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I am studying the various types of convergence for random variables, in particular how convergence in probability implies convergence in distribution.

Let $(X_n)_n$ be a sequence of random variables and let us assume that $X_n$ converges in probability to $X$. After some computations, we can write: $$F_X(t-\epsilon) - P(|X_n - X|> \epsilon) \le F_{X_n}(t) \le F_X(t+\epsilon) + P(|X_n - X|> \epsilon) $$ where $F_X(t)$ is the CDF of $X$ and $\epsilon > 0$.

At this point I don't understand why several authors (e.g. example of proof) use $\lim \limits_{n \to \infty} \mbox{sup}$ and $\lim \limits_{n \to \infty}\mbox{inf}$ to show that the limit of $F_{X_n}(t)$ is bounded by $F_X(t-\epsilon)$ and $F_X(t+\epsilon)$ for $n \to \infty$; in particular they write: $$F_X(t-\epsilon) \le \lim \limits_{n \to \infty} \mbox{inf } F_{X_n}(t) \le \lim \limits_{n \to \infty} \mbox{sup } F_{X_n}(t) \le F_X(t+\epsilon)$$ and then they conclude the proof by resorting to the continuity of $F_X(t)$.

My question is, why is it not sufficient to just compute the limits of the elements of the inequality above obtaining $F_X(t-\epsilon) \le \lim \limits_{n \to \infty} F_{X_n}(t)\le F_X(t+\epsilon)$ ? In particular we can define: $$a_n := F_X(t-\epsilon) - P(|X_n - X|> \epsilon)$$ $$b_n :=F_X(t+\epsilon)+ P(|X_n - X|> \epsilon)$$ for all $n$, then knowing that $a_n \le F_{X_n}(t) \le b_n$, from one of the comparison theorems for sequences and the converge in probability of $X_n$, we obtain: $$\lim \limits_{n \to \infty} a_n = F_X(t-\epsilon) \le \lim \limits_{n \to \infty} F_{X_n}(t)\le \lim \limits_{n \to \infty} b_n = F_X(t+\epsilon)$$

Am I missing anything?

Thanks a lot.

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This because $\lim_{n}F_{X_{n}}(t)$ does not have to exist for all $t$.

Example. Let $X_{n} \sim U(0,1/n)$. Then, $X_{n}$ converges in probability to a random variable $X$ that is almost surely zero. To see this, let $\epsilon>0$ and note that for all $n\geq1/\epsilon$ $$ \mathbb{P}\left(\left|X_{n}-X\right|>\epsilon\right) =\mathbb{P}\left(X_{n}>\epsilon\right)=0. $$ However, $$ F_{X_{n}}(t)=0\cdot\boldsymbol{1}_{(-\infty,0)}(t)+tn\cdot\boldsymbol{1}_{[0,1/n)}(t)+1\cdot\boldsymbol{1}_{[1/n,\infty)}(t) $$ and hence $\lim_n F_{X_n}(0)$ does not exist.


Addendum

Definition. Let $(X_{n})_{n}$ be a sequence of real-valued random variables. $X_{n}$ converges in distribution to some real-valued random variable $X$ if $F_{X_{n}}(t)\rightarrow F_{X}(t)$ for all points $t$ at which $F_{X}$ is continuous.

In the proof you provided, you established the inequality $$ F_{X}(t-\epsilon)\leq\liminf_{n}F_{X_{n}}(t)\leq\limsup_{n}F_{X_{n}}(t)\leq F_{X}(t+\epsilon) $$ for all points $t$. Now, fix your attention to a point $t$ at which $F_{X}$ is continuous. In this case, you can take the limit as $\epsilon\rightarrow0$ in the above chain of inequalities to get $$ F_{X}(t)\leq\liminf_{n}F_{X_{n}}(t)\leq\limsup_{n}F_{X_{n}}(t)\leq F_{X}(t). $$ Since the leftmost and rightmost sides of the inequality are the same, it follows that $$ \liminf_{n}F_{X_{n}}(t)=\limsup_{n}F_{X_{n}}(t), $$ from which you can conclude that the limit $\lim_{n}F_{X_{n}}(t)$ exists.

Note that the above argument only works because we used a point $t$ at which $F_X$ is continuous!

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    $\begingroup$ Hello and thanks for your answer :) I get your point but then I am misunderstanding the definition of convergence in distribution. How should we read it, 1) if $\lim \limits_{n \to \infty} F_{X_n}(t)$ exists then the equality $\lim \limits_{n \to \infty} F_{X_n}(t)=F_X(t)$ must hold or 2) $\lim \limits_{n \to \infty} F_{X_n}(t)$ must exist and $\lim \limits_{n \to \infty} F_{X_n}(t)=F_X(t)$ must hold? Until now I have been thinking we are in case 1, so for me the existence of such limit was an assumption. But if we are in case 2, then we have to prove its existence. $\endgroup$ – zeroKnowl Feb 24 at 9:53
  • $\begingroup$ @zeroKnowl: It's actually neither (1) nor (2). The limit has to hold at points of continuity of $F_X$. I added the definition and some further elaboration to my answer. $\endgroup$ – parsiad Feb 24 at 20:44
  • $\begingroup$ Thanks a lot for the time you spent on this answer, everything is clear now :) $\endgroup$ – zeroKnowl Feb 25 at 19:31

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