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I've been given the following problem:

For $n \in \mathbb{N}^{*}$, let $X_{n}$ be a random variable such that $\mathbb{P} \left[ X_{n} = \frac{1}{n} \right] = 1 - \frac{1}{n^{2}}$ and $\mathbb{P} \left[ X_{n} = n \right] = \frac{1}{n^{2}}$. Does $X_{n}$ converge in probability?

And the definition I've been given for convergence in probability of a random variable is:

Let $\left( T_{n} \right)_{n \geq 1}$ be a sequence of r.v. and $T$ a r.v. ($T$ may be deterministic). Then:

$T_{n}\overset{\mathbb{P}}{\underset{n \rightarrow \infty}\rightarrow} T$ if and only if $\mathbb{P} \left[ \left| T_{n} - T \right| \geq \epsilon \right] {\underset{n \rightarrow \infty}\rightarrow} 0$, for all $\epsilon > 0$.

But I don't understand how to apply this definition to the problem. I'm not even sure what I'm supposed to be seeing if $X_{n}$ is converging to. We have no $X$ as far as I can see. Can anyone give me some insight on this?

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3 Answers 3

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Fix any number $\epsilon>0$. Then, by density of rationals, $\exists n\in \mathbb{Z}^+$, such that $\epsilon>1/n$. Then, $\forall m\ge n$, $$P(X_m>\epsilon)\le P(X_m> 1/m )=\frac{1}{m^2}\stackrel{m\to \infty}{\to} 0$$ Hence $X_m\stackrel{p}{\to}0 $.

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$X_n$ converges to $0$ in probability as $n\to\infty$. To see that, use Markov's inequality, i.e. $$ P(X_n\ge\varepsilon)\le\frac{\operatorname EX_n}{\varepsilon}. $$

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  • $\begingroup$ I understand how the math works out for that, but how does that fit into the definition I've been given for convergence in probability? Is it that the T in the definition can just be anything and we have to figure out what it is? $\endgroup$ Sep 16, 2016 at 7:36
  • $\begingroup$ Good observation (+1). $\endgroup$
    – Surb
    Sep 16, 2016 at 7:39
  • $\begingroup$ @bones_mccoy $T$ in the definition can be any random variable. In this particular case, $T$ is equal to $0$ almost surely. So we want to show that the probability $P(|X_n-0|\ge\varepsilon)=P(X_n\ge\varepsilon)$ goes to $0$ as $n\to\infty$ for each $\varepsilon>0$. We have to figure out what the limit is. But we might suspect that the limit is $0$ because $X_n=n^{-1}$ with probability that gets to $1$ and $X_n=n$ with probability that gets to $0$. Of course, that's only a guess and we need to prove that. We use Markov's inequality to make this argument rigorous. I hope this helps. $\endgroup$
    – Cm7F7Bb
    Sep 16, 2016 at 7:48
  • $\begingroup$ If I was also asked whether $X_{n}$ converged in $L^{2}$ would I still see whether or not it converges to 0 or is it possible for $X_{n}$ to converge to 0 in probability and something else in $L^{2}$? $\endgroup$ Sep 16, 2016 at 10:58
  • $\begingroup$ @bones_mccoy Convergence in $L^2$ implies convergence in probability. So if a sequence converges to some limit in $L^2$, it must also converge to the same limit in probability. $\endgroup$
    – Cm7F7Bb
    Sep 16, 2016 at 11:24
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Let $m\in\mathbb N^*$. As you can see, $$\mathbb P\left\{X_n> \frac{1}{m}\right\}=0,$$ when $n>m$. Therefore, if $n\to \infty $, you get $$\lim_{n\to \infty }\mathbb P\left\{X_n>\frac{1}{m}\right\}=0,$$ and thus, $$\forall m\in\mathbb N^*,\lim_{n\to \infty }\mathbb P\left\{X_n>\frac{1}{m}\right\}=0.$$

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