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Let $X_1,X_2,...$ independent random variables, with $P(X_n=n)=1/{n^2}$ and $P(X_n=1/n)=1-1/{n^2}$ . Show that $X_n\to0$ in probability.

I got this problem in an old probability test but I'm not sure how to use the definition of convergence in probability:

$X_n\to X$ in probability, if $$P(\omega \in \Omega | \lim_{n\to \infty}X_n(\omega))=1$$

p.s.: I'm using Sheldon Ross' book "A First Course in Probability", but it doesn't have a convergece topic. Is there a better book?

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There must be a typo with the statement $P(X_n=1-1/{n^2})$ –  Learner Nov 26 '12 at 5:32
    
There was, thanks! –  Fernando H. M. Bastos Nov 26 '12 at 5:38

1 Answer 1

up vote 4 down vote accepted

We say that $X_n \rightarrow 0$ in probability if $$P ( X_n \geq \epsilon) \rightarrow 0$$ for every $\epsilon$.

So fix any $\epsilon$.

Pick $N$ so that $\frac{1}{N} < \epsilon$, then for $n \geq N$ can you find a bound on the probability that $X_n \geq \epsilon$?

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