I am reading a proof of recurrence of Brownian Motion from the book of Morters and Peres. I have a question about a particular step in the proof of neighborhood recurrence for Brownian Motion in dimension 2.

He has proven that if the Brownian Motion starts at $x\in {\Bbb R}^2$ then with probability 1 it enters any annulus in finite time. I have no problem with the way he proves this. Using this fact, he begins to prove the neighborhood recurrence for dimension 2. The proof is as follows.

By the above property and shift-invariance the stopping time $ t_1$ = inf $\{t > 0: B(t) ∈ B(x, ε)\}$ is almost surely finite. Using the strong Markov property at time $t_1+1$ we see that this also applies to $t_2$ = inf$\{t > t_1 + 1: B(t) ∈ B(x, ε)\}$, and continuing like this, we obtain a sequence of times $t_n ↑ \infty$ such that, almost surely, $B(t_n ) ∈ B(x, ε)$ for all $n ∈ {\Bbb N}$. Taking an intersection over a countable family of balls $(B(x_i, ε_i): i = 1, 2, . . .)$, forming a basis of the Euclidean topology, implies that in $d = 2$ Brownian motion is neighbourhood recurrent.

My issue is with the application of the Strong Markov Property (SMP) in this proof. I assume we are applying the SMP for the stopping time $t_1+1$. So now we consider the Brownian Motion starting at $t_1+1$ i.e we will consider the process $\{B^{(1)}(t)=B(t+t_1+1)-B(t_1+1):t\geq0\}$. Since $\{B^{(1)}(t):t\geq0\}$ is a Brownian Motion starting at the origin, with probability 1, it will enter the ball $B(x,\epsilon)$ in finite time. However, $B^{(1)}$ only measures the displacement of $B(t)$ after the stopping time $t_1+1$. How does $B^{(1)}(t)\in B(x,\epsilon)$ (or in fact $B^{(1)}(t)$ being in any other open ball) produce a new time point, say $t'\geq t_1+1$ such that $B(t')\in B(x.\epsilon)$ ?

Once this issue is resolved, the rest of the proof I am fully on board with. It's just this part that is bugging me. I can intuitively see what the author is getting at but I'm looking for some rigorous justification. Clearly there is a step here but I cannot figure it out. Any help would be great. Thanks.


Do you refer to corollary 3.19 in this book? I believe that you miss-understand the content of it. What he states in corollary 3.19 is in fact exactly what you're are asking for. Corollary 3.19 states essentially that $P_x( T_r<\infty)=1$ for all $x\in \mathbb{R}^2$. The definition of $T_r$ is

$$T_r:=\inf\{t>0 :| B(t)|=r\}$$ That is; No matter what $x$ we start the brownian motion, it returns to the circle of radius $r$ around 0!

EDIT: Looking at your question again, i realize that you may ask because you are strictly using the form of SMP given in theorem 2.16 stating that the displaced process $B_{T+t}-B_{t}$ is again a brownian motion started at $0$. However I believe that the book refers to the form given in remark 2.17 with $$f(F)=1\{ \exists t>0 : F(t)\in B(x,\varepsilon)\} $$ where the "1" notation means the indicator function. Using the tower property this gives

$P( \exists t>0 : B(t+t_1)\in B(x,\varepsilon)\}) =\mathbb{E}f(B_{t+t_1})\stackrel{2.17}{=}\mathbb{E}f(B_{t})\stackrel{3.19}{=} 1$

  • $\begingroup$ Okay I get it now. And yes that book is the one I am following. I also want to clarify one other thing since you brought it up. Corollary 3.19 does assume that we have a fixed starting point, right? Please correct me if I am wrong. $\endgroup$ – Rahul Raphael Kanekar Mar 18 at 16:48
  • 1
    $\begingroup$ Well yes $P_x$ is the probability law under which the brownian motion starts in the fixed point $x$ but if we define a brownian motion started randomly at $X\sim \nu$ as $X+B^0_t$, where $B^0_t$ is a brownian motion started at $0$ then the 2d-part of the corollary still holds since $P(T^r<\infty)=\int P_x(T^r<\infty) d\nu(x)=\int 1 d\nu(x) = 1$ $\endgroup$ – Conformal Mar 18 at 18:34
  • $\begingroup$ If you are otherwise satisfied with the answer, would you mind accept it? Otherwise elaborate on whats missing. $\endgroup$ – Conformal Mar 22 at 14:43

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.