If $(X_n)$ is independent with $X_n$ Bernoulli of parameter $\frac1n$, how to show that $X_n$ doesn't converge almost surely to $0$?

Question

If $X_n \sim Bernoulli(\frac{1}{n})$ independently, then I know that because $\sum_{i=1}^{\infty} P(X_n = 1) = \infty$. Then by Borel-Cantelli, I know that $P(\limsup_{n \to \infty}\{X_n=1\}) = 1$. This means that the event $\{X_n = 1\}$ happens infinitely often with probability 1. However, I do not know how to rigorously use this to prove that $X_n$ doesn't converge to $0$ almost surely. I know that the definition of almost sure convergence is:

$$ P(\{\omega \in \Omega: \lim_{n \to \infty}X_n(\omega) = X\}) = 1 $$

How can I use the above to show that this doesn't hold?

Answer 1

Note that $$\mathbb P\left(\left\{\omega\in\Omega:\lim_{n\to\infty} X_n(\omega)=X(\omega)\right\} \right)=1 $$ is equivalent to $$\mathbb P\left(\left\{\omega\in\Omega:\liminf_{n\to\infty} |X_n(\omega)-X(\omega)|<\varepsilon \right\}\right)=1, \text{ for all }\varepsilon > 0. $$ This follows directly from the definition of convergence of a sequence of real numbers: $$\omega\in\left\{\lim_{n\to\infty} X_n= X\right\}\implies \omega\in\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\{|X_n-X|<\varepsilon \}, \text{ for all }\varepsilon > 0.$$ Now, since \begin{align} \left(\limsup_{n\to\infty}\ \{X_n=1\}\right)^c &= \left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty \{X_k=1\} \right)^c\\ &= \bigcup_{n=1}^\infty\bigcap_{k=n}^\infty \{X_k=0\}\\ &= \liminf_{n\to\infty}\ \{X_n=0\}, \end{align} we see that $$\mathbb P\left(\liminf_{n\to\infty}\ \{X_n=0\}\right)=0. $$ From this the result follows immediately.

Answer 2

Let us call $A$ the event $\{\limsup_{n \to \infty}X_n=1\}$ (shorthand notation for $\{\omega : \limsup_{n \to \infty}X_n(\omega)=1\}$). You have already proved that $P(A)=1$.

Let us call $B$ the event $\{\lim_{n \to \infty}X_n=0\}$ (shorthand notation for $\{\omega : \lim_{n \to \infty}X_n(\omega)=0\}$). Almost sure convergence is equivalent (according to your definition) to $P(B)=1$

But the events $A$ and $B$ are disjoint (this is not probability anymore, just the mere deterministic fact that a sequence that has infinitely many ones can't converge to zero).

So $B \subset A^C$ so $P(B) \leq P(A^C) = 1-P(A) = 1-1 = 0$. Hence $P(B)=0$ so the definition of almost sure convergence is not met.

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That was the developed argument, once you get more accustomed to probability you'd say : Almost surely the sequence has infinetely many ones, so almost surely it does not converge to 0, so there can't be almost sure convergence.

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Last remark, you haven't just proven no almost sure convergence, but the stronger statement of almost-sure non convergence. Try to wrap your head around these two statements, and to write their formal definition yourself as a good exercise.