1.4. Expected time to escape 31
Then just as in the one-dimensional case we can see that f(x) = eA(x)
satisfies
(1.20) f(x) = 0, x ∂A,
(1.21) Lf(x) = −1, x A.
We can argue the same as in the one-dimensional case that there is
at most one function satisfying these equations. Indeed, if f, g were
two such functions, then L[f g] 0 with f g 0 on ∂A, and only
the zero function satisfies this.
Let f(x) = |x|2 = x1 2 + · · · + xd. 2 Then a simple calculation shows
that Lf(x) = 1. Let us consider the process
Mn = |Sn∧TA
|2
(n TA).
Then, we can see that
E[Mn+1 | S0,...,Sn] = Mn,
and hence Mn is a martingale. This implies
E[Mn] = E[M0] =
|S0|2,
E[n TA] = E[|Sn∧TA
|2]

|S0|2.
In fact, we claim we can take the limit to assert
E[TA] = E[|STA
|2]

|S0|2.
To prove this we use the monotone convergence theorem, see Exercise
1.6. This justifies the step
lim
n→∞
E[|STA
|2
1{TA n} = E[|STA
|2].
Also,
E |STA
|2
1{TA n} P{TA n} max
x∈A
|x|2
0.
This is a generalization of a formula we derived in the one-dimensional
case. If d = 1 and A = {1,...,N 1}, and
E[|STA
|2]
= N
2
P{STA = N | S0 = x} = N x,
E[TA] = E[|STA
|2]

x2
= x (N x).
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