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Let $X$, $X_1,X_2,...$ be random variables defined on $(\Omega,\mathcal{F},\mathbb{P})$, why is the following bi-conditional statement for convergence in probability true? $$\lim\limits_{n\rightarrow\infty}\mathbb{E}(\min(|X_n-X|,1))=0\Leftrightarrow\forall\epsilon>0,\lim\limits_{n\rightarrow\infty}\mathbb{P}(|X_n-X|\geq\epsilon)=0$$

I tried splitting the expectation into regions $|X_n-X|>1$ and $|X_n-X|\leq1$ and tried to use Markov's inequality but seems to get nowhere. I am thinking this must be quite trivial but I keep getting nowhere with it, so please any help will really be appreciated and greatly needed, thanks in advance.

user152874
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1 Answers1

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"$\Leftarrow$" Use that $$\mathbb{E}\min(|X_n-X|,1) \leq \int_{|X_n-X| < \epsilon} |X_n-X| \, d\mathbb{P} + \int_{|X_n-X| \geq \epsilon} 1 \, d\mathbb{P}.$$

"$\Rightarrow$": Show that $$\mathbb{P}(|X_n-X| \geq \epsilon) = \mathbb{P}(\min(|X_n-X|,1) \geq \epsilon)$$ for all $\epsilon \in (0,1)$ and apply Markov's inequality.

saz
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  • Thanks, but I am still not sure as for the "$\Leftarrow$", I had already tired the hint you gave and it yields $\mathbb{E}\min(|X_n-X|,1) \leq \epsilon\mathbb{P}(|X_n-X|<\epsilon)+\mathbb{P}(|X_n-X|\geq\epsilon)\xrightarrow[\infty]{n}\epsilon(1) +0$, and my confusion is whether the $\epsilon$ is fixed or arbitrary as we used it in splitting the area of inetgration, or if $\epsilon$ is arbitrary then we have a non-negative number less than any positive number so it must be $0$, is that correct? Thanks again. – user152874 Mar 14 '17 at 10:22
  • @user152874 Yes, $\epsilon>0$ is arbitrary. If a non-negative number $x$ satisfies $x \leq \epsilon$ for any $\epsilon>0$, then obviously $x=0$. Applying this for $x:= \limsup_{n \to \infty} \mathbb{E}\min(|X_n-X|,1)$ gives the desired resilt. – saz Mar 14 '17 at 10:28
  • Why do you need to take the $\limsup$ and not just the $\lim$? – user152874 Mar 14 '17 at 12:34
  • @user152874 At this point of the proof, we don't know whether the limit exists. The estimate $\mathbb{E}\min(|X_n-X|,1) \leq \epsilon$ implies $$0 \leq \limsup_{n \to \infty} \mathbb{E}\min(|X_n-X|,1) \leq \epsilon$$ and therefore $$\limsup_n \mathbb{E}\min(|X_n-X|,1)=0.$$ Thus, $$0 \leq \liminf_n \mathbb{E}\min(|X_n-X|,1) \leq \limsup_n \mathbb{E}\min(|X_n-X|,1) = 0.$$ This implies that $$\liminf_n \dots = \limsup_n \dots = 0.$$ Thus, we conclude that $\lim \dots$ exists and equals $0$. – saz Mar 14 '17 at 12:53
  • (Recall that for any sequence $(a_n)$, the limit $\lim_n a_n$ exists if and only if $\liminf_n a_n = \limsup_n a_n$.) – saz Mar 14 '17 at 12:54
  • Thank you again for the "remedial" lesson. I appreciate and thank you for your patience and time in answering me. – user152874 Mar 15 '17 at 15:21
  • @user152874 You are welcome. If you find the answer helpful, you can accept/upvote it by clicking on th tick/up arrow next to it. – saz Mar 15 '17 at 16:00