5
$\begingroup$

$X_1,\dots ,X_n$ are i.i.d random variables with $P(X_1=1)=p$, $P(X_1=-1)=1-p$, $p\neq\frac{1}{2}$. And $S_n=\sum_{k=1}^nX_k$.

I need to show that $P(\limsup_{n\to\infty}\{S_n=0\})\in\{0,1\}$.

It seems to me that $\limsup_{n\to\infty}\{S_n=0\}$ is not in tail sigma-algebra. Without this, i don't know how to proceed with it.

$\endgroup$
3
$\begingroup$

One can do things explicitly here, without $0$-$1$ laws, and show that the event in question has probability $0$.

Observe that $\mathbb{P}(S_n = 0) = 0$ if $n $ is odd, and if $n= 2k$ with $k\in \mathbb{N}$, then $$ (1) \qquad \mathbb{P}(S_{2k} = 0) = \left(\begin{array}{c} 2k \\ k \end{array} \right) p^{k}(1-p)^{k}. $$

Now, using Stirling's formula for factorials, we get $$ \left(\begin{array}{c} 2k \\ k \end{array} \right) = \frac{(2k)! }{(k!)^2} \leq C \frac{\sqrt{2\pi 2k } (2k)^{2k} e^{-2k} }{2\pi k \ k^{2k} e^{-2k} } = C \frac{4^k}{\sqrt{\pi k}}, $$ where $C>0$ is a constant independent of $k$. From this, getting back to $(1)$ we obtain $$ \qquad \mathbb{P}(S_{2k } = 0) \leq C \frac{1}{\sqrt{\pi k}} 4^k p^k (1-p)^k, \qquad k = 1,2,... $$ It is easy to see that $p(1-p) < 1/4$ where $0\leq p \leq 1$ and $p\neq 1/2$, and hence $$ \sum_{k=1}^{\infty}\mathbb{P}(S_k = 0) < \infty. $$ This implies, by Borel-Cantelli, that the event $ \limsup \{{S_k = 0} \}$ has probability $0$.

$\endgroup$
  • 1
    $\begingroup$ Addendum: There is yet another quick way to see that $\mathbb{P} (\limsup \{S_n = 0\} ) = 0$. Observe, that the strong law of large number applies here, implying that $$\frac{S_n}{n} \to 2p-1 \text{ a.s. } .$$ But since $2p-1 \neq 0$, it follows that $\mathbb{P}(\limsup \{S_n = 0\} ) = 0$ as otherwise would have convergence of $S_n/n$ to $0$ with a positive probability, which is a contradiction. $\endgroup$ – Hayk May 14 '18 at 8:35
5
$\begingroup$

One approach would be using the Hewitt-Savage 0-1 law, which is a generalization of Kolmogorov's 0-1 law. Roughly speaking we introduce a concept called exchangeable $\sigma$-field $\mathcal{E}$, and show that event $\limsup_{n\rightarrow \infty}\{S_n = 0\}$ or $\{S_n = 0, i.o.\}$ belongs to $\mathcal{E}$. The Hewitt-Savage 0-1 law claims that if $X_i$ are i.i.d. and $A\in \mathcal{E}$, then $P(A) =0$ or $1$. You can see that detailed explanation at Page 153-154 of Probability: Theory and Examples, Ed 4.1 by Rick Durrett. (see it here online).

Roughly speaking, for our case, an event $A$ is in $\mathcal{E}$ if the occurence of $A$ does not change if we exchange the value of finitely many $X_i$'s. Thus it can be seen that the tail $\sigma$-field $\mathcal{T}\subset \mathcal{E}$, indicating that this result is more general than the Kolmogorov's 0-1 law.

Actually this can generalize to any Borel set $B$, because $\{S_n \in B, i.o.\}\in \mathcal{E}$.

Here $\{S_n = B, i.o.\}\in \mathcal{E}$ because exchanging values of finitely many $X_i$'s does not affect $\{S_n = B, i.o.\}$.

$\endgroup$

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.