98 CHAPTER 3. FIRST STEP ANALYSIS FOR STOCHASTIC PROCESSES
(Check for yourself that with this expression 0≤qT 0 ≤1 as it should be a
for a probability.)
We should show that the solution is unique. So supposerT 0 is another
solution of the difference equations. Given an arbitrary solution of (3.1), the
two constantsAandBcan be determined so that (3.2) agrees withrT 0 at
T 0 = 0 andT 0 =a. (The reader should be able to explain why by reference
to a theorem in Linear Algebra!) From these two values, all other values
can be found by substituting in (3.1) successivelyT 0 = 1, 2 , 3 ,...Therefore,
two solutions which agree forT 0 = 0 andT 0 = 1 are identical, hence every
7.2 Solution of the Black-Scholes Equation
The solution breaks down ifp=q= 1/2, since then we do not get two
linearly independent solutions of the difference equation (we get the solution
1 repeated twice). Instead, we need to borrow a result from differential equa-
tions (from the variation-of-parameters/reduction-of-order set of ideas used
to derive a complete linearly independent set of solutions.) Certainly, 1 is
still a solution of the difference equation (3.1). A second linearly independent
solution isT 0 , (check it out!) and the general solution isqT 0 =A+BT 0. To
satisfy the boundary conditions, we must putA= 1, andA+Ba= 0, hence
qT 0 = 1−T 0 /a.
We can consider a symmetric interpretation of this gambling game. In-
stead of a single gambler playing at a casino, trying to make a goalabefore
being ruined, consider two gamblers Alice and Bill playing against each other.
Let Alice’s initial capital beT 0 and let her play against adversary Bill with
initial capitala−T 0 so that their combined capital isa. The game continues
until one gambler’s capital either is reduced to zero or has increased toa,
that is, until one of the two players is ruined.
Corollary 1.pT 0 +qT 0 = 1
Proof.The probabilitypT 0 of Alice’s winning the game equals the probability
of Bill’s ruin. Bill’s ruin (and Alice’s victory) is therefore obtained from our
ruin formulas on replacingp,q, andT 0 byq,p, anda−T 0 respectively. That
is, from our formula (forp 6 =q) the probability of Alice’s ruin is
qT 0 =(q/p)a−(q/p)T^0
(q/p)a− 1and the probability of Bill’s ruin is
pT 0 =(p/q)a−(p/q)a−T^0
(p/q)a− 1