Brownian motion and harnack inequality for ... - Math.caltech.edu

224 M. AIZENMAN AND B. SIMON
Proof: We begin by computing
E( T 2 t; b(r) E A ) =I E,J T 2 t; b(r) E A)#(x)d”x.
This is precisely (see [31])
with xA the indicator function of A; thus the conditional distribution of b(t)
subject to T h t is exactly #(x)d”x. It follows by the Markov property that
E(f(b( 7’)); T L I)
is independent of r, so that the claimed independence is proven. Moreover,
taking A = s2 in the above equation, we find that
from which (3.4) immediately follows.
P( T 2 t ) = e-a‘,
Remark. A similar argument shows that the probability distribution for
{ b( T - s )JoCs~ I conditional on T 2 t is independent of T. This fact should be
remembered in thinking about the constructions below.
For the case where !2 is a ball, the E-distribution of b( T) is obviously du(y)
by symmetry. The formula for more general !2 is discussed extensively in
Appendix 3. If yo E as2 and, neary,, as2 is a smoothly embedded submanifold of
R”. then the E-distribution of the exit place for b( T) is
(3.5)
(2cx-l- a+ du(y)
an
fory near yo. Here n is an inward pointing normal and du the usual surface area.
When as2 is everywhere smooth, the normalization condition
follows from
J(2a)f g du(y) = 1
(3.7)