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

226 M. AIZENMAN AND B. SIMON
Now introduce a random variable f with the distribution ae-a,lds. f will be
chosen independently - of q. Notice that the joint probability distribution with
respect to E,. of q(s,), * - - , q(s,,) and T 2 s, is
Q.~,(Y*YI). . Q,,-,, , ( Y ~ - I * Y ~ ) ~ * I * * d~n.
with the initial
distribution du(y)-the normalized surface measure on aQ, and an independent
variable f. The basic duality result is:
By e we shall denote the expectation of the Markov process { i,.}
THEOREM 3.3. The E-probability distribution of T and { b(T - s ) ) ~ ~ , is , ~
identical to the k distribution of f and { q(s))o, ,,( i.
Before proving this theorem, we want to note several things. First, the fact
that we took a ball for Q obscures somewhat the general formula for the correct
initial distribution for q(0). Obviously, we must take the distribution of b( T), i.e.,
(2a)-'(a\l/(y)/an,,)du(y) for general "nice" Q.
Secondly, we note that there is a very close relation between I and h-
processes of Doob [I I]. In place of Q = JI-'P\l/, Doob considers h-' Ph with h
an excessive function; JI is not an excessive function but it is close enough for
considerable formal connections. Indeed, various authors (see [ 191, [21]) have
discussed reversal of processes at an exit time in terms of h-processes.
Proof of Theorem 3.3: Since T and f have identical distributions, it suffices
to fix 05 sI 5 * - * 5 s, and prove the equality of the distributions of
{q(sI), - , q(s,)), conditioned on T L s,,, and of { b(T - sI), * * * , b(T - s,,)},
conditioned on T 1 s,,. Let F be a continuous function on 52" and fix s 2 s, and
y > 0. We claim that
where
since the left side of (3.13) is, by the Markov property for Brownian motion (P,
being the kernel of exp( -sHi'j),