# Prove that $X$ is a.s. constant and $Y_n$ converges to a constant a.s.

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.

• $t\mapsto P(X\le t)$ is increasing. So there must be some $t_0$ where it jumps from $0$ to $1$, right? The function is also right-continuous... Oct 11, 2017 at 15:28
• amsmath: How do we prove it is right continuous? Oct 11, 2017 at 15:33
• 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/… Oct 11, 2017 at 15:34
• 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. Oct 11, 2017 at 15:46
• It looks like this problem is setting you up to use the Kolmogorov 0-1 law. Oct 11, 2017 at 16:11

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, $$\mu ^{-1}Y=\lim _n\left (\dfrac{S_n}{n}\right )^{-1}\left (\dfrac{S_n}{y_n}\right )=\lim _n\dfrac{n}{y_n}.$$
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 $$\mathsf{P}(Z\le t)\in\{0,1\}$$.