Let $\{X_{n}\}$ be a sequence of r.v. such that $$ X_{n} \stackrel{a.s.}{\to} X $$ According to wikipedia https://en.wikipedia.org/wiki/Convergence_of_random_variables $$ P[\omega\in\Omega: X_{n}(\omega) \to X(\omega)] = 1 $$ is equivalent to $$ P[\underset{n\to\infty}{\liminf} \{\omega\in\Omega:|X_{n}(\omega) - X(\omega)|<\varepsilon \}] = 1 $$ for any $\varepsilon$. Could someone prove it?

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    $\begingroup$ You have a notation and a definition of a concept. What's there to prove? $\endgroup$ – Eric Towers Apr 14 '18 at 20:11
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    $\begingroup$ @EricTowers I think they're probably looking for a proof that this is equivalent to $P(X_n\to X) = 1.$ $\endgroup$ – spaceisdarkgreen Apr 14 '18 at 20:16
  • $\begingroup$ I have edited the post $\endgroup$ – Mr.M Apr 15 '18 at 15:14

There is indeed something to prove here.

Claim 1:

If $\{B_k\}_{k=1}^{\infty}$ is an infinite sequence of events such that $P[B_k]=1$ for all $k \in \{1, 2, 3,...\}$, then $$P[\cap_{k=1}^{\infty} B_k] = 1$$

Proof: Computing the probability of the complement of $\cap_{k=1}^{\infty} B_k$ gives:
$$ P[(\cap_{k=1}^{\infty} B_k)^c] = P[\cup_{k=1}^{\infty} B_k^c] \leq \sum_{k=1}^{\infty} P[B_k^c] = \sum_{k=1}^{\infty} 0 = 0 $$ where we have used the union bound. $\Box$

The definition of $\liminf$ of a sequence of sets $\{A_n\}_{n=1}^{\infty}$ is: $$ \liminf_{n\rightarrow\infty} A_n = \cup_{M=1}^{\infty} \cap_{n=M}^{\infty} A_n$$ For each $\epsilon>0$ and each $n \in \{1, 2, 3, ...\}$, define $A_n(\epsilon)$ as the following event: $$ A_n(\epsilon) = \{|X-X_n|< \epsilon\}$$ Define events $B(\epsilon)$ by $$ B(\epsilon) =\liminf_{n\rightarrow\infty} A_n(\epsilon) = \cup_{M=1}^{\infty} \cap_{n=M}^{\infty} \{|X-X_n|<\epsilon \} $$ The event $B(\epsilon)$ is equivalent to the event that there exists an integer $M>0$ such that $|X-X_n|<\epsilon$ for all $n \geq M$.

We want to prove that: $$\boxed{P[X_n\rightarrow X]=1 \quad \mbox{ if and only if } \quad P[B(\epsilon)]=1 \quad \forall \epsilon>0} $$

Claim 2:

$\{X_n\rightarrow X\} = \cap_{k=1}^{\infty} B(1/k)$.

Proof: This claim says that $X_n\rightarrow X$ if and only if for each integer $k$ there is an integer $M>0$ such that $|X_n-X|<1/k$ for all $n \geq M$. This is true by definition of a limit. $\Box$

Claim 3

a) $P[X_n\rightarrow X] = P[\cap_{k=1}^{\infty} B(1/k)] \leq P[B(1/i)]$ for all $i \in \{1, 2, 3, ...\}$.

b) $P[X_n\rightarrow X] = 1$ if and only if $P[B(1/i)]=1$ for all $i \in \{1, 2, 3, ...\}$.

c) $P[X_n\rightarrow X] = 1$ if and only if $P[B(\epsilon)]=1$ for all $\epsilon>0$.

Proof: Part (a) follows immediately from Claim 2 and the fact that for every $i \in \{1, 2, 3, ...\}$ we have $\cap_{k=1}^{\infty} B(1/k) \subseteq B(1/i)$.

Part (b) follows from (a) together with Claim 1, specifically, if $P[B(1/i)]=1$ for all $i \in \{1, 2, 3, ...\}$ then $P[\cap_{i=1}^{\infty} B(1/i)]=1$.

To prove part (c), assume $P[B(\epsilon)]=1$ for all $\epsilon>0$. Then $P[B(1/i)]=1$ for all positive integers $i$, and by part (b) we conclude $P[X_n\rightarrow X]=1$. Now supose $P[X_n\rightarrow X] = 1$. Then from part (b) we know $P[B(1/i)]=1$ for all $i \in \{1, 2, 3, ...\}$. Fix $\epsilon>0$. Choose $i \in \{1, 2, 3, ...\}$ such that $1/i\leq \epsilon$. Then $$ B(1/i)\subseteq B(\epsilon) \implies \underbrace{P[B(1/i)]}_{1} \leq P[B(\epsilon)]$$ and so $P[B(\epsilon)]=1$. $\Box$

  • $\begingroup$ Note: An additional equivalence that I find more useful is: $P[X_n\rightarrow X]=1$ if and only if for all $\epsilon>0$ we have $$ \lim_{n\rightarrow\infty} P[\cup_{m=n}^{\infty} \{|X_m-X|>\epsilon\}]=0$$ since from this we see that "with probability 1" convergence is more stringent than "in probability" convergence, which means $\lim_{n\rightarrow\infty} P[|X_n-X|>\epsilon]=0$. $\endgroup$ – Michael Apr 15 '18 at 23:32

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