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Let $X_n$ be a sequence of random variables. Let $a$ be a negative real number. If $X_n\overset{p}\to a$, is it true that $\lim_{n\to\infty} P(X_n > 0) = 0$? I think I came up with a proof but it's a little bit long (and in fact, not sure about "exchanging limit and sum" the last step).

Wlog, assume $a=-1$. Since $X_n\overset{p}\to -1$, then for any $\epsilon >0$, $\lim P(X_n > \epsilon - 1) = 0$. Let $\epsilon = 1 + \frac{1}{k}$, we have $\lim_n P(X_n > \frac{1}{k}) = 0$ for any $k$.

For any fixed $n$, $P(X_n > 0) = P(\cup_{k=1}^\infty X_n > \frac{1}{k})$ by continuity of probability. Then we have $P(X_n > 0) = \sum_{k=1}^\infty P(X_n > \frac{1}{k})$. So $\lim P(X_n >0) = \lim_n \sum_{k=1}^\infty P(X_n > \frac{1}{k})=\sum_{k=1}^\infty \lim_n P(X_n > \frac{1}{k}) = \sum_k 0 = 0$

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  1. $\{X_n>0\}\subseteq \{|X_n -a|>|a|\}$
  2. $P(X_n>0) \le P(|X_n -a|>|a|) \to 0$ by definition of convergence in probability.
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  • $\begingroup$ Since $a$ is negative, isn't it the case that $\mathbb{P}\left(\left|X_n - a\right| > a\right) = 1$? $\endgroup$ – 3x89g2 Dec 10 '17 at 7:46
  • $\begingroup$ It is, this was a typo, now corrected (the correct event is $\{|X_n-a|>|a|\}$, namely distance between $X_n$ and $a$ is larger than the norm of $a$). Thanks! $\endgroup$ – Fnacool Dec 13 '17 at 18:44

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