18

I'm trying to understand this proof (also in the image below) that proves if $X_{n}$ converges to some constant $c$ in distribution, then this implies it converges in probability too.

Specifically, my questions about the proof are:

  1. How are they getting $\lim_{n \to \infty} F_{X_{n}}(c+\frac{\epsilon}{2}) = 1$?

  2. Why do they state the conclusion at the end in this way? They're basically saying that knowing $\lim_{n \to \infty}P(|X_{n} - c| \geq \epsilon) \geq 0$ allow you to conclude that $\lim_{n \to \infty}P(|X_{n} - c| \geq \epsilon) = 0$ but the real reason we can conclude this is because of the whole body of the proof above, right?

enter image description here

User1865345
  • 679
  • 1
  • 5
  • 18
A user
  • 1,036

2 Answers2

15

If a sequence of random variables $X_n$ converges to $X$ in distribution, then the distribution functions $F_{X_n}(x)$ converge to $F_X(x)$ at all points of continuity of $F_X$.

In this case $X=c$, so $F_X(x)=0$ if $x<c$ and $F_X(x)=1$ if $x\geq c$. $F_X$ is continuous everywhere except at $x=c$, hence $$ \lim_{n\to\infty}F_{X_n}\Big(c+\frac{\varepsilon}{2}\Big)=F_X\Big(c+\frac{\varepsilon}{2}\Big)=1 $$ as claimed.

For the second part, the argument has shown that the limit $\leq 0$, and the point the book is making (somewhat clumsily) is that the limit is of course non-negative, so these two facts imply that the limit is zero.

carmichael561
  • 54,793
  • why do they divide by 2 instead of just saying $F_{X_{n}}(c+\epsilon)$ – A user Mar 28 '16 at 14:38
  • I meant to say: why do they divide by 2 instead of just saying $F_{X_{n}}(c+\epsilon) = 1$? Isn't this an equivalent statement, and then there wouldn't be the need to do the last few steps? – A user Mar 28 '16 at 14:51
  • 2
    The issue is $\mathbb{P}(X_n\geq c+\varepsilon)=1-\mathbb{P}(X_n<c+\varepsilon)$, which is not necessarily equal to $1-F_{X_n}(c+\varepsilon)$. – carmichael561 Mar 28 '16 at 15:23
  • Hmm, why is it not necessarily equal? Because we have $1 - P(X_{n} < c + \epsilon)$ instead of $1 - P(X_{n} \leq c + \epsilon)$? In general, why are we dividing $\epsilon$ by 2? – A user Mar 28 '16 at 16:11
  • Yes, the = sign is the important part. After all, $\mathbb{P}(X_n=c+\varepsilon)$ could be non-zero. Dividing by 2 is just a convenient way to choose a slightly smaller point. – carmichael561 Mar 28 '16 at 18:54
  • The clumsiness can be fixed if we cite the Squeeze Theorem. – G.O.F. Oct 28 '22 at 05:09
0

$\lim_{n\to \infty}P(|X_n−c|\ge \epsilon )\ge 0$ means that probability cannot be negative. And since $\lim_{n\to \infty} P(|X_n−c|\ge \epsilon )\le 0$ thus, $\lim_{n\to \infty} P(|X_n−c|\ge \epsilon )=0$.

Siong Thye Goh
  • 153,832
Valeron
  • 88
  • hmm probability is always nonnegative right? – Siong Thye Goh Sep 20 '21 at 17:27
  • Yes, probability is always nonnegative. Maybe in Physics or other disciplines. But I've never seen a negative probability. – Valeron Sep 20 '21 at 17:38
  • so the deduction form the first line is confusing to me. We can just state directly that the probability is nonnegative right. Also, where does the second line comes from? – Siong Thye Goh Sep 20 '21 at 17:49
  • 1
    The first line comes from “axioms of probability" that is the probability of any event must be nonnegative. The second line comes from the proof shared by the user that is limn→∞P(|Xn−c|≥ϵ) ≤ limn→∞P(|Xn>c+ϵ/2) = 0. – Valeron Sep 20 '21 at 17:59