3
$\begingroup$

The problem (not homework) is

Consider the probability space $([0,1], B_{[0,1]}, P)$ where $B_{[0,1]}$ is the Borel set and $P$ is Lebesgue measure on $[0,1]$. For any integer $n>0$, there exist $m$ and $k$ such that $n = 2^m-2+k$ and $0\leq k \leq 2^{m+1.}$ Define

$$X_n(\omega)=\left\{\begin{array}{clcr} 1, & \mbox{if } \frac{k}{2^m}\leq \omega\leq\frac{k+1}{2^m}\\0, & \mbox{if } \mbox{ otherwise} \end{array}\right.$$ for any $\omega\in[0,1]$.

Question: is the statement ``$X_n \to 0$ a.s.'' correct?

I know the solution can be found in p.36, Exercise 47, Mathematical Statistics: Exercises and Solutions by Jun Shao. That is, $X_n(\omega)$ does not converge to $0$ a.s. The reason is the following.

For any fixed $\omega \in[0,1]$ and m, there exists $k$ with $1\leq k\leq 2^m$ such $\frac{k-1}{2^m}\leq \omega \leq \frac{k}{2^m}$. Let $n_m =2^m-2+k$. Then $X_{n_m}(\omega)=1$. Since m is arbitrary selected we can find an infinite sequence $\{n_m\}$ such that $X_{n_m}(\omega)=1$. This implies $X_n(\omega)$ does not converge to 0. Since $\omega$ is arbitrary, $X_n$ does not converge to 0 a.s.

However, I got a different answer. My answer is :

Since $P(|X_n| > \epsilon) \leq \frac{1}{2^m}$ for any $\epsilon >0$, $\sum_n P(|X_n| > \epsilon) < \infty$. By the first Borel-Cantelli lemma, $P(\{|X_n| > \epsilon\} \mbox{ i.o.}) = 0$. Since $\epsilon$ is arbitrary, $X_n \to 0$ a.s.

If my answer is wrong, I'm curious why the 1st Borel Cantelli lemma does not work or where I have done something wrong.

$\endgroup$
5
  • $\begingroup$ What is the reasoning behid your last sentence "Since $\epsilon$ is arbitrary, $X_n \to 0$ a.s"? $\endgroup$
    – user10525
    Sep 12, 2012 at 10:13
  • $\begingroup$ @Procrastinator, Since $P(\{|X_n| > \epsilon\} i.0) = 0$ implies $P(\{|X_n| \leq \epsilon\} i.0) = 1$ for any $\epsilon >0$, I conclude my last sentence. $\endgroup$
    – Jim
    Sep 12, 2012 at 10:23
  • 2
    $\begingroup$ Note that $2^m=n+2-k$, then the convergence of the series is not ensured because the summation is over $n$. Otherwise, you would be correct, see. $\endgroup$
    – user10525
    Sep 12, 2012 at 11:04
  • $\begingroup$ You left out the important but obvious fact that the limit is taken as n goes to infinity. $\endgroup$ Sep 12, 2012 at 11:04
  • 3
    $\begingroup$ One of the proofs that the harmonic series diverges is that it is at least $1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 + ...$ with $2^{n-1}$ copies of $1/2^n$. This series of reciprocals of powers of $2$ diverges in a similar way. $\endgroup$ Sep 12, 2012 at 13:59

1 Answer 1

2
$\begingroup$

Shao is correct. As Procrastinator points out m has a special form. So you have only shown convergence to 0 along a particular subsequence and not every subsequence.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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