9
$\begingroup$

These days I've been learning the properties of Brownian sample paths(Chapter 2 in Le Gall's Brownian Motion, Martingales, and Stochastic Calculus). As he mentioned in Proposition 2.14:

If $B=(B_t)_{t\geq0}$ is a Brownian motion, then we have a.s. for every $\epsilon>0$, $$\sup_{0\leq s\leq\epsilon}B_s>0,\qquad\inf_{0\leq s\leq\epsilon}B_s<0,$$ which means $B$ attains zero infinitely many times on $s\in[0,\epsilon]$ almost surely.

This property reminds me a problem I met several weeks ago, which goes as follows:

Suppose $(\epsilon_n)_{n\geq1}$ is a sequence of i.i.d. random variables and the common law is Bernoulli: $$\mathbb{P}[\epsilon_1=1]=\mathbb{P}[\epsilon_1=-1]=1/2.$$ Consider the random series $f(x)=\sum_{n=1}^\infty \epsilon_nx^n$. Show that the random series attains zero infinitely many times on $x\in[0,1)$ almost surely.

I had some idea on this problem: the series $f(x)$ must vibrate a lot in the left of $x=1$. All we need to prove is that for all $0<c<1$, we can find a zero of $f(x)$ in the interval $[c, 1)$.

BTW, I want to know some thing about the proposition above: does there exist some other stochastic processes having the similar property?

Any help would be appreciated.

$\endgroup$
  • $\begingroup$ I think this question will require use of a $0-1$ law. $\endgroup$ – rubikscube09 May 11 at 1:56
  • 1
    $\begingroup$ @rubikscube09 I also think so. Just as the proof of Proposition 2.14 mentioned above, where Le Gall used the Blumenthal's zero-one law, which is related to the simple Markov property of Brownian motion. And the simple Markov property is established under an important property of Gaussian random variable: If two Gaussian random variables have zero variance then they are independent. However, in this problem, I can't see a similar property. So I got stuck. $\endgroup$ – Feng Shao May 11 at 2:29
5
$\begingroup$

Lemma 1: One of the following three statements holds true:

  • $\lim_{x \uparrow 1} f(x)=\infty$ a.s.
  • $\lim_{x \uparrow 1} f(x) = - \infty$ a.s.
  • $\limsup_{x \uparrow 1} f(x) = \infty$ and $\liminf_{x \uparrow 1} f(x) = -\infty$ a.s.

Proof: Denote by $\mu$ the distribution of $\limsup_{x \uparrow 1} f(x)$, i.e. $$\mu(B) := \mathbb{P} \left( \limsup_{x \uparrow 1} \sum_{n=1}^{\infty} \epsilon_n x^n \in B \right).$$

If we set

$$\tau := \inf\{N \in \mathbb{N}; \sum_{n=1}^N \epsilon_n = 1\}$$

then $\tau< \infty$ almost surely and

$$\xi_n := \epsilon_{n + \tau(\omega)}, \qquad n \geq 1$$

defines a sequence of iid Bernoulli random variables. In particular, $(\xi_n)_{n \in \mathbb{N}}$ equals in distribution $(\epsilon_n)_{n \in \mathbb{N}}$, and so

$$\mu(B) = \mathbb{P} \left( \limsup_{x \uparrow 1} \sum_{n=1}^{\infty} \xi_n x^n \in B \right) \tag{1}$$

for all $B$. Moreover, we have

\begin{align*} \limsup_{x \uparrow 1} \sum_{n =1}^{\infty} \xi_n x^n &= \limsup_{x \uparrow 1} \sum_{n=\tau+1}^{\infty} \epsilon_n x^n \\ &= - \sum_{n=1}^{\tau} \epsilon_n + \limsup_{x \uparrow 1} \sum_{n=1}^{\infty} \epsilon_n x^n \\ &= -1 + \limsup_{x \uparrow 1} \sum_{n=1}^{\infty} \epsilon_n x^n. \end{align*}

Combining this with $(1)$ we get

$$\mu(B) = \mu(B+1)$$

for any Borel set $B$. The only finite measure on $\mathcal{B}(\mathbb{R})$ which is invariant under (non-trivial) translations is the trivial measure, and therefore we conclude that $\mu(\mathbb{R})=0$. The same reasoning works for $\liminf_{x \uparrow 1} f(x)$ (because of symmetry), and this finishes the proof of the lemma.


Lemma 2: $\limsup_{x \uparrow 1} f(x) = \infty$ and $\liminf_{x \uparrow 1} f(x)= - \infty$ almost surely.

Proof: The sequence $(-\epsilon_n)_{n \in \mathbb{N}}$ equals in distribution $(\epsilon_n)_{n \in \mathbb{N}}$, and therefore the random variables

$$\limsup_{x \uparrow 1} \sum_{n =1}^{\infty} \epsilon_n x^n$$

and

$$\limsup_{x \uparrow 1} \sum_{n=1}^{\infty} (-\epsilon_n) x^n = - \liminf_{x \uparrow 1} \sum_{n=1}^{\infty} \epsilon_n x^n$$

have the same distribution. Now the assertion follows from Lemma 1.

Corollary: $f$ has infinitely many zeros in $(0,1)$ with probability $1$.

Proof: As already noted by the OP, it suffices to show that for any $c \in (0,1)$ there exists with probability $1$ some $x^* \in (c,1)$ such that $f(x^*)=0$. Fix $c \in (0,1)$. By Lemma 2, we can find (with probability $1$) some $x_1 \in (c,1)$ and $x_2 \in (c,1)$ such that $f(x_1)>1$ and $f(x_2)<-1$. Since $f$ is continuous on $(0,1)$ this implies, by the intermediate value theorem, that there exists $x^* \in (x_1,x_2) \subseteq (c,1)$ such that $f(x^*)=0$.

Remark: In this paper you can find some more general statements on the behaviour of random series.

$\endgroup$
  • $\begingroup$ Thanks for your contribution! In your proof of Lemma 1, you said that $\xi_n$ is a sequence of iid Bernoulli random variables. I got stuck here and then I tried to prove this statement. Fix $A\in\mathcal{F_\tau}$ and $n_1<n_2<\cdots<n_k$, and let $F$ be a bounded continuous function from $\mathbb{R}^k$ into $\mathbb{R}_+$. Then $E[1_AF(\xi_{n_1},\cdots,\xi_{n_k})]=\sum_{m=0}^\infty E[1_{A\cap\{\tau=m\}}F(\epsilon_{n_1+m},\cdots,\epsilon_{n_k+m})]=\sum_{m=0}^\infty P(A\cap\{\tau=m\})E[F(\epsilon_{n_1},\cdots,\epsilon_{n_k})]=P(A)E[F(\epsilon_{n_1},\cdots,\epsilon_{n_k})]$. Is it enough? $\endgroup$ – Feng Shao May 12 at 5:41
  • 1
    $\begingroup$ @FengShao Yes, actually, it's enough to consider $A=\Omega$. What we use here is essentially the strong Markov property of the random walk, i.e. if $S_n := \sum_{i=1}^n \epsilon_i$ is the simple random walk associated with the steps $(\epsilon_i)_i$ then $$T_n := S_{n+\tau}-S_{\tau} = \sum_{i=1}^{n} \epsilon_{\tau+i}$$ is a simple random walk for any stopping time $\tau$; in particular the steps of $(T_n)_n$ are iid Bernoulli $\endgroup$ – saz May 12 at 5:52

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.