SØK460/ECON460 Finance Theory, Fall semester, 2002
Proof of equation (6), p. 177 in Cox and Rubinstein
The equation to prove is


!
n

1 X
n!
C0n (S0 ) = n 
pj (1 − p)n−j max[0, uj dn−j S0 − K] ,
r
j=0 j!(n − j)!
(1)
where I have added subscripts to show that the option expires at t = n, we want its price
at t = 0, which depends on the share price at t = 0. The variable t here denotes any point
in time, not the time left until an expiration date.
The proof by induction (see Sydsæter, Matematisk analyse, 6th ed., Vol. 1, 1994, appendix
A.5, or Hammond and Sydsæter, Mathematics for Economic Analysis, 1995, appendix B.5)
starts by showing that the formula holds when n = 1. This is easy from what we have
been through, and is left to the reader. (Remember that 0! = 1! = 1.)
Then we must show that if the formula is valid for some arbitrary n − 1, it will also be
valid for n. This can be done as follows: Assume the formula holds at t = 1 for an option
with expiration at t = n. Then we shall see that it also holds at t = 0 for the same option
(i.e., expiring at t = n). At t = 1 there are only n − 1 periods left until expiration, and (1)
looks like this:

C1n (S1 ) =
X
1 n−1
rn−1  j=0

!

(n − 1)!
pj (1 − p)n−1−j max[0, uj dn−1−j S1 − K] .

j!(n − 1 − j)!
(2)
From this formula and our usual method, based on the existence of a replicating portfolio,
we shall find the value of the same option at t = 0. This turns out to be
C0n (S0 ) =

X
1  n−1
= n p
r
j=0
+ (1 − p)
1
[pC1n (uS0 ) + (1 − p)C1n (dS0 )]
r
!
(n − 1)!
pj (1 − p)n−1−j max[0, uj+1 dn−1−j S0 − K]
j!(n − 1 − j)!
n−1
X

!

(n − 1)!
pj (1 − p)n−1−j max[0, uj dn−j S0 − K] ,

j!(n − 1 − j)!
j=0
(3)
where I have entered the two possible values of S1 , i.e., uS0 and dS0 .
In order to complete the proof, we must show that this can be rearranged to get (1).
Multiply the factors p and 1 − p into the sums, and change summation variable in the first
sum from j to i = j + 1. This yields
n
1 X
(n − 1)!
C0n (S0 ) = n
pi (1 − p)n−i max[0, ui dn−i S0 − K]
r i=1 (i − 1)!(n − i)!
(
!
1
+
n−1
X
j=0

!

(n − 1)!
pj (1 − p)n−j max[0, uj dn−j S0 − K] .

j!(n − 1 − j)!
(4)
Observe that the first term of the second sum is (1 − p)n max[0, dn S0 − K], while the
last term of the first sum is pn max[0, un S0 − K]. These are the first and the last terms
of the sum in (1). It remains to show that the n − 1 terms from the beginning of the
first sum and the n − 1 terms at the end of the last sum of (4) are equal to the n − 1
terms in the middle of the sum in (1). These terms contain common factors of the form
pj (1 − p)n−j max[0, uj dn−j S0 − K]. Preceding these there is
(n − 1)!
(j − 1)!(n − j)!
in one case and
(n − 1)!
j!(n − 1 − j)!
in the other case. It is easy to show that the sum of these two fractions is
n!
,
j!(n − j)!
so we find that

1 
C0n (S0 ) = n (1 − p)n max[0, dn S0 − K] + pn max[0, un S0 − K]
r
+
n−1
X
j=1


!
n!
pj (1 − p)n−j max[0, uj dn−j S0 − K] ,

j!(n − j)!
which equals (1), q.e.d.
2
(5)