1
$\begingroup$

It is stated in this Wikipedia article that convergence almost surely may be restated using the notion of the limit inferior of a sequence of sets, in that:

$$ \operatorname{Pr}\Big( \omega \in \Omega : \lim_{n \to \infty} X_n(\omega) = X(\omega) \Big) = 1. $$

can be restated as:

$$ \operatorname{Pr}\Big( \liminf_{n\to\infty} \big\{\omega \in \Omega : | X_n(\omega) - X(\omega) | < \varepsilon \big\} \Big) = 1 \quad\text{for all}\quad \varepsilon>0. $$

I am wondering why this is the case. I know that generally, it is NOT the case that:

$$ \omega\in\left\{\lim_{n\to\infty} X_n= X\right\}\iff \omega\in\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\{|X_k-X|<\varepsilon \}, \text{ for all }\varepsilon > 0. $$

So why are the two statements above equal?

$\endgroup$
  • 3
    $\begingroup$ "I know that generally, it is NOT the case that: [...]" Why do you think that this is not the case...? $\endgroup$ – saz Jan 15 '17 at 20:05
1
$\begingroup$

Fix $\varepsilon>0$. Denote $$A=\{\omega \in \Omega:\lim_{n \to \infty}X_n(\omega)=X(\omega)\},$$ $$A_n^\varepsilon=\{\omega \in \Omega \, : \,|X_k(\omega)-X(\omega)|< \varepsilon\text{ for all } k \ge n\}.$$ Then the original statement can be written as $P(A)=1$, and the second statement can be written either as $\lim\limits_{n \to \infty} P(A_n^\varepsilon)=1$ or as $P(A^\varepsilon)=1$, where $A^\varepsilon \equiv \lim\limits_{n \to \infty}A_n^\varepsilon=\bigcup\limits_{n=1}^{\infty}A_n^\varepsilon$. (Notice that $\{A_n^\varepsilon\}$ is an increasing sequence of events. By the continuity theorem for monotone events, $P(A^\varepsilon)=\lim\limits_{n \to \infty} P(A_n^\varepsilon)$.)

  1. Suppose $P(A)=1$. Note that $A\subseteq A^\varepsilon$. Indeed, in detail, $A^\varepsilon$ is the following set of outcomes: \begin{align*} \quad \ A^\varepsilon & = \{\omega \in \Omega: \exists \,n_\varepsilon(\omega) \text{ such that } \omega \in A_k^\varepsilon, \ \forall k\, \ge n_\varepsilon(\omega)\}&\\ & =\{\omega \in \Omega: \exists \,n_\varepsilon(\omega) \text{ such that } |X_k(\omega)-X(\omega)| \le \varepsilon, \ \forall \, k \ge n_{\varepsilon}(\omega)\}& %& = \{\omega \in \Omega: \exists n_\varepsilon(\omega) \text{ such that } \omega \in A_k^\varepsilon, \ \forall k\ge n_\varepsilon(\omega)\}& \end{align*} Let $\omega \in A$. Then for a chosen $\varepsilon>0$, $\exists \, n_\varepsilon(\omega)$ such that $\forall \,(k \ge n_\varepsilon(\omega)) \ |X_k(\omega)-X(\omega)|\le\varepsilon$. Clearly then, $\omega \in A^\varepsilon$.
    The fact that $A\subseteq A^\varepsilon$ implies that $P(A^\varepsilon) \ge P(A)=1$ and, thus, because probability cannot be greater than $1$, $P(A^\varepsilon) = 1$.

  2. Suppose now that $P(A^\varepsilon)=1$ for any $\varepsilon>0$. Take $\varepsilon=1,\frac{1}{2},\frac{1}{3},\ldots$ Then $A^1\supset A^\frac{1}{2} \supset A^\frac{1}{3}\supset \ldots$ $-$ a decreasing sequence of events, for which $\lim\limits_{m \to \infty}A^\frac{1}{m}=\bigcap\limits_{m=1}^{\infty}A^\frac{1}{m}$. By the continuity theorem, $$P\left(\bigcap\limits_{m=1}^{\infty}A^\frac{1}{m}\right) = P\left(\lim\limits_{m \to \infty}A^\frac{1}{m}\right)=\lim\limits_{m \to \infty}P\left(A^\frac{1}{m}\right)=1.$$ Now notice that $A=\bigcap\limits_{m=1}^{\infty}A^\frac{1}{m}$ and, thus, $P(A)=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.