1. ## random variable convergence

Hello,

question is about about one point in proposition saying that convergence of random variable almost surely implies convergence in probability.

A sequence $\displaystyle (X_n),n\in \mathbb{N}$ of random variables is said to converge almost surely to the random variable X if $\displaystyle P(\omega\mid X_n\to X, n\to \infty)=1$, which is equivalent to $\displaystyle \forall \varepsilon \forall \delta \exists N : P(\bigcap_{n=N}^\infty\[\omega: \mid {X_n - X}\mid< \varepsilon])\geq 1-\delta$.

In proof of this proposition (in book of John B.Thomas) it is said that:
$\displaystyle P(\bigcap_{n=N}^\infty\[\omega: \mid {X_n - X}\mid< \varepsilon])\geq 1-\delta$ is equivalent to $\displaystyle P(\bigcap_{n=N}^\infty\[\omega: \mid {X_n - X}\mid\geq \varepsilon])< \delta$

Although, using De Morgan's law and property of pobability measure I get that
$\displaystyle P(\bigcap_{n=N}^\infty\[\omega: \mid {X_n - X}\mid< \varepsilon])\geq 1-\delta$ is equivalent to $\displaystyle P(\bigcup_{n=N}^\infty\[\omega: \mid {X_n - X}\mid\geq \varepsilon])< \delta$,
which is essentially different statement.

Can anybody comment this?

2. It looks like a typo to me.

For me, the easiest thing to remember is that $\displaystyle X_n \to X$ almost surely if and only if for every $\displaystyle \epsilon > 0$ we have $\displaystyle P([\omega: |X_n - X| \ge \epsilon \mbox{ for infinitely many$n$}]) = P(\limsup [\omega: |X_n -X| \ge \epsilon]) = 0$ (the first equality being by definition). Then the result follows from

$\displaystyle \limsup P([\omega: |X_n - X| \ge \epsilon]) \le P(\limsup [\omega: |X_n - X| \ge \epsilon])$.