2
$\begingroup$
  1. Let $X$ be a random variable with the property that $P(X\leq t)$ is either $0$ or $1$. Prove that $X$ is a.s. constant.

  2. Suppose $\{X_n\}$ is an i.i.d. of random variables and $\{y_n\}$ is a sequence of positive numbers that tends to $+\infty$ as $n\rightarrow \infty$. The sequence $\displaystyle Y_n=\frac{X_1+X_2+...+X_n}{y_n}$ converges a.s. Prove it converges to a constant a.s.

For 1, what I need is there is a constant $C$ such that $P(X=C)=1$. Intutively it seems correct, but I have hard time proving it rigourously. Any hint/help is appreciated.

For 2. what I have is there is some function $Y$ such that $P(\lim_{n\rightarrow \infty} Y_n(\omega)=Y(\omega))=1$, and what I need to prove is $Y(\omega)$ is independent of $\omega$. But stuck on that. Any help is appreciated.

$\endgroup$
8
  • 2
    $\begingroup$ $t\mapsto P(X\le t)$ is increasing. So there must be some $t_0$ where it jumps from $0$ to $1$, right? The function is also right-continuous... $\endgroup$
    – amsmath
    Oct 11, 2017 at 15:28
  • $\begingroup$ amsmath: How do we prove it is right continuous? $\endgroup$
    – Extremal
    Oct 11, 2017 at 15:33
  • 1
    $\begingroup$ Call that function $f$. We need the right-continuity for having $f(t_0) = 1$, where $t_0$ is the point where it jumps. The right-continuity follows from the continuity of measure: math.stackexchange.com/questions/234292/… $\endgroup$
    – amsmath
    Oct 11, 2017 at 15:34
  • 1
    $\begingroup$ Exactly. Now,$$\{X\neq t_0\} = \bigcup_n\{X > t_0+\frac 1 n\}\,\cup\,\bigcup_n\{X\le t_0-\frac 1 n\}.$$Show that all these sets have measure zero and use continuity of measure from below. $\endgroup$
    – amsmath
    Oct 11, 2017 at 15:46
  • 2
    $\begingroup$ It looks like this problem is setting you up to use the Kolmogorov 0-1 law. $\endgroup$
    – Michael
    Oct 11, 2017 at 16:11

2 Answers 2

1
$\begingroup$

Let $S_n=X_1+\cdots +X_n$. If $EX_1=0$, $$ \dfrac{S_n}{y_n}\to 0 $$ a.s. (see Chung's A Course in Probability Theory, p.132, Corollary). Assume that $\mu =EX_1\neq 0$ and let $Y=\lim _n(S_n/y_n)$. By the SLLN, $$ \mu ^{-1}Y=\lim _n\left (\dfrac{S_n}{n}\right )^{-1}\left (\dfrac{S_n}{y_n}\right )=\lim _n\dfrac{n}{y_n}. $$

$\endgroup$
0
$\begingroup$

You may use the Kolmogorov's 0-1 law. Let $Y_n\xrightarrow{a.s.} Y$ and $Z=\limsup_{n\to\infty}{Y_n}$. Then since $y_n\to\infty$, for any $t\in \mathbb{R}$, the event $\{Z\le t\}$ is a tail event. Now $$ \mathsf{P}(Y\le t)=\mathsf{P}(Y\le t,Z=Y)+\mathsf{P}(Y\le t,Z\ne Y)=\mathsf{P}(Z\le t) $$ and $\mathsf{P}(Z\le t)\in\{0,1\}$.

$\endgroup$
1
  • $\begingroup$ I'm unfamiliar with lot of terms here. What does "exchangeable" means and what is Hewitt-Savage 0-1 law? $\endgroup$
    – Extremal
    Oct 11, 2017 at 16:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .