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.

  • $\begingroup$ In Durrett's notation, you have $R=1+T_0\circ \theta_1.$ $\endgroup$ – user940 Apr 16 '16 at 22:30
  • $\begingroup$ I think part of my confusion is about the $\theta_1$ function. Does $T_0 \circ \theta_1 = \inf\{s>0 : B_{s+1}=0\}$? $\endgroup$ – Brownianmotionhurtsmyhead Apr 16 '16 at 22:36
  • 1
    $\begingroup$ Yes. Carefully read the definitions of $\theta_1$ and $T_0$ to see this. $\endgroup$ – user940 Apr 16 '16 at 22:40
  • $\begingroup$ Thank you, this comment actually clarified a lot of my confusion about this section! $\endgroup$ – Brownianmotionhurtsmyhead Apr 16 '16 at 22:47

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 $$P_x(R>1+t) =\int_{-\infty}^\infty p_1(x,y)P_y(T_0>t)\,dy. $$

  • $\begingroup$ Could you please be more specific about how the Markov property was applied to get that first equality? $\endgroup$ – Brownianmotionhurtsmyhead Apr 16 '16 at 22:23
  • $\begingroup$ 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 $\mathcal{F_s}$. $\endgroup$ – Brownianmotionhurtsmyhead Apr 16 '16 at 23:01
  • $\begingroup$ I am using $Y=1_{\{T_0>t\}}$, for which $Y\circ\theta_1=1_{\{T_0\circ \theta_1>t\}}=1_{\{R>1+t\}}$. $\endgroup$ – John Dawkins Apr 16 '16 at 23:06
  • $\begingroup$ Thank you very much, I learned a lot from your answer! $\endgroup$ – Brownianmotionhurtsmyhead Apr 16 '16 at 23:23

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.