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Consider the following two statements about a random sequence $X_n$:

(1) $X_n \stackrel{a.e.}{\rightarrow} X$.

(2) $\mathrm{P}\{|X_n-X|>\epsilon, \ i.o.\} = 0, \ \forall \epsilon>0$.

(a.e. stands for "almost everywhere" and i.o. stands for "infinitely often".)

What is the difference between statements (1) and (2), or are they equivalent? Does one imply the other?

I appreciate any help!

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    $\begingroup$ In (2) you shouldn't write a limit since the event $\{ \vert X_n-X\vert>\varepsilon,\, i.o\}$ does not depend on $n$. $\endgroup$
    – Etienne
    Jul 18, 2013 at 9:17
  • $\begingroup$ You're right Etienne. I edited (2) accordingly. $\endgroup$
    – user86780
    Jul 18, 2013 at 11:02

1 Answer 1

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Ok, we first show that an event in $(2)$ is very similar to the convergence, so that $(2)$ follows from $(1)$. Moreover, you can express a convergence as a monotonic intersection of such events as in $(2)$. That would imply that $(1)$ follows from $(2)$. Now some technical stuff:

Note that $\{|X_n - X|>\varepsilon \text{ i.o.}\} = \{|X_n - X|\leq\varepsilon \text{ e.a.}\}$ where $\mathrm{e.a.}$ stands for eventually always: $$ \{\omega:\exists N(\omega,\varepsilon) \text{ such that }|X_n(\omega) - X(\omega)|\leq \varepsilon \text{ for all }n\geq N(\omega,\varepsilon)\}. $$ Now let $A_m:=\{\omega:|X_n - X|\leq\frac1m \text{ e.a.}\}$ so that $P(A_m) = 1$ and $A_{m+1}\subseteq A_m$. As a result, since $$ \bigcap_mA_m = \{\omega:|X_n - X|\leq \frac1m \text{ e.a. for any }m \} = \{\omega:X_n \to X\} $$ we obtain that $(2)$ implies $$ P\{\omega:X_n \to X\} = P\left(\bigcap_mA_m\right) = \lim_m P(A_m) = 1 $$ so that $(2)\implies (1)$. The reverse direction is much easier to show.

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  • $\begingroup$ Thanks Ilya! My intuition was also saying that (1) and (2) should be equivalent. $\endgroup$
    – user86780
    Jul 18, 2013 at 11:00

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