Let $(B_t)$ a Brownien motion. Let $\tau_a=\inf\{t\geq 0\mid B_t=a\}$ (with $a\neq 0$) and $T_{a,b}=\tau_a\wedge \tau_b$. Suppose $x\in[a,b]$ and $B_0=x$. Let $h>0$ very small. Why do me have $p\{B_{T_{x+h,x-h}}=x\pm h\}=\frac{1}{2}$ ? The argument is the symmetry but I do not understand.
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The symmetry alluded to is that if $B_0=x$ then $W_t=2x-B_t$ defines another Brownian motion $(W_t)$ also starting from $x$. (The restriction that $h$ is "very small", whatever this would mean, is irrelevant.) – Did Jun 02 '15 at 09:11