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Let $T=\inf_t\{|X_t|=x\}$, where $(X_t)$ is a standard Brownian motion. Show that there is positive constants $c$ and $b$ for which $P(T >t) \le ce^{-bt}$.

I try the following: Let $T_x=\inf_t\{ X_t=x\}$, then $P(T>t)=P(\min(T_x,T_{-x})>t)=P(T_x>t,T_{-x}>t)$. Using the reflection principle, we have $P(T_x>t,T_{-x}>t) \le P(T_x>t) = 2P(X_t>|x|)$, but this doesn't highlight the constant $c$ and $b$.

1 Answers 1

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First note that the function $z\mapsto\mathrm P_z(T\gt 1)$ is continuous and that $\mathrm P_z(T\gt1)\lt1$ for every $|z|\leqslant x$. Hence, there exists $b\gt0$ such that $\mathrm P_z(T\gt1)\leqslant\mathrm e^{-b}$ for every $|z|\leqslant x$. (One can replace the continuity argument by the remark that $\mathrm P_z(T\gt1)$ is maximal for $z=0$ and that $\mathrm P_0(T\gt1)\lt1$.)

Now, for every $|z|\leqslant x$, conditioning on $X_n$, $ \mathrm P_z(T\gt n+1)=\mathrm E_z(\mathrm P_{X_n}(T\gt1);T\gt n)\leqslant\mathrm E_z(\mathrm e^{-b};T\gt n)=\mathrm e^{-b}\mathrm P_z(T\gt n), $ hence $\mathrm P_z(T\gt n)\leqslant\mathrm e^{-bn}$ for every $|z|\leqslant x$ and every nonnegative integer $n$. It follows that $\mathrm P_z(T\gt t)\leqslant c\mathrm e^{-bt}$ for every $|z|\leqslant x$ and every nonnegative real number $t\geqslant0$, with $c=\mathrm e^b$.


Edit This is to answer a question raised in the comments.

Introduce $u(t,z)=\mathrm P_z(T\gt t)$ for $t\geqslant0$ and $|z|\leqslant x$. Then $u(0,z)=1$ for every $|z|\lt x$, $u(t,x)=u(t,-x)=0$ for every $t\geqslant0$, and $ \partial_tu=\tfrac12\partial_{zz}u, $ on $t\gt0$, $|z|\lt x$. Heuristics leading to this so-called backward equation are as follows.

Fix $t\gt0$ and $|z|\lt x$ and consider some small positive $s$. Then $u(t+s,x)$ corresponds to the fact that the Brownian motion $B$ starting from $z$ does not exit $[-x,x]$ during the time interval $(0,s)$, and that, starting from the random point $B_s$ it does not either exit $[-x,x]$ during a new time interval of length $t$, namely $[s,t+s]$.

Assume that, when $s$ is small, $B_s$ is so close to $x$ most of the time that one can omit the first condition. Then $ u(t+s,z)\approx\mathrm E_z(u(t,B_s))=\mathrm E(u(t,z+\sqrt{s}Z)), $ where $Z$ is a standard gaussian random variable. Now, $ u(t,z+\sqrt{s}Z)=u(t,z)+\partial_zu(t,z)\sqrt{s}Z+\tfrac12\partial_{zz}u(t,z)sZ^2+o(s), $ and $\mathrm E(Z)=0$ and $\mathrm E(Z^2)=1$. Integrating the expansion above and pretending that the stochastic $o(s)$ term, when integrated, gives a deterministic $o(s)$ term, one gets $ u(t+s,z)=u(t,z)+\partial_zu(t,z)\sqrt{s}\cdot0+\tfrac12\partial_{zz}u(t,z)s\cdot1+o(s), $ that is, $ u(t+s,z)-u(t,z)=\tfrac12\partial_{zz}u(t,z)s+o(s), $ which, in the limit $s\to0^+$, yields Kolmogorov backward equation $\partial_tu=\frac12\partial_{zz}u$.

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    Our textbook refers to these as the heat equations (Introduction to Stochastic Process, Lawler). The precision you add on this in your post, helps me better understand these equations. Thank you very much Didier.2012-03-09