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I have a question on the meaning of this claim:

Consider a sequence of real valued random variables $\{X_n\}_{n\in \mathbb{N}}$ and some constants $a,b$. Then we have

$$ \sqrt{n}a\leq X_{n}\leq \sqrt{n}b \text{ }\text{ with probability approaching $1$ as $n\rightarrow \infty$} $$

Specifically, I am confused on the meaning of "with probability approaching $1$ as $n\rightarrow \infty$" and an help to clarify it would be extremely helpful.

I know that writing $X_n$ converges to $X$ with probability approaching $1$ as $n\rightarrow \infty$ is equivalent to $X_n\rightarrow_{a.s.}X$, i.e., $$ \mathbb{P}(\{\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow \infty} X_n(\omega)=X(\omega)\})=1 $$ but the claim above contains inequalities. Do they mean $$ \mathbb{P}(\{\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow \infty} [X_n(\omega)-\sqrt{n}b]\leq 0\})=1 $$ and $$ \mathbb{P}(\{\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow \infty} [X_n(\omega)-\sqrt{n}a]\geq 0\})=1 $$ ?

BCLC
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Star
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2 Answers2

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It means that $$ \lim _{n\to\infty} P(\{\omega\mid \sqrt{n}a\leq X_n(w)\leq \sqrt{n} b\})=1 $$

Sri-Amirthan Theivendran
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  • Thank you. Then I guess that also my point about equivalence between "$X_n$ converges to $X$ with probability approaching $1$ as $n \rightarrow \infty$" and almost sure convergence was wrong? – Star May 01 '18 at 16:48
  • @CGT See my answer – BCLC May 01 '18 at 16:58
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No. It's this one: $$ \lim _{n\to\infty} P(C_n)=1 $$

where

$$C_n := \{ \sqrt{n}a\leq X_n(w)\leq \sqrt{n} b \}$$

$$= \{ \sqrt{n}a\leq X_n(w)\} \cap \{X_n(w)\leq \sqrt{n} b \}$$

$$:= A_n \cap B_n$$

Thus, $$C_n \subseteq A_n, C_n \subseteq B_n$$

$$\to P(C_n) \le P(A_n), P(C_n) \le P(B_n)$$

$$\to \lim_n P(C_n) \le \lim_n P(A_n), \lim_n P(C_n) \le \lim_n P(B_n)$$

Thus when you say

$$ \mathbb{P}(\{\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow \infty} [X_n(\omega)-\sqrt{n}b]\leq 0\})=1 $$ and $$ \mathbb{P}(\{\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow \infty} [X_n(\omega)-\sqrt{n}a]\geq 0\})=1 $$

, you are saying something weaker namely merely that $\lim_n P(A_n) = 1 = \lim_n P(B_n)$. Here, it could be that $\lim_n P(C_n) < 1$.

Intuitively, I guess that there isn't a common way that $n$ approaches $\infty$. Do you know/Have you heard of Cauchy principal value ?

Anyhoo, a similar problem arises in comparing

$$P(\bigcap_{n \in T} A_n) = 1$$

$$P(A_n) = 1 \ \forall n \in T$$

Do you know modification vs indistinguishable?

BCLC
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  • There is another difference with what I wrote: in my case the limit is inside the probability, in your case it is outside. Could you comment on that too? – Star May 01 '18 at 17:04
  • In my case I mimic almost sure convergence (limit inside the probability). In your case you have a sequence of probabilities. – Star May 01 '18 at 17:06
  • @CGT Oh, right. Then in that case yours is worse actually, no offense: It is more wrong to say $$P(\lim_n [X_n - \sqrt{nb}] \le 0) = 1$$ than it is to say $$P(\lim_n [X_n - \sqrt{nb} \le 0]) = 1$$. At least the latter has an inclusion relationship with $$\lim_nP( [X_n - \sqrt{nb} \le 0]) = 1$$. The former: I'm not sure the limit even exists. – BCLC May 01 '18 at 17:30
  • Thanks but your first two probabilities are the same. Could you explain better? – Star May 01 '18 at 17:33
  • @CGT Which two probabilities? $$P(\lim_n [X_n - \sqrt{nb}\color{red}{]} \le 0) \ne P(\lim_n [X_n - \sqrt{nb} \le 0 \color{red}{]})$$ $$'\lim_n [X_n - \sqrt{nb}\color{red}{]} \le 0'$$ is the event that the random variable $$'\lim_n [X_n - \sqrt{nb}\color{red}{]}'$$ is less than or equal to zero. $$'\lim_n [X_n - \sqrt{nb} \le 0 \color{red}{]}'$$ is the limit of the events $[X_n - \sqrt{nb} \le 0 \color{red}{]}$ – BCLC May 01 '18 at 17:35