# How to use the Markov property of Brownian motion

This is a problem from Durrett's probability with examples, exercise 8.2.1. It is not homework. The exercise states: Let $T_0 = \inf\{s > 0 : B_s = 0\}$ and let $R = \inf\{t > 1 : B_t = 0\}$. Use the Markov property at time 1 to get $$P_x(R>1+t) = \int_{-\infty}^\infty p_1(x,y)P_y(T_0>t)dy.$$ Here, $P_x$ is the Wiener measure associated to Brownian motion started at $x$ and I think $p_1(x,y) = \frac{1}{\sqrt{2\pi t}}e^{-(x-y)^2/2t}$ is the density of Brownian motion started at $x$ but I'm not sure.

I cannot even see where to begin. I don't see how $P_y(T_0>t)$ will be involved. I don't see what function $Y$ to apply the Markov property to. Some explanation of how to approach this would be much appreciated.

• In Durrett's notation, you have $R=1+T_0\circ \theta_1.$ – user940 Apr 16 '16 at 22:30
• I think part of my confusion is about the $\theta_1$ function. Does $T_0 \circ \theta_1 = \inf\{s>0 : B_{s+1}=0\}$? – Brownianmotionhurtsmyhead Apr 16 '16 at 22:36
• Yes. Carefully read the definitions of $\theta_1$ and $T_0$ to see this. – user940 Apr 16 '16 at 22:40
Consider $\tilde B_s:=B_{s+1}$, $s\ge 0$, and notice that $R-1=\inf\{s>0:\tilde B_s=0\}=T_0(\tilde B)$. Now condition on where $B$ is at time $1$: the density of $B_1$ is $y\mapsto p_1(x,y)$ (as you surmise) and the conditional distribution of $\tilde B$, given that $B_1=y$, is $P_y$. Putting these thoughts together, \eqalign{ P_x(R>1+t) &=\int_{-\infty}^\infty p_1(x,y)P_x(T_0(\tilde B)>t|B_1=y)\,dy\cr &=\int_{-\infty}^\infty p_1(x,y)P_y(T_0>t)\,dy.\cr }
More detail on the above calculation. Using the shift operator, $\{R>1+t\}=\theta_1^{-1}(\{T_0>t\})$. The Markov property (8.2.2) tells us that \eqalign{ P_x(R>1+t; B_1\in C) &=P_x(\theta_1^{-1}\{T_0>t\}; B_1\in C)\cr &=E_x[\varphi(B_1); B_1\in C]\cr &=\int_C \varphi(y)p_1(x,y)\,dy, } where $\varphi(y)=P_y(T_0>t)$, and $C$ is any Borel subset of $\Bbb R$. In particular, taking $C=\Bbb R$ we get $$P_x(R>1+t) =\int_{-\infty}^\infty p_1(x,y)P_y(T_0>t)\,dy.$$
• I'm still confused, since you seem to be using a slightly different statement of the Markov property (probably equivalent but I'm too confused to see it). Durrett states it in the form: if $s \ge 0$ and $Y$ is a bounded function from $C([0,\infty),\mathbb{R})$ to $\mathbb{R}$, then $E_x(Y \circ \theta_s | \mathcal{F}_s) = E_{B_s}[Y]$. I don't quite see what $Y$ should be here, or at what stage we are conditioning on the $\sigma$-field $\mathcal{F_s}$. – Brownianmotionhurtsmyhead Apr 16 '16 at 23:01
• I am using $Y=1_{\{T_0>t\}}$, for which $Y\circ\theta_1=1_{\{T_0\circ \theta_1>t\}}=1_{\{R>1+t\}}$. – John Dawkins Apr 16 '16 at 23:06