I came across a problem:

Let $X_1,X_2,...$ be independent random variables with uniform(0,1) distribution from probability space $(\Omega,\cal F,P)$. Prove that:


where the intersection is taking over all increasing sequences $(n_1,n_2,...)\in\mathbb N^\mathbb N$.

I have managed to show that $P(\omega\in\Omega: \{X_1(\omega),X_2(\omega),X_3(\omega)...\} \text{is dense in (0,1)})=1$ which was said to be related to solving the problem above. I don't see how the result I proved related to the problem and have no lead to start..I can also see it's kinda related to SLLN? But not sure...

  • $\begingroup$ You have $\lim_{n\rightarrow\infty}$ but then use $k$ rather than $n$. Are they the same thing? $\endgroup$ – Henry Oct 7 '14 at 6:54

Consider the subset of $\Omega$ defined as $$A=\bigcap_{(n_k)}\left\{\omega\in\Omega\,;\,\lim_k\frac1k(X_{n_1}(\omega)+\cdots+X_{n_k}(\omega))=\frac12\right\}.$$ Almost surely with respect to $\omega$, $X_n(\omega)\leqslant.1$ for infinitely many $n$, say every $n$ in the increasing sequence $(n_k)$. This is not enough to show that $\frac1k(X_{n_1}(\omega)+\cdots+X_{n_k}(\omega))$ converges but enough to be sure that it does not converge to $1/2$ since $\frac1k(X_{n_1}(\omega)+\cdots+X_{n_k}(\omega))\leqslant.1$ for every $k$. Thus, $\omega$ is not in $A$.

This proves that $A\subset B$ where $B$ is an event and $P(B)=0$, not that the (uncountable) intersection $A$ is an event. Here is a more elaborate approach, which settles this measurability problem.

Fix $\omega$ and consider the set $K(\omega)=\{n\mid X_n(\omega)\ne\frac12\}$, then one of the following happens:

$\quad$ (i) $K(\omega)$ is finite

$\quad$ (ii) There exists $x\gt\frac12$ such that $U_x(\omega)=\{n\mid X_n(\omega)\gt x\}$ is infinite

$\quad$ (iii) There exists $x\lt\frac12$ such that $L_x(\omega)=\{n\mid X_n(\omega)\lt x\}$ is infinite

In cases (ii) and (iii), consider the sequence $(n_k)$ which enumerates $U_x(\omega)$ or $L_x(\omega)$. Then $\frac1k(X_{n_1}(\omega)+\cdots+X_{n_k}(\omega))$ does not converge to $\frac12$, either because it does not converge, or because it converges to another value. In case (i), $\omega$ is in $A$.

To sum up, $A=\{K\lt\infty\}=\{S\lt\infty\}$, where $S=\sum\limits_n\mathbf 1_{X_n\ne1/2}$. Each $\mathbf 1_{X_n\ne1/2}$ is a random variable hence $A$ is indeed an event.

  • $\begingroup$ You might be in the same class as the user asking this. $\endgroup$ – Did Oct 7 '14 at 9:00
  • $\begingroup$ Thanks Did. Can you tell me what is the intuition of this problem? $\endgroup$ – Victor Oct 7 '14 at 13:04
  • $\begingroup$ Intuition: there are values of the sequence everywhere in (0,1) hence choosing only the values in some subset I, after they were all generated, yields abnormal empirical means. $\endgroup$ – Did Oct 7 '14 at 14:53
  • $\begingroup$ I have one more question. Is the set A you defined is same as the one in the problem? Why is it an uncountable union then? $\endgroup$ – Victor Oct 8 '14 at 7:35
  • $\begingroup$ It's a countable intersection over N^N, right? $\endgroup$ – Victor Oct 8 '14 at 7:36

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