There is a sequence of random variables defined by $$Y_n = \Big(\Big|{1-\frac\Theta \pi}\Big|\Big)^n$$ where $\Theta\sim\mathrm{unif}[0,2\pi].$ I have shown that the sequence converges to $0$ in probability. Applying Borel Cantelli Lemma by taking $A_n = \{|Y_n|>0\}$ and then evaluating $\sum_{n=1}^\infty P(A_n)$, I get it to be $\infty$. Since the random variables are not independent, I can't conclude anything from this result. How do I prove/disprove convergence almost surely?


$Z=(1-\Theta/\pi)$ is almost surely in $(-1,1),$ in which case $Z^n\to 0.$ Thus $Y_n = Z^n \to 0$ almost surely.

  • $\begingroup$ Thank you for you answer. $P(\lim_{n\to\infty}Y_n = 0) = 1$ implies a.s. convergence right? In this case $Y_n \to 0$ a.s. as $n \to \infty$. Does that guarantee a.s. convergence? $\endgroup$ – learner Nov 3 '18 at 6:34
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    $\begingroup$ @learner If it almost surely converges to zero, then it almost surely converges. More formally, $P(Y_n \mbox{ converges}) \ge P(Y_n\to 0) = P(Z^n\to 0) \ge P(|Z|<1) = 1.$ $\endgroup$ – spaceisdarkgreen Nov 3 '18 at 6:41
  • $\begingroup$ thank you so much for clarifying. $\endgroup$ – learner Nov 3 '18 at 6:50

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