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).