Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a sequence of random variables $X_1,X_2,...$ defined on the same probability space $(\Omega,\mathcal{B},P)$.

part 1: Verify that $P(\limsup_{n\to\infty}X_n>x)=0$ if and only if $\lim_{n\to\infty}P(\bigcup_{m=n}^\infty\{X_m>x+\frac1k\})=0$ for all $k\geq 1$

part 2: Suppose $E(X_n)\leq x$ for all $n$ and $\sum_{n=1}^{\infty}\var(X_n)<\infty$.

Prove that $P(\limsup_{n\to\infty}X_n\leq x)=1$

I am guessing that these proofs require the use of Boole and/or Chebychev, but I have always struggled with these.

Help is appreciated!

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted
  1. The key is to see that $$\lim_{n\to +\infty}P\left(\bigcup_{j\geq n}\{X_j>x+k^{-1}\}\right)=P\left(\bigcap_{n\geq 1}\bigcup_{j\geq n}\{X_j>x+k^{-1}\}\right).$$

    Assume that $P(\limsup_{n\to +\infty}X_n>x)>0$. As the sequence $P(\limsup_{n\to +\infty}X_n>x+k^{—1})$ is increasing, we can find a $k$ such that $P(\limsup_{n\to +\infty}X_n>x+k^{—1})>0$. Let $A:=\{\omega,\limsup_{n\to +\infty}X_n(\omega)>x+k^{—1}\}$. If $\omega\in A$, then we can find $I(\omega)\subset \Bbb N$ infinite such that $X_i(\omega)>x+k^{—1}$ for all $i\in I(\omega)$ hence $\omega\in\bigcap_{n\geq 1}\bigcup_{j\geq n}\{X_j>x+k^{—1}\}$. This proves that the set $\bigcap_{n\geq 1}\bigcup_{j\geq n}\{X_j>x+k^{—1}\}$ has a positive probability.

    Conversely, assume that the latter set has a positive probability for some $k\geq 1$. For $\omega$ in this set we have a $\limsup_{j\to +\infty} X_j(\omega)\geq x+k^{—1}>x$.

  2. By the first question, we have to show that for each $k\geq 1$, we have $P\left(\bigcup_{j\geq n}\{X_j>x+k^{-1}\}\right)\to 0.$ We can write, as $EX_j\leq x$, $$\bigcup_{j\geq n}\{X_j>x+k^{-1}\}\subset \{(X_j-EX_j)^2\geq (x+k^{—1}-EX_j\})^2,$$ hence $$P\left(\bigcup_{j\geq n}\{X_j>x+k^{-1}\}\right)\leq \sum_{j\geq n}\frac 1{(x+k^{—1}-EX_j\})^2}\operatorname{Var}(X_j)=k^2\sum_{j\geq n}\operatorname{Var}(X_j).$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.