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Suppose that $X_n$ is a sequence of random variables and that I want to show that $X_n$ goes to $-\infty$ in probability, in other words. I am wondering how this is defined by using the definition of convergence in probability. Usually if we have a target $X$, then $X_n \to X$ in probability means that for all $\epsilon >0$,

$$ \lim_{n \to \infty}P(|X_n - X|\geq \epsilon) = 0 $$

In this case, does it suffice to show that for all $K>0$

$$ \lim_{n \to \infty}P(X_n \geq -K) = 0 $$ ? Thanks!

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    Yes I think that's it. – 2017-02-16
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    Is there an intuitive explanation why? What would I replace $X$ with above to make it mesh with the normal definition? Thanks! – 2017-02-18
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    Well, if the $X_n$ were to take values in any general metric space $(E,d)$ then I believe you could define convergence in probability as $X_n \to^p X$ iff $P(d(X_n,X)>\epsilon) \to 0$ as $n \to \infty$. And I think that this notion will only depend on the topology, not on the metric $d$. In particular, if we define a metric on $[-\infty, +\infty]$ by $d(x,y)=|\tan^{-1}x-\tan^{-1}y|$ then it induces the usual topology on the subspace $\Bbb R$, and moreover, $\lim_{n\to \infty} P(X_n>-K) =0$ for all $K$ is equivalent to $\lim_n P(d(X_n,-\infty)>\epsilon)=0$ for all $\epsilon$. – 2017-02-18

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