1.2. Boundary value problems 15

Recall that if V1, V2 are events, then P(V1 | V2) denotes the conditional
probability of V1 given V2. It is defined by
P(V1 | V2) =
P(V1 V2)
P(V2)
,
assuming P(V2) 0.
We can give a gambling interpretation to this by viewing Sn as
the number of chips currently held by a gambler who is playing a
fair game where at each time duration the player wins or loses one
chip. The gambler starts with x chips and plays until he or she has
N chips or has gone bankrupt. The chance that the gambler does
not go bankrupt before attaining N is F (x). Clearly, F (0) = 0 and
F (N) = 1. Suppose 0 x N. After the first game, the gambler
has either x 1 or x + 1 chips, and each of these outcomes is equally
likely. Therefore,
(1.8) F (x) =
1
2
F (x + 1) +
1
2
F (x 1), x = 1,...,N 1.
One function F that satisfies (1.8) with the boundary conditions
F (0) = 0,F (N) = 1 is the linear function F (x) = x/N. In fact,
this is the only solution as we shall now see.
Theorem 1.4. Suppose a, b are real numbers and N is a positive
integer. Then the only function F : {0,...,N} R satisfying (1.8)
with F (0) = a and F (N) = b is the linear function
F0(x) = a +
x(b a)
N
.
This is a fairly easy theorem to prove. In fact, we will give several
proofs. This is not just to show off how many proofs we can give! It
is often useful to give different proofs to the same theorem because
it gives us a number of different approaches for trying to prove gen-
eralizations. It is immediate that F0 satisfies the conditions; the real
question is one of uniqueness. We must show that F0 is the only such
function.
Proof 1. Consider the set V of all functions F : {0,...,N} R
that satisfy (1.8). It is easy to check that V is a vector space, i.e., if
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