- #1
Gregg
- 459
- 0
Homework Statement
We have ##\mathbb{P}(X_n = 1) = p_n ## and ##P(X_n=0) = 1-p_n ## the question is about almost sure convergence. i.e. does ## X_n \overset{a.s.}{\longrightarrow} 0 ## if ##p_n = 1/n##?
Homework Equations
##X_n \overset{a.s.}{\longrightarrow } X ## if ## \mathbb{P}( \omega \in \Omega : X_n(\omega) \to X(\omega) \text{ as } n\to \infty) = 1 ##
The Attempt at a Solution
I don't think I understand this properly. Looking at my attempt I've tried a quick ##\epsilon -\delta## setting ##\epsilon = 2/N ## and having ## |1/n| < \epsilon ## for ##n>N##
I don't think this is what it's asking. Can I say that ## X(\omega) = 0 ## "clearly" and then that ##\mathbb{P}( \omega \in \Omega : |X_n(\omega) - X(\omega)| > \epsilon \text{ i.o. }) = 0## ?
Where i.o. means infinitely often.