; Formal proof of the Black-Scholes options pricing _2001_
Documents
User Generated
Resources
Learning Center

# Formal proof of the Black-Scholes options pricing _2001_

VIEWS: 42 PAGES: 7

• pg 1
```									          An Elementary Proof of Black-Scholes
Eric Hillebrand
CeVis – Center for Complex Systems and Visualization∗

March 3, 2001

Abstract
This paper presents a formal proof of the Black-Scholes options pricing
formula using arguments of elementary calculus and elementary stochas-
tics only.

The most common way to teach the Black-Scholes option pricing formula
to students who are not trained in stochastic analysis is by the Cox-Ross-
Rubinstein (1979) binomial tree approach. The convergence of the binomial
distribution to the normal distribution is often assumed to be known from ele-
mentary statistics.
We will present an approach to show the Black-Scholes formula that is like-
wise elementary but leads all the way through to the explicit formula. The
central idea is to introduce the problem that an expression of the mean return
must be found that restricts the discounted expected value of the stock price to
its start value. This problem can be solved by means of elementary stochastics
as we restrict the consideration to the expected value. The solution to this
problem is then applied to the whole process and this didactic ‘trick’ allows us
o
to shortcut the use of the Itˆ and Girsanov theorems.
The proof is not constructive inasmuch as it does not provide insight into the
mechanics of the hedge portfolio. The reader I have in mind is a student who
knows the binomial tree approach up to the pricing formula with the binomial
distribution and who wants to know how the actual Black-Scholes formula can be
proven without having to proof the Central Limit Theorem. The proof presented
o
here does that and it gives elementary insight into why the Itˆ correction comes
Besides fundamental calculus, the reader should have an understanding of
the normal distribution, namely its probability density function
1        (X − α)2
ϕ(α,γ 2 ) (X) = √   exp −                  ,
2π         2γ 2
∗ Universit¨tsallee 29, D-28359 Bremen, Germany, Fon: +49-421-218-4822, Fax: +49-421-
a
218-4236, e-mail: erhil@cevis.uni-bremen.de, erhil@aol.com

1
its cumulative distribution function

Φ(α,γ 2 ) (X) =       ϕ(α,γ 2 ) (X)dX,

and its standardisation and symmetry properties. The concept of the expected
value, EX = Xϕ(α,γ 2 ) dX for X ∼ N (α, γ 2 ), should be known. From the
viewpoint of mathematical ﬁnance, it is necessary to know about the role of
the exponential function for continuously compounded returns and that the
“fair price” of a claim is the discounted expected value of the stream of cash
ﬂows. The standard geometric Brownian motion model for stock prices should
be known inasmuch as returns are modelled by a drift term µt and a diﬀusion
term σWt , where W is standard Brownian motion. The model can be motivated,
for instance, by the discrete random walk.
The considered market consists of two assets, one risk free bond and one
stock which pays no dividend. There is a constant risk free interest rate r
that is the return on the bond, a constant mean return µ that is the return on
the stock and a volatility σ of the stock price. All returns are understood as
continuously compounded. The stock price S is then modelled by

S(t) = ceµt+σWt ,                             (1)

where W is standard Brownian motion. Let c denote the start value of the
stock price process, i.e. S(0) =: c > 0. The market is arbitrage free, that is,
there is no self-ﬁnanced portfolio with gains G greater than or equal to zero
and a nonzero probability of positive gains (P(G(t) > 0) > 0). Self-ﬁnanced
means that the total value of the portfolio at its setup time is zero, or in other
words, all purchases are made with credited money. The market is assumed
to be complete: Every claim in this market can be hedged using a portfolio of
the stock and the bond. There are no transaction costs. We will restrict the
presentation to the case of a European call.
The pay-oﬀ function C of a European call with expiry in t = T and strike
price q is given as:

(S(T ) − q)+ ,          t=T
C(t) =                                       ,            (2)
0,              t ∈ [0, T )

where (S(T ) − q)+ means S(T ) − q when this diﬀerence is positive and else zero.
The fair price V (0) of the option at time zero is its expected value discounted
to t = 0:
V (0) = e−rT E(S(T ) − q)+ .

Proposition 1 The discounted expected value of the stock price must be con-
stant:
e−rt ES(t) = c, c = S(0) > 0.

2
Proof. Assume e−rt ES(t) < c. Consider the claim C(τ ) as in (2) with q = 0
and expiry in τ = t . Then C(t) = S(t). As the market is complete, this claim
exists and can be hedged. Its fair price is given by its discounted expected value

C(0) = e−rt EC(t) = e−rt ES(t).

Buy such a claim at time zero for the price e−rt ES(t) and short one stock for
S(0). According to the assumption, S(0) − e−rt ES(t) > 0 and the cash balance
is positive at time zero. Cover at time t when claim and stock are worth the
same. The result is an arbitrage, contradicting the assumption that the market
is arbitrage free. In the case of e−rt ES(t) > c do it the other way around: Sell
the claim and buy the stock.

Problem 1 Inserting the stock price model of Eq. (1) into the required condi-
tion of Proposition 1, the problem is to ﬁnd a mean return µ that satisﬁes
!
ES(t) = Eceµt+σWt = cert .

Write the mean return µ = r + θ. Thus the problem can be restated to the search
of θ ∈ R that fulﬁls
!
ES(t) = Ece(r+θ)t+σWt = cert .

Remark. We mostly think of the mean return as consisting of the risk-free in-
terest rate and a risk premium, which should be positive as long as the volatility
σ is not zero. This mean return is not the µ in our Problem 1. Theta is not the
risk premium. On the contrary, we look for a correcting additive term θ that
restricts the expected value of the stock price to the start value c plus risk-free
o
interest. Hence, θ will be negative. It is the so-called Itˆ correction.

Lemma 1 If a random variable X is normally distributed with mean α and
variance γ 2 , then
γ2
EeX = eα+     2   .

This is a standard result. The proof is given in the Appendix for the sake of
completeness.
We can use the result of Lemma 1 directly to solve Problem 1. As the
random variable (µt + σWt ) is normally distributed with mean µt and variance
σ 2 t (because increments in Wt are normally distributed with zero mean and
variance t and µt is deterministic), it follows that
σ2                         σ2
ES(t) = Eceµt+σWt = ceµt+      2   t
= ce(r+θ)t+    2   t
.

The condition of Proposition 1 thus translates to
σ2
t !
ES(t) = ce(r+θ)t+    2       = cert

and hence θ = −σ 2 /2, µ = r − σ 2 /2 and the stock price process reads

3
σ 2 )t+σW
S(t) = ce(r−              2        t
.

We are now able to derive the Black-Scholes formula. All that is needed are
elementary manipulations of the normal distribution functions.

Lemma 2 If a random variable X is normally distributed with mean α and
variance γ, then
c                                       c
γ2        log      q   + α + γ2                      log   q   +α
E(ceX − q)+ = ceα+         2   Φ                                      − qΦ
γ                                  γ

where Φ denotes the cumulative distribution function of the standard normal
distribution.

The proof in the Appendix is taken from Irle (1998).

Proposition 2 (Black-Scholes 1973)
At time zero the fair price V (0) of the European call with strike-price q and
expiration in t = T in the market setting outlined above is
c               σ2                                    c            σ2
−rT                  +
log   q   + (r +      2 )T            −rt
log   q   + (r −   2 )T
V (0) = e        E(S(T )−q) = cΦ                          √                      −e       qΦ                    √             ,
σ T                                                   σ T

where c denotes the price of the underlying stock in t = 0, that is c = S(0), r
is the risk-free interest rate, σ is the volatility of the underlying stock, and Φ is
the cumulative distribution function of the standard normal distribution.

Proof. According to our stock-price model, we have
σ2
S(T ) = ce(r−             2   )T +σWT
.
2                                                                                      2
The exponent (r − σ )T +σWT is normally distributed with mean (r − σ )T and
2                                                  2
2
variance σ 2 T as (r− σ )T is deterministic and σWT is normally distributed with
2
2
zero mean and variance σ 2 T . Hence, we can apply Lemma 2 with α = (r− σ )T2
σ2                    γ2
and γ 2 = σ 2 T such that α + γ 2 = (r +                    2 )T   and α +        2    = rT . Thus,

c            σ2                             c              σ2
2
(r− σ )T +σWT        +       rT
log       q   + (r +   2 )T                  log      q   + (r −     2 )T
E(ce        2          −q) = ce          Φ                    √                   −qΦ                     √               ,
σ T                                         σ T
or,

4
c            σ2                               c            σ2
−rT              +
log   q   + (r +   2 )T            −rt
log   q   + (r −   2 )T
e      E(S(T ) − q) = cΦ                 √             −e             qΦ               √
σ T                                           σ T

References.

Black, F. / Scholes, M. 1973. The Pricing of Options and Corporate
Liabilities. Journal of Political Economy 81: 637–652.

Cox, J.C. / Ross, S.A. / Rubinstein, M. 1979. Option Pricing: A Sim-
pliﬁed Approach. Journal of Financial Economics 7: 229–263.

Irle, A. 1998. Finanzmathematik: Die Bewertung von Derivaten. Stuttgart:
Teubner.

Proof of Lemma 1.
1 − (X−α)2
EeX     =         eX √ e 2γ 2 dX
2π
1 X− (X−α)2
=         √ e       2γ 2 dX
2π
Fact:
(X − α)2     γ2   (X − α − γ 2 )2
X−            =α+    −                 .                                       (3)
2γ 2       2        2γ 2
From the Fact it follows that
1 X− (X−α)2                     1 α+ γ 2 − (X−α−γ 2 )2
EeX =     √ e     2γ 2 dX            =   √ e 2           2γ 2     dX
2π                               2π
γ2    1 − (X−α−γ 2 )2
= eα+ 2   √ e         2γ 2    dX
2π
2 2
Now √1 exp(− (X−α−γ ) ) is the probability density function ϕ(α+γ 2 ,γ 2 ) of
2π           2γ 2
a normally distributed random variable (X + γ 2 ) ∼ N (α + γ 2 , γ 2 ) and hence

1 − (X−α−γ 2 )2
√ e     2γ 2     dX = 1.
2π
γ2
=⇒        EeX = eα+      2   .

5
Proof of Lemma 2.
1 − (X−α)2
E(ceX − q)+           =                      (ceX − q) √ e 2γ 2 dX
2π
{X∈R|ceX >q}

1 X− (X−α)2
= c                        √ e     2γ 2 dX
2π
{X∈R|ceX >q}

1 − (X−α)2
−q                      √ e 2γ 2 dX.                                (4)
2π
{X∈R|ceX >q}

Again, the integral in the second summand is a cumulative distribution function
and noticing that ceX > q ⇐⇒ X > log q we can write
c

1 − (X−α)2                                           q
√ e 2γ 2 dX                  = P X > log
2π                                                   c
q
{X∈R|X>log   c}

X −α      log q − α
c
= P            >
γ           γ
log q − α
c
=     1−Φ               .
γ

As the standard normal distribution is symmetric (1 − Φ(x) = Φ(−x)), we have
q                            c
log   c   −α              log      q   +α
1−Φ                              =Φ                        .
γ                            γ

Regarding the ﬁrst summand in Eq. (3) recall Eq. (2):

(X − α)2     γ2   (X − α − γ 2 )2
X−             2
=α+    −                 .
2γ         2        2γ 2
Thus we have:
c
X      +        α+ γ
2
1 − (X−α−γ 2 )2                       log   q   +α
E(ce − q) = ce           2                          √ e     2γ 2     dX − qΦ                               .
2π                                          γ
q
{X∈R|X>log      c}

With the same argument as used in the proof of Lemma 1 we interpret the
integral in the ﬁrst summand as the cumulative distribution function of the
random variable (X + γ 2 ) and write
∞
1 − (X−α−γ 2 )2
√ e     2γ 2     dX                =              ϕ(α+γ 2 ,γ 2 ) dX
q
2π                                           q
{X∈R|X>log   c}                                           log   c
q
= P(X + γ 2 > log )
c

6
X −α      log q − γ 2 − α
c
= P           >
γ              γ
log q − γ 2 − α
c
=    1−Φ
γ
c
log   q   + α + γ2
=    Φ                              .
γ

Finally,
c                                   c
γ2       log   q   + α + γ2                  log   q   +α
E(ceX − q)+ = ceα+    2   Φ                          − qΦ                        .
γ                              γ

7

```
To top
;