Introduction to Financial Mathematics
November 12, 2004
Notes on Option Valuation
1. Flat dollars. Assume a universal fixed interest rate r, so that ert dollars at time t can
be freely exchanged for 1 dollar at time 0 and vice versa.
Definition: At time t,
1 flat dollar = ert dollars.
Equivalently: At time t,
e-rt flat dollars = 1 dollar.
Note:
(a) The exchange rate between flat dollars and (ordinary) dollars varies with time.
(b) A guaranteed flat dollar at time t is worth the same as a flat dollar now. So,
working with flat dollars is like assuming that r = 0. But we’re not really making
that assumption—we’re just doing our calculations in another currency, and we
can translate our results into ordinary dollars whenever we like.
2. Geometric Brownian Motion (GBM):
Let:
S(t) = stock price at time t
L(t) = ln S(t)
R(t1, t2) = L(t2) – L(t1) = change in log price from time t1 to time t2
(if t2 ≥ t1)
t = t2 – t1
Then GBM assumes that
R(t1, t2) is normally distributed with parameters
mean
variance
std. dev.
=  t
= σ2 t
=  t
where t = t2 – t1.
and
 = “drift parameter” and
1
σ = “volatility parameter”
both depend on the particular stock.
(Also, R’s for non-overlapping intervals are independent.)
Typical test question:
Suppose that a stock follows GBM with parameters  = 0.01 and σ = 0.20, and
suppose that the price at time t = 14 is S(14) = $ 40.0. What is the probability that
S(30) > $60.00 ?
Solution.
We want:
P ( S(30) > 60 )
= P( L(30) > ln 60 )
= P( L(30) – L(14) > ln(60) – L(14) ).
Note that L(14) = ln(S(14)) = ln(40), and L(30) – L(14) is R(14,30).
So,
P ( S(30) > 60 )
= P( R(14,30) > ln(60)-ln(40) ).
Now, R(14,30) is normally distributed with mean
t =  * 16 = 0.16
and standard deviation
 t = σ * sqrt(16) = σ * 4 = 0.80.
So we can standardize:
P ( S(30) > 60 )
= P( R(14,30) > ln(60)-ln(40) )
 R(14,30)  0.16 (log 60  ln 40)  0.16 

= P

0.80
0.80


 (ln 60  ln 40)  0.16 
= 1 

0.80


= 1   .3068
= 0.3795.
3. Expected value of S(t):
Assuming GBM with parameters  and σ, the expected value of S(t) (as viewed
from time 0) is given by
    t .
E ( S(t) ) = S (0) e
1
2
2
2
4. Royalty Story.
Suppose S is lognormal with underlying parameters  and σ, and suppose that
the random variable X is given by
X=
S if S ≥ K, and
0 if S < K.
Then
E ( X ) = E(S ) P(S  Ke ) .
2
We can prove this by doing an integral.
5. Royalty Story applied to GBM.
Suppose S(t) obeys GBM with parameters  and σ. Then S(T) is lognormal
with underlying parameters T and standard deviation  T . Suppose X is a
random variable defined by
X =
S(T) if S(T) ≥ K, and
0
if S(T) < K.
Then
E(X) = E(S (T )) P(S (T )  Ke T ) .
2
6. Black-Scholes with a risk-neutrality assumption.
We’ll continue to use flat dollars.
For now, let’s assume that every participant in the market is “risk-neutral”, which
means that they value every asset according to the expected value of its future
payoff.
Assume that a stock price S(t) follows GBM with parameters  and σ, and assume
that the initial price is a known constant,
S(0) = S0.
(Thus, L(0) = ln(S0) is also known.)
The risk-neutrality assumption forces  = – (1/2) σ2. That’s because the time-t
payoff from owning a share of stock is S(t), and the expected value of that is
3
    t .
1
2
S (0) e
2
But if everyone values the stock by its expected value, that means that S(0) must
    t . That forces  + (1/2)σ2 = 0, so  = – (1/2) σ2.
be equal to S (0) e
1
2
2
A call option. Assume that a certain call option contract gives the holder the
right to buy one share for K at time T, if that is favorable. Thus, the payoff at
time T is
X =
S(T) – K
0
if S(T) ≥ K, and
otherwise.
The Black-Scholes Formula. With the risk-neutrality assumption, the value of
this option is equal to the expected value of its time-T payoff. That’s
V
=
E (X)
=
E ( X1 ) – E ( X2 )
where X1 = S(T) if S(T) ≥ K, 0 otherwise
and
X2 = K if S(T) ≥ K, 0 otherwise.
Now E ( X2 ) is just K P(S(T)≥K), and E(X1) is given by the royalty formula
applied to S(T). So:
V
=
E(S (T )) P(S (T )  Ke T ) – K P(S(T)≥K)
=
S0 P( S (T )  Ke T ) – K P(S(T)≥K)
2
2
(since with risk-neutrality, S0 must equal E(S(T))).
We calculate:
S(T) ≥ K means L(T) ≥ ln(K)
means R(0,T) ≥ ln(K) – ln(S0)
and R(0,T) is normal with
mean T = -(1/2)σ2T
stdev  T
4
so
P(S(T) ≥ K) =
=
=
P ( R(0,T) ≥ ln(K) – ln(S0) )
 ln(K)-ln(S0 )  12  2T 
1 

 T


 -ln(K)+ln(S0 )  12  2T 

.
 T


The other probability is the same, with an extra σ2T term…
P(S (T )  Ke
 2T
 -ln(K)+ln(S0 )  12  2T 
) = 
.
 T


So, we have:
 -ln(K)+ln(S0 )  12  2T 
 -ln(K)+ln(S0 )  12  2T 
V = S0  
–
K




 T
 T




which simplifies to
  S0 

  S0 

 ln  

 ln  

K 1
K 1





V = S0 
 2 T – K
 2 T 
  T

  T









(1)
This is the Black-Scholes formula for the value of a call option, at least in the case
of flat dollars.
Note that  doesn’t appear. But that’s just because our risk-neutral assumption
tells us that  = -(1/2)σ2, and we have substituted that value for  along the way.
7. Black-Scholes in ordinary dollars.
The dollar amounts in (1) are V, S0, and K. We have assumed that they are given
in flat dollars. But outside this class, people will give us S0 and K in ordinary
dollars, and they will expect us to compute V in ordinary dollars.
In the case of V and S0, it makes no difference. These quantities are measured at
time 0, when flat dollars are the same as ordinary dollars.
5
But the strike price refers to a payment at time T. If the strike price is K ordinary
dollars, then it is equal to Ke-rT flat dollars.
So, we can apply equation (1) to determine V, but we must use Ke-rT in place of K
wherever it appears. With this change, (1) becomes
  S0 

  S0
 ln    rT

 ln 
K
K



rT
1
V = S0  
 2 T  – Ke  



 T










  rT

 12  T 

T


(2)
which is the usual form of the Black-Scholes formula for call options.
Typical test question: Given S0, K, T, σ, and (if we’re not using flat dollars) r,
calculate V.
WITHOUT RISK NEUTRALITY…
8. The no-arbitrage and replicating-portfolio arguments.
Assume a simpler stock price model: The price at time 0 is S, and at time 1 it is
either S1 or S2:
S1
S
S2
We will always assume that S1 > S > S2.
Suppose that we have some option contract (put or call) with payoff C1 if the
stock goes up, and C2 if the stock goes down. What is C, the market value of the
option at time 0?
“Theorem”:
 S  S2 
 S1  S 
C 
 C1  
 C2 .
 S1  S2 
 S1  S2 
(3)
Proof 1 (outline): If C had any other value, then it would be possible to create
some mix of stocks and options that guarantees a profit, regardless of which
branch occurs. That’s an arbitrage opportunity, and it’s impossible in actual
markets. So, C must have the value given. //
6
Proof 2. Consider two portfolios:
Portfolio 1. One option.
Portfolio 2. x shares stock,
y dollars (cash).
We can choose x and y so that, at time 1, the two portfolios end up with the same
value regardless of which branch occurs. To do this, we must solve
x S1 + y = C1
x S2 + y = C2.
Since the two portfolios have the same payoff (as each other) regardless of what
occurs, they must have the same value at time 0 (“law of one price”). The value
at time 0 of portfolio 2 is xS + y, and the value at time 0 of portfolio 1 is just C.
So,
C = xS + y.
If we solve for x and y and substitute, we get (3). //
(The word “theorem” is in quotes because we haven’t been clear about our
assumptions—we made them up as we went along. The assumption in proof 1 is
the no-arbitrage assumption, and the assumption in proof 2 is the law of one price.
If you accept either of these assumptions, you must believe the theorem.)
9. How to think about formula (3):
Write
p1 
S  S2
,
S1  S2
p1 
S1  S
.
S1  S2
Then it’s easy to prove that p1 and p2 are both non-negative, and that their sum
is 1. That is, they look like probabilities.
With these definitions, (3) becomes
C  p1C1  p2C2 .
That is, C is the expected value of the future payoff, calculated using these
“probabilities.”
But notice that we did not assume that C was determined as an expected value—
this was a consequence of our reasoning. And, the p’s aren’t really the
7
probabilities of the branches in any realistic sense; they are just values that fell out
of our calculations. In this case, the p’s are determined by S, S1, and S2.
10. Put-call parity.
Let
C = value of call option with strike price K, expiration T.
P = value of (European) put option with strike price K, expiration T.
S0 = current price of one share.
Consider these portfolios:
Portfolio 1. K dollars.
Portfolio 2.
+1 put option
– 1 call option
+1 share stock.
At time T, either the put or call option will be exercised, and the share of stock
will be sold for K, so we will wind up with K dollars regardless of S(T). So, at
time T, the portfolios have the same value. So, they must have the same value at
time 0. So,
P – C + S0 = K.
This is usually written
C – P = S0 – K.
If we use ordinary dollars, we have to translate K dollars to flat dollars and the
formula is
C – P = S0 – Ke-rT.
(4)
This is the put-call parity formula. Since we can use it to get the value of a put
option from the value of a call option, we don’t need a separate Black-Scholes
formula for put options. (We could work one out if we needed to.)
11. The arbitrage theorem.
See handout.
We usually assume that markets allow no arbitrage opportunities. In this case, the
arbitrage theorem says that…
8
In an arbitrage-free market, there exists a set of probabilities for the possible
outcomes (called “risk-neutral probabilities”) such that every asset is priced
at the expected value of its future payoffs (using these probabilities).
12. The binomial tree model.
Consider the following model of stock prices from time 0 to time T.
(a) Choose n, the number of time steps.
(b) Set  = T / n, the length of a time step.
(c) The initial stock price, S(0) = S0, is a known constant.
(d) In each time step of length , if the stock price at the start of the
time step is S, it either goes up to Su or down to Sd, where
u=1+  
d=1–  
(and σ is the volatility parameter, depending on the stock).
(e) The probability q of the price going up at each step is…well, it
doesn’t matter.
In the limit of large n, this model approaches GBM with parameters  and σ.
( depends on q, but neither  nor q matters.)
According to the arbitrage theorem, we can find a set of “risk-neutral
probabilities” for the branches, such that all assets are priced according to their
expected values. The same set of probabilities must apply to the stock itself and
all options. But the only set of probabilities that makes the stock price look like
expected values is
p = ½ on every branch.
So, for valuing options, use p = ½ everywhere.
(Note: If we had wanted to approximate a GBM with  = - ½ σ2, we would have
chosen p = ½ everywhere.)
9
We don’t really care about the binomial model for its own sake. We care about
its limit for large n, since in the limit, results for the binomial model approach the
corresponding results for GBM.
13. How to evaluate any (European) contingent claim:
A contingent claim is a contract whose payoff at time T depends only on the stock
price S(T). Call and put options are contingent claims.
To evaluate a contingent claim:
(a) Build a tree representing the binomial model.
(b) Calculate the value of the contract for each possible stock price at time T.
(c) Work backwards to time 0, calculating the contract value at each node as the
average of the values at the two successor nodes.
(If some of the inputs are in ordinary dollars, translate them to flat dollars for this
calculation.)
14. How to evaluate any (American) contingent claim:
An American contingent claim can be exercised early. If it is exercised at time t,
the payoff depends on the stock price S(t) at that time.
To evaluate an American contingent claim:
(a) Build the tree.
(b) Calculate the “exercise value” at every node of the tree—the value, in flat
dollars, of exercising the claim at that node.
(c) For time T, value = exercise value.
(d) For each earlier node, calculate
“wait value” = average of the values at the two successor nodes
value at the node = larger of exercise value and wait value.
(e) The value of the option is the value calculated at the time-0 node.
10