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