Brownian motion: hitting time of double barrier vs hitting time of single barrier 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 $b$.
 A: 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 $\partial_tu=\frac12\partial_{zz}u$.
