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Convergence with Probability 1

Consider random variables X_1,X_2,\dots, and X on (\Omega, \mathscr{A}, \mathbb{P}). We say that X_n converges with probability 1 to X if

\displaystyle\mathbb{P}\left(\lim_{n\to\infty} X_n=X\right)=1.\quad\quad(1)

An equivalent condition for convergence wp1 is

\displaystyle\lim_{n\to\infty} \mathbb{P}(|X_m-X|<\varepsilon,\, \forall\, m\geqslant n)=1,\quad \forall\varepsilon>0.\quad\quad(2)

This facilitates comparison with convergence in probability.

The equivalence is proved by simple set-theoretic arguments.

\begin{aligned}\displaystyle\left\{\omega\mathrel{}\middle|\mathrel{}\lim_{n\to\infty}X_n(\omega)=X(\omega)\right\}&=\left\{\omega\mathrel{}\middle|\mathrel{}\forall\varepsilon>0,\, \forall m\geqslant n,\, |X_m-X|<\varepsilon\right\}\\\displaystyle&=\lim_{\varepsilon\to0}\bigcup_{n=1}^\infty\left\{\omega\vphantom{\lim_{n}}\mathrel{}\middle|\mathrel{}\, |X_m(\omega)-X(\omega)|<\varepsilon,\, \forall\, m\geqslant n\right\}.\end{aligned}

The set sequence S(n)=\left\{\omega\vphantom{\lim_{n}}\mathrel{}\middle|\mathrel{}\, |X_m(\omega)-X(\omega)|<\varepsilon,\, \forall\, m\geqslant n\right\} is increasing, hence \displaystyle\bigcup_{n=1}^\infty S(n)=\lim_{n\to\infty}S(n). Therefore,

\displaystyle\left\{\omega\mathrel{}\middle|\mathrel{}\lim_{n\to\infty}X_n(\omega)=X(\omega)\right\}=\lim_{\varepsilon\to0}\lim_{n\to\infty}\left\{\omega\vphantom{\lim_{n}}\mathrel{}\middle|\mathrel{}\, |X_m(\omega)-X(\omega)|<\varepsilon,\, \forall\, m\geqslant n\right\}.\quad(3)

By the continuity theorem for probability functions, this implies

\displaystyle\mathbb{P}(\lim_{n\to\infty}X_n=X)=\lim_{\varepsilon\to0}\lim_{n\to\infty}\mathbb{P}\left(|X_m-X|<\varepsilon,\, \forall\, m\geqslant n\right).

Let \varepsilon\to0, we yield (1) from (2).

Likewise,

\displaystyle\left\{\omega\mathrel{}\middle|\mathrel{}\lim_{n\to\infty}X_n(\omega)=X(\omega)\right\}\subset\bigcup_{n=1}^\infty\left\{\omega\vphantom{\lim_{n}}\mathrel{}\middle|\mathrel{}\, |X_m(\omega)-X(\omega)|<\varepsilon,\, \forall\, m\geqslant n\right\},\quad \forall\varepsilon>0,

which implies

\displaystyle\mathbb{P}(\lim_{n\to\infty}X_n=X)\leqslant\lim_{n\to\infty}\mathbb{P}\left(|X_m-X|<\varepsilon,\, \forall\, m\geqslant n\right).

So we yield (2) from (1).

Lemma (Borel-Cantelli Lemma I): For a sequence of events A_1,A_2,\dots, if \displaystyle\sum_{n=1}^\infty\mathbb{P}(A_n)<\infty, then \displaystyle\mathbb{P}\left(\limsup_{n\to\infty}A_n\right)=0.

It can be simply proved by the continuity property of \mathbb{P},

\displaystyle\mathbb{P}\Big(\limsup_{n\to\infty}A_n\Big)=\mathbb{P}\Big(\lim_{k\to\infty}\bigcup_{n=k}^{\infty}A_n\Big)=\lim_{k\to\infty}\mathbb{P}\Big(\bigcup_{n=k}^{\infty}A_n\Big)\leqslant\lim_{k\to\infty}\sum_{n=k}^{\infty}\mathbb{P}\left(A_n\right)=0.

With this lemma, it can be easily seen that: if \displaystyle\sum_{n=1}^\infty\mathbb{P}(|X_n-X|>\varepsilon)<\infty\, (\forall\varepsilon>0), then X_n\stackrel{wp1}{\rightarrow}X.

\displaystyle\mathbb{P}\Big(\limsup_{n\to\infty}\, \{\mathrel{}\omega\mathrel{}|\mathrel{}|X_n(\omega)-X(\omega)|>\varepsilon\}\Big)=0,\\\mathbb{P}\Big(\liminf_{n\to\infty}\, \{\mathrel{}\omega\mathrel{}|\mathrel{}|X_n(\omega)-X(\omega)|<\varepsilon\}\Big)=1.

The latter implies

\displaystyle\lim_{k\to\infty}\mathbb{P}\Big(\bigcap_{n=k}^\infty\, \{\mathrel{}\omega\mathrel{}|\mathrel{}|X_n(\omega)-X(\omega)|<\varepsilon\}\Big)=\lim_{k\to\infty} \mathbb{P}(|X_n-X|<\varepsilon,\, \forall\, n\geqslant k)=1.

Therefore, X_n\stackrel{wp1}{\rightarrow}X.