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Let $(Y_n)$ real valued random variables, such that $Y_n\to c \in \mathbb R$ in distribution. It has been shown that $Y_n \to c$ in probability. I want to prove that $Y_n \to c $ a.s. does not hold! Therefore consider $Y_n \sim Ber_{1/n}\,$, i.e. $\mathbb P(Y_n=0)=1-\frac1n$ and $\mathbb P(Y_n=1)=\frac1n$. Then $\mathbb E(Y_n)=\frac1n$. I will use Markow's inequality two times.

$Y_n$ converges in probability to zero $$\lim_{n\to\infty}\mathbb P(\vert Y_n - 0\vert \ge \epsilon)\le \lim_{n\to\infty}\frac{\mathbb E(Y_n)}{\epsilon}=\lim_{n\to\infty}=\frac{\frac1n}{\epsilon}=0.$$

$Y_n$ does not converge a.s. to zero. I will use Borel Cantelli,i.e.$$\sum_{n=1}^\infty \mathbb P(\vert Y_n-0\vert \ge \epsilon) \le\sum_{n=1}^\infty\frac{\mathbb E(\vert Y_n\vert)}{\epsilon}=\sum_{i=1}^\infty\frac{\frac1n}{\epsilon}=\infty$$

Are we supposed to assume $\epsilon=1$ and is this attempt fine? If yes, Borel Cantelli provides $Y_n$ does not almost surely converges to zero. Some comments on this attempt are welcomes!

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  • $\begingroup$ Your reasoning for the second part is not correct. Note that the statement $$\sum_n \mathbb{P}(|Y_n-0| \geq \epsilon) \leq \infty$$ is trivial; it doesn't allow you to conclude anything. If you want to apply the Borel-Cantelli lemma to prove that the sequence does not converge almost surely to $0$, you have to show that $$\sum_n \mathbb{P}(|Y_n-0| \geq \epsilon) \color{red}{=} \infty$$ for some $\epsilon>0$. $\endgroup$
    – saz
    Dec 16 '17 at 16:28
  • $\begingroup$ @saz I missed it. Knew what to show but did not realise that I may not use markow's .., question to first part: We are allowed to assume $\epsilon =1 $ or? $\endgroup$
    – user409387
    Dec 16 '17 at 17:16
  • $\begingroup$ and $\sum P (\vert Y_n \vert \ge \epsilon)=\sum \frac1n = \infty $right? $\endgroup$
    – user409387
    Dec 16 '17 at 17:33
  • $\begingroup$ Re your first comment: Why do you think that it suffices to check $\epsilon=1$? Have a look at the very definition of convergence in probability! Re your 2nd comment: Yes, that's true for $\epsilon \in (0,1]$. $\endgroup$
    – saz
    Dec 16 '17 at 18:30
  • $\begingroup$ @saz my first part would not hold for something like $\epsilon=\frac1n$ that's what is bothering me.. $\endgroup$
    – user409387
    Dec 16 '17 at 18:57
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It seems you don't apply Borel-Cantelli correctly. Consider the events {$Y_n=1$} for $n \in \mathbb{N}$. They are independent and

$$\sum_{n=1}^\infty \mathbb{P}(Y_n=1) = \sum_{n=1}^\infty \frac{1}{n} = \infty$$

Thus by Borel-Cantelli, $\limsup_{n \to \infty} Y_n = 1$.

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  • $\begingroup$ I can show $\sum P(\mid Y_n -0 \mid \ge \epsilon)= \infty$ for $ \epsilon =1$ . Then we have shown $Y_n$ does not converge to zero a.s. Therefore $\sum P(\mid Y_n \mid \ge 1)= \sum \frac1n = \infty$ And that is what I did in the comments! $\endgroup$
    – user409387
    Dec 18 '17 at 18:04
  • $\begingroup$ Oh, I understand your move now. $\endgroup$
    – Leon
    Dec 18 '17 at 21:16

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