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i stacked in one example.

We have independent sequence of random variables $X_{n}$, where $X_{n}$ has Bernoulli distribution with parameter $\frac{1}{n}$, so $X_{n} \sim Bernoulli(\frac{1}{n})$ and then $P(X_{n} = 1) = \frac{1}{n}$ and $P(X_{n} = 0) = 1-\frac{1}{n}$. Show, that this convergence does not apply $X_{n} \xrightarrow{a.s} 0 $.

I know, that converges a.s is defined like this:

$P(\lim_{n\to\infty} |X_{n} - X| = 0) = 1 $.

So in my case i need to prove that:

$P(\lim_{n\to\infty} |X_{n}| = 0) \neq 1 $.

So i know that $\lim_{n\to\infty} X_{n} = 1 $, if $ X_{n} = 0 $ and $\lim_{n\to\infty} X_{n} = 0 $, if $ X_{n} = 1 $. And then $P(\lim_{n\to\infty} |X_{n}| = 0) = 1 $. But this is wrong. Can you please tell me where i made a mistake ? Thank you!

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  • $\begingroup$ $\lim_{n \to \infty} X_n \to 0$ does not make sense. It should be $\lim_{n \to \infty} X_n = 0$. Do not use an arrow when you have taken a limit. $\endgroup$
    – Xiaomi
    Commented Oct 21, 2018 at 12:29

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Your mistake is taking limits of random variables. $\lim_{n \to \infty} X_n = 1$ does not make sense. $X_n$ are random. You cannot just assert the limit is 1 or 0. If you do take a limit you need to state that it is almost surely or with probability 1. But even then, what you write really doesn't make sense.

Let $0<\epsilon <1,$. Note that $X_n$ are independent, so if you can show that

$$\sum_n P(|X_n - 0|> \epsilon) = \infty$$

then it will follow by the second Borel-Cantelli theorem that

$$P(|X_n - 0|>\epsilon \text{ i.o }) = 1$$

And hence $X_n$ does not converge to $0$ almost surely. But clearly

$$\sum_n P(|X_n - 0|>\epsilon) = \sum_n P(X_n = 1) = \sum_n \frac{1}{n} = \infty$$

Hence $X_n$ does not converge to $0$ almost surely.

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