2

This article says https://en.wikipedia.org/wiki/Set-theoretic_limit#Almost_sure_convergence.

The event that a sequence of random variables $Y_1, Y_2, \dots$ converges to another random variable $Y$ is formally expressed as $\{{\limsup _{n\to \infty }|Y_{n}-Y|=0\}}$. It would be a mistake, however, to write this simply as a limsup of events. That is, this is not the event $ \limsup _{n\to \infty }\{|Y_{n}-Y|=0\}$ !

I was wondering if the second expression is simply incorrect notation because it did not have a pair of parentheses around the whole expression. Both expressions mean the same thing to me.

user10354138
  • 33,239
johnson
  • 482

2 Answers2

3

Remember the event $$ \{\limsup_{n\to\infty}\lvert Y_n-Y\rvert=0\} $$ is $$ \{\omega\in\Omega : \limsup_{n\to\infty}\lvert Y_n(\omega)-Y(\omega)\rvert=0\}. $$ But the limsup of events $$ \limsup_{n\to\infty}\{\lvert Y_n-Y\rvert=0\} $$ is $$ \{\omega\in\Omega : \lvert Y_n(\omega)-Y(\omega)\rvert=0 \text{ infinitely often}\} $$ which is not the same. For example, for some $\omega\in\Omega$ we could have $Y_n(\omega)-Y(\omega)=n^{-1}$ and this $\omega$ would be in the first, but clearly $\lvert Y_n(\omega)-Y(\omega)\rvert$ is never zero so is not in the second.

user10354138
  • 33,239
2

The first expression concerns a $\limsup$ of functions.

The second expression concerns a $\limsup$ of sets (which again is a set).

$$\omega\in\{\limsup_{n\to\infty}|Y_n-Y|=0\}\iff\limsup_{n\to\infty}|Y_n(\omega)-Y(\omega)|=0\iff\lim_{n\to\infty}Y_n(\omega)=Y(\omega)\tag1$$ and:$$\omega\in\limsup\{|Y_n-Y|=0\}\iff\{n\in\mathbb N|Y_n(\omega)-Y(\omega)|=0\}\text{ is infinite}\tag2$$

Note that $(1)$ does not imply $(2)$ and also $(2)$ does not imply $(1)$.

If e.g. $Y_n(\omega)=\frac1n$ and $Y(\omega)=0$ then $(1)$ is true and $(2)$ is not true.

If e.g. $Y_n(\omega)=0=Y(\omega)$ for $n$ odd, and $Y_n(\omega)=1$ for $n$ even then $(2)$ is true and $(1)$ is not true.

drhab
  • 151,093
  • 1
    @johnson Yes ${\limsup_{n\to\infty}|Y_n-Y|=0}$ is a set all right. I am not denying that but only state that in this expression we deal with a limsup of functions (between the brackets {}). – drhab Jun 23 '19 at 09:09
  • I thought that the first expression is also a lim inf of sets. What about this expression from https://en.wikipedia.org/wiki/Convergence_of_random_variables $\operatorname {Pr} {\Big (}\liminf {n\to \infty }{\big {}\omega \in \Omega :|X{n}(\omega )-X(\omega )|<\varepsilon {\big }}{\Big )}=1\quad {\text{for all}}\quad \varepsilon >0. $ Is this a lim inf of functions or a lim inf of sets. – johnson Jun 23 '19 at 09:10
  • 1
    Between the brackets () we have a $\liminf$ on sets here. – drhab Jun 23 '19 at 09:15
  • But isn't the complement of that equal to this $\lim_{n\to\infty}\mathbb{P}[\omega:\sup_{k>n}|X_k(\omega) - X(\omega)|>\epsilon] = 0$ for some $\epsilon>0$ which is equal to the first expression. So a lim inf on sets becomes a lim sup on functions? – johnson Jun 23 '19 at 09:38
  • The complement of set $\liminf\left{ \left|X_{n}-X\right|<\epsilon\right} $ is set $\limsup\left{ \left|X_{n}-X\right|\geq\epsilon\right} $ so that $$P\left(\liminf\left{ \left|X_{n}-X\right|<\epsilon\right} \right)=1\iff P\left(\limsup\left{ \left|X_{n}-X\right|\geq\epsilon\right} \right)=0$$ – drhab Jun 23 '19 at 09:57
  • Here $\limsup\left{ \left|X_{n}-X\right|\geq\epsilon\right} =\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}\left{ \left|X_{k}-X\right|\geq\epsilon\right} $ . Setting this as $\bigcap_{n=1}^{\infty}B_{n}$ we have $B_{1}\supseteq B_{2}\supseteq B_{3}\supseteq\cdots$ so that we may conclude that $P\left(\limsup\left{ \left|X_{n}-X\right|\geq\epsilon\right} \right)=\lim_{n\to\infty}P\left(\bigcup_{k=n}^{\infty}\left{ \left|X_{k}-X\right|\geq\epsilon\right} \right)=0$. – drhab Jun 23 '19 at 09:58
  • but lim sup ${|X_n-X|\ge \epsilon}$ is a lim sup on sets right? – johnson Jun 23 '19 at 10:03
  • I phrased my question a bit more clearly here. https://math.stackexchange.com/questions/3271494/equivalence-of-a-lim-sup-of-functions-with-a-lim-sup-of-sets – johnson Jun 23 '19 at 10:18