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Consider a sequence of real-valued random variables $\{X_n\}_{\forall n \in \mathbb{N}}$ and a real-valued random variable $X$

All r.v. are defined on the probability space $ (\Omega, \mathcal{F}, P)$

Could explain what is the relation (equivalent, one implies the other, etc) between

$$ (1) \hspace{1cm}X_n \rightarrow_{a.s.} X \text{ as $n\rightarrow \infty$} $$

and $$ (2) \hspace{1cm}X_n =X \text{ with probability approaching 1 as $n\rightarrow \infty$} $$


Some considerations:

Using the definitions,

(1) $P(\omega \in \Omega \text{ s.t. } \lim_{n\rightarrow \infty}X_n(\omega)=X(\omega))=1$

(2) $\lim_{n\rightarrow \infty} P(\omega \in \Omega \text{ s.t. } X_n(\omega)=X(\omega))=1$

So (1) has the limit inside, (2) has the limit outside.

(2) may look very similar to $X_n\rightarrow_pX$, where $X_n\rightarrow_pX$ means that $\forall \epsilon>0$ $\lim_{n\rightarrow \infty} P(\omega \in \Omega \text{ s.t. } X_n-X\leq \epsilon)=1$

What can we deduce from here?

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In fact, $(2)$ is stronger than convergence in probability. (Assume $(2)$ and take $\epsilon > 0$. Then $\{X_n = X\} \subseteq \{|X_n - X| < \epsilon\}$.)

Under the given conditions, there are no implications between these statements. Take $(\Omega, \mathcal{F}, P) = ([0,1], \mathcal{B}([0,1]), \lambda)$. i.e. the Borel sets on $[0,1]$ equipped with the Lebesgue measure.

  1. when both statements are true: simply take $X_n = c$, i.e. be a constant sequence of constant random variable.
  2. $(1)$ holds but $(2)$ does not: adapt the textbook example for . Take $X_n(\omega) = \omega^n$ and $X \equiv 0$.
    • $X_n \overset{a.s.}{\to} X$
    • $\forall n \in \Bbb{N}, P(X_n = X) = P(\{0\}) = 0$
  3. $(2)$ holds but $(1)$ does not: adapt the classic "sliding bump" functions for showing that convergence in measure doesn't implies convergence almost everywhere. Take $Y_{m,n} = 1_{\left[\frac{m-1}{n}, \frac mn \right]}$ for $m \in \{1,\dots,n\}$ and $n \in \Bbb{N}$. i.e. $Y_{m,n}$ is a "bump $\left[\frac{m-1}{n}, \frac mn \right]$ of width $\frac1n$ sliding from left to right as $m$ runs from $1$ to $n$". Enumerate $(Y_{m,n})$ as $\{Y_{1,1},Y_{2,1},Y_{2,2}, \dots\}$ and denote this sequence of random variables as $X_k$.
    • $X_k$ oscillates almost surely as every point on $[0,1]$ is "visited by infinitely many bumps". For all $\omega \in [0,1]$, choose $n \in \Bbb{N}$. As the "bumps" $\left[\frac{m-1}{n}, \frac mn \right]$ cover the whole space $[0,1]$, there exists an $m$ so that $Y_{m,n}(\omega) = 1$. Therefore, $Y_{m,n}(\omega) = 1$ infinitely often. Idem for $Y_{m,n}(\omega) = 0$, so $(1)$ doesn't hold.
    • $P(Y_{m,n} = 0) = \dfrac{n-1}{n} \xrightarrow[n\to\infty]{} 1$, so $(2)$ follows.

Remarks: $(2) \implies (1)$ is "partially true", in the sense that $(2)$ implies convergence in probability implies convergence of a subsequence almost surely.

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  • $\begingroup$ GNU Supporter, $(1)$ is read as which, if any? 1. $P(\lim[X_n=X])=1$ 2. $P([\lim X_n]=X)=1$ $\endgroup$
    – BCLC
    May 3 '18 at 12:32
  • $\begingroup$ @BCLC I don't understand what you mean by $[X_n = X]$. I read $(1)$ like OP does: "$P(\omega \in \Omega \mid \lim \limits_{n \to \infty} X_n(\omega)=X(\omega)) = 1$". $\endgroup$ May 3 '18 at 13:03
  • $\begingroup$ GNU Supporter, So you read $(1)$ as 2. K. got it. By $[X_n = X]$, I mean it like $[X_n \le X]$ and $[X_n \ge X]$ in the important inequalities $\endgroup$
    – BCLC
    May 3 '18 at 13:07
  • $\begingroup$ @BCLC I still don't understand this notation, which doesn't appear in the capture photo you provided. Does [] means the indicator function of an event? I do see some author using {}, but never seen one using [] $\endgroup$ May 3 '18 at 13:26
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    $\begingroup$ @BCLC No, as $P(\lim X_n =X) = 1$ is already clear. {} is usually used to denote events, so $\{\lim X_n\}$ will create confusion as $\limsup X_n$ is a random variable $\endgroup$ May 3 '18 at 13:50
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TL;DR Not a full answer but no inclusion for similar reasons to Is it correct to say that ($\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) \supseteq \limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}$?


I like to this of this in terms of Fatou's lemma and important inequalities:

Fatou's Lemma:

$$ \mathbb P\left(\liminf A_n\right) \leq \liminf\mathbb P\left(A_n\right)\leq \limsup\mathbb P\left(A_n\right) \leq \mathbb P\left(\limsup A_n\right). $$

Important inequalities:

Williams - Probability with Martingales


enter image description here


Deduced similarly:

(iii) If $\liminf x_n > z$, then

$ \ \ \ \ \ \ \ (x_n > z)$ eventually (that is, for infinitely many n)

(iv) If $\liminf x_n < z $, then

$ \ \ \ \ \ \ \ (x_n < z)$ infinitely often (that is, for infinitely many n)


$(2)$ is read as $\lim P(X_n = X) = 1$. By Fatou's Lemma, we have equality to 1 for the middle and RHS equality but there's no guarantee that the LHS equality equals 1 too. Even if LHS equals 1, $P(\lim[X_n=X])=1$ doesn't imply $P([\lim X_n]=X)=1$. Besides constructing counterexamples, you can also try to change equalities to inequalities with $\varepsilon$ and try, to no avail, to use the important inequalities to prove the implication.

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  • $\begingroup$ This is really (and rather desperately) spamming the site. $\endgroup$
    – Did
    May 4 '18 at 19:05

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