1.2. Boundary value problems 17
where
A =










−1
1
2
0 0 · · · 0 0
1
2
−1
1
2
0 · · · 0 0
0
1
2
−1
1
2
· · · 0 0
.
.
.
0 0 0 0 · · · −1
1
2
0 0 0 0 · · · 1
2
−1










, w =











F (0)
2
0
0
.
.
.
0

F (N)
2










.
If we prove that A is invertible, then the unique solution is v =
A−1w.
To prove invertibility it suffices to show that Av = 0 has a unique
solution and this can be done by an argument as in the previous proof.
Proof 4. Suppose F is a solution to (1.8). Let Sn
be the random
walk starting at x. We claim that for all n, E[F (Sn∧T )] = F (x).
We will show this by induction. For n = 0, F (S0) = F (x) and
hence E[F (S0)] = F (x). To do the inductive step, we use a rule for
expectation in terms of conditional expectations:
E[F (S(n+1)∧T )] =
N
y=0
P{Sn∧T = y} E[F (S(n+1)∧T ) | Sn∧T = y].
If y = 0 or y = N and Sn∧T = y, then S(n+1)∧T = y and hence
E[F (S(n+1)∧T ) | Sn∧T = y] = F (y). If 0 y x and Sn∧T = y, then
E[F (S(n+1)∧T ) | Sn∧T = y] =
1
2
F (y + 1) +
1
2
F (y 1) = F (y).
Therefore,
E[F (S(n+1)∧T )] =
N
y=0
P{Sn∧T = y} F (y) = E[F (Sn∧T )] = F (x),
with the last equality holding by the inductive hypothesis. Therefore,
F (x) = lim
n→∞
E[F (Sn∧T )]
= lim
n→∞
N
y=0
P{Sn∧T = y} F (y)
= P{ST = 0} F (0) + P{ST = N} F (N)
= [1 P{ST = N}] F (0) + P{ST = N} F (N).
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