methods are needed. $\endgroup$
    – user76844
    Dec 11 '14 at 14:43
  • $\begingroup$ Oh, my mistake. Sorry. $\endgroup$
    – Clement C.
    Dec 11 '14 at 14:45
  • $\begingroup$ By $Y_{(n-1)}$ you mean $Y_{n-1}$ or something else? $\endgroup$
    – Jimmy R.
    Dec 11 '14 at 14:59
  • $\begingroup$ Edited. I meant $Y_{n-1}$ only. $\endgroup$ Dec 11 '14 at 15:10
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The events $\{Y_n=0\}$ are independent, thus $P(Y_n=0)=1-a_n \implies \sum P(Y_n=0)=\sum(1-a_n) = \infty - \infty$ but this is indeterminate, and thus we cannot apply the second lemma diretly to the event of interest.

However, since the set of events $\{\{Y_n=0\},\{Y_n \neq 0\}\}$ partition the sample space, we can conclude that $\{Y_n\neq 0\}$ are also independent. Thus, we can apply the second lemma to this series to get:

$$ \sum P(Y_n \neq 0) = \sum a_n = \infty \implies P(\lim \sup Y_n \neq 0) = 1 \implies P(\lim \sup Y_n=0)=0 \;\;\;\square$$

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