%bs.tex: The Black-Scholes Equation WITHOUT Tears (or Probability!)
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{\bf The Black-Scholes Equation WITHOUT Tears (or Probability!) }
\bigskip
\centerline
{\it  By Doron ZEILBERGER}


{\bf The Fair price of a European Call Option (Black-Scholes Equation)}

The Crown jewel of mathematical finance is

{\bf The Black-Scholes Equation}([BS])

Suppose that a certain stock costs now $S_0$ dollars, its {\it volatility} is $\sigma$, and the {\it interest rate}
is $r$. The {\it fair price} of an {\it option} to buy that stock, for the price of $K$ dollars,  at
time $T$, is
$$
S_0 \, \Phi \left ( \frac{ \log \frac{S_0}{K} + (r +\frac{1}{2}\sigma^2)T }
{\sigma \sqrt{T}} \right )
\, - \,
K \, e^{-rT} \, \Phi \left ( \frac{ \log \frac{S_0}{K} + (r -\frac{1}{2}\sigma^2)T }
{\sigma \sqrt{T}} \right ) \quad ,
\eqno(BS)
$$
where
$$
\Phi(z) := \int_{-\infty}^z \frac{1}{\sqrt{2\pi}} e^{-x^2/2} dx \quad  .
$$

The purpose of the present, purely expository, {\it classroom note} is to derive it {\it from scratch}, 
{\bf without any probability theory}, following the beautiful
Cox-Ross-Rubinstein[CRR] discrete-time model, but in a way that should be understood
by the proverbial smart middle-school student.
The only prerequisite is the ability to solve two linear equations with two unknowns.
I was greatly helped by Alison Etheridge's wonderful book[E].

Let $S_T$ be the price of the stock at time $T$. Note that if $S_T<K$, it
would be stupid for the buyer to {\it exercise} the option, since he can just buy 
the stock in the free market (rather than pay more for it), and he definitely 
lost  the money spent buying the option. 
But if $S_T>K$, then he would exercise it, and by reselling it, would make
$S_T-K$ dollars. If $S_T-K$ (adjusted) is larger than the price that he paid for the 
option then he came out ahead. Regarding the seller, he can ``take a chance'',
pocket what he got from the buyer, and do nothing. If  $S_T<K$ he lucked-out,
but if $S_T>K$ then he would have to buy the stock for $S_T$ dollars, and sell it
to the buyer, at a loss, for $K$ dollars, that is equivalent to paying the buyer $S_T-K$ dollars.
If this  (adjusted) amount is less than what he got for the
option, then he  still came out ahead, but if it is more, then he lost.

But doing nothing is only {\bf one} option, (no pun intended!).

For the sake of simplicity, we will only consider the case where the interest rate, $r$, is zero.
The extension to general $r$ is straightforward.

{\bf Let's Bet!}

A stock {\it today} costs $12$ dollars. It is known that {\it tomorrow}
it is either going to cost $6$ dollars or $24$ dollars. 

The {\it seller} offers the {\it buyer}, if he pays him {\bf 8} dollars.

$\bullet$ $3$ dollars if the stock would go down;

$\bullet$ $15$ dollars if the stock would go up. 

Schematically:

{\obeylines
 \obeyspaces
             $\quad\quad 12$
           $\quad /$    $\quad \backslash$
   
        $6\, [3]$     $24 \,[15]$
}

Let's assume that the {\bf seller does not buy any stock}. Then this deal is a {\bf gamble} whether the
stock would go up or down, and the actual prices, $6$ and $24$, are {\bf irrelevant}.

If the stock would go down the buyer would lose $5$ dollars
(and the seller would make $5$ dollars), but if the stock would go up, the buyer would
make $7$ dollars (and the seller would lose $7$ dollars).

If the buyer estimates that the {\it probability} that the stock would go down is $q$,
his expected gain is $q(-5)+(1-q)7=7-12q$, so he would agree to the deal
if $q<\frac{7}{12}$, if he is {\bf rational}.

But the {\it fair price} has {\bf nothing to do} with what the buyer (or seller) think
about the future of the stock. It is determined by the

{\bf AXIOM of OPTION PRICING}: If you want to have a chance to make money,
you should be willing to risk losing money! In other words:

{\bf IT SHOULD BE IMPOSSIBLE TO HAVE A STRATEGY 
THAT WOULD LET YOU MAKE MONEY (AT LEAST SOMETIMES) AND NEVER LOSE MONEY}

It turns out that $8$ dollars is too high of a price for the above deal.

Here is how to figure out the {\it fair price} of the above deal.

We assume that the seller of this bet has no money of his own, and will get in jail if
he can not fulfill his obligations to the buyer. So he has to design a {\bf mixed portfolio},
{\it only} using the money that he would get from the buyer of the bet.

{\bf Buy $a$ units of stock and $b$ dollars in cash} ($b$ can be negative, so he can borrow).

The value of this portfolio {\bf today} is $12a+b$ (since one unit of stock costs $12$ dollars today)

The value of this portfolio {\bf tomorrow} is

$\bullet$ if the price of the stock went down to $6$ dollars: $6\,a \, + \, b$

$\bullet$ if the price of the stock went up to $24$ dollars: $24\,a \, + \, b$.

In order to {\it meet his obligations} (and not get arrested), he needs to make sure that

$$
6a+b \geq 3 \quad, \quad 24a+b \geq 15 \quad \quad .
$$
But if any of the inequalities is {\bf strict} then he made money!, (at least sometimes).
And that is not allowed, so to {\bf barely} meet his obligations he has to solve the
{\bf system of two linear equations} in the {\bf two unknowns}, $a$ and $b$:
$$
\{ 6a+b =3 \quad, \quad 24a+b=15 \, \} \quad ,
$$
whose solution is $a=\frac{2}{3}$, $b=-1$. 
The cost of this portfolio {\it today} is
$$
12a+b= 12 \cdot \frac{2}{3}  + (-1) =7 \quad .
$$

So the price must be at most $7$ dollars. What if it is less?
Then the {\it buyer} can {\it borrow} $\frac{2}{3}$ units of stock, and
lend $1$ dollar, and buy the deal, and come out ahead, forbidden!

So the {\bf fair}, {\it arbitrage-free}, price of the bet is {\bf exactly} seven dollars.

Note that the mixed portfolio `buy $\frac{2}{3}$ units of stock and borrow $1$ dollar', the
so-called {\bf hedging portfolio}, is {\bf only} of theoretical interest. If the
seller of the above deal would follow it, he is guaranteed to {\bf never} make any money, so
in real life he would use a different portfolio, based on his belief of the future market,
and take a risk of losing money. But the price of seven dollars {\it guarantees} that
neither buyer nor seller would be able to have a possibility of making money, without ever
losing it, and hence is considered the {\it fair} price.


{\bf The General Single Period Binary Case}


A stock {\it today} costs $S_0$ dollars. It is known that {\it tomorrow}
it is either going to go down to $S_d$ dollars or go up to $S_u$ dollars. 

The {\it seller} offers the {\it buyer}, the following deal.

$\bullet$ $C_d$ dollars if the stock would go down;

$\bullet$ $C_u$ dollars if the stock would go up. 

Schematically:

{\obeylines
 \obeyspaces
             $\quad\quad S_0$
           $\quad /$    $\quad \backslash$
   
        $S_d\, [C_d]$     $S_u \,[C_u]$
}

Here is how to figure out the {\it fair price} of that deal.

We assume that the seller has no money, and will get in jail if
he can not fulfill his obligations. 
All he has is the money that he got from the buyer. How much money should the buyer give the seller for the
above deal?

The seller has to design a {\bf mixed portfolio}.

{\bf Buy $a$ units of stock and $b$ dollars in cash} ($b$ can be negative, so he can borrow).

The value of this portfolio {\bf today} is $a\cdot S_0+b$ 
(since one unit of stock costs $S_0$ dollars today).

The value of this portfolio {\bf tomorrow} is

$\bullet$ if the price of the stock went down to $S_d$ dollars: $a \cdot S_d +b$

$\bullet$ if the price of the stock went up to $S_u$ dollars: $a \cdot S_u+b$.

In order to {\it meet his obligations} (and not get arrested), we need

$$
a S_d+b \geq C_d \quad, \quad aS_u +b \geq C_u \quad \quad .
$$
But if any of the inequalities is {\bf strict} then he would make money!, (at least sometimes),
without any risk of losing money,
and that is {\bf forbidden}, so to {\bf barely} meet his obligations he has to solve the
{\bf system of two linear equations and two unknowns}
$$
a S_d+b \,= \, C_d \quad, \quad aS_u +b \, = \, C_u \quad  ,
$$
whose solution is 
$$
a= \frac{C_u - C_d}{S_u -S_d} \quad , \quad 
b= \frac{C_d S_u - C_u S_d}{S_u -S_d} \quad.
$$
The cost of this portfolio {\it today} is
$$
 \frac{C_u - C_d}{S_u -S_d} \cdot S_0 \,+ \, \frac{C_d S_u - C_u S_d}{S_u -S_d}
$$
$$
= \frac{   (S_0 -S_d)\, C_u \, + \, (S_u -S_0) \,C_d }{S_u -S_d} \quad .
$$

We just proved (using Algebra 1), the

{\bf Fundamental Lemma of Option Pricing} 

Suppose that the price of a certain stock today is $S_0$ dollars, and it is known that tomorrow it is either
going to cost $dS_0$ ($d<1$) or $uS_0$ ($u>1$).

The fair price of a {\bf deal} to be paid $C_d$ dollars if the stock goes down
and $C_u$ dollars if the stock goes up is
$$
\frac{  (1 -d) \, C_u \, + \, (u -1) \,  C_d } {u-d} \, =  \,  \frac{1-d}{u-d} \, C_u  \, + \, \frac{u-1}{u-d} \, C_d   \quad .
$$
Let's abbreviate
$$
\alpha:= \frac{1-d}{u-d} \quad , \quad \beta :=  \frac{u-1}{u-d} \quad  .
$$
Then the fair price is
$$
 \alpha  C_u \, + \, \beta C_d  \quad.
$$

{\bf What if there are Two Days?}

Suppose that today's stock costs $S_0$ dollars, and tomorrow it would
go either down to $S_d$ or up to $S_u$. If tomorrow it went down to
$S_d$, then the day after tomorrow it may either go further down to
$S_{d,d}$, or pick up, and go up to $S_{d,u}$. If tomorrow it went up to $S_u$,
then the day after tomorrow it may drop to $S_{u,d}$ or go further up to
$S_{u,u}$. Assume that the seller offers the buyer the following conditional promises

$\bullet$ $C_{d,d}$ dollars if after two days the price of the stock is $S_{d,d}$

$\bullet$ $C_{d,u}$ dollars if after two days the price of the stock is $S_{d,u}$

$\bullet$ $C_{u,d}$ dollars if after two days the price of the stock is $S_{u,d}$

$\bullet$ $C_{u,u}$ dollars if after two days the price of the stock is $S_{u,u}$

Then to figure out the fair price, one uses {\bf backwards induction}.
First figure out the fair price tomorrow under the assumption that the stock went down,
get a certain number (using the fundamental lemma), let's call it $C_d$.
Similarly for $C_u$, and then use the formula once again to get the fair price {\bf today}.

Now let's make the simplifying assumption that if the price of the stock
today is $S$, then the next day it is either $dS$ or $uS$, and $d$ and $u$ are
always the same from day to day. Hence
$$
S_{d,d}=S_0 d^2 \quad, \quad S_{d,u}=S_{u,d}=S_0 ud \quad, \quad
S_{u,u}=S_0 u^2  \quad .
$$
Also let's assume that the payoff,
at the end, only depends on the price of the stock, not on the history, so
denoting $C_{u,u}=C_0, C_{u,d}=C_{d,u}=C_1, C_{d,d}=C_2$, it is
immediate that the fair price, using the above formula is
$$
 \alpha (\alpha C_0 + \beta C_1) +\beta (\alpha C_1 + \beta C_2) \, = \,
 \alpha^2 C_0  + 2\alpha \beta C_1 + \beta^2 C_2 \quad .
$$
If there are three days, then the price is
$$
\alpha \cdot \left ( \alpha^2 C_0  + 2\alpha \beta C_1 + \beta^2 C_2 \right ) +
\beta \cdot \left ( \alpha^2 C_1  + 2\alpha \beta C_2 + \beta^2 C_3 \right ) \quad
$$
$$
=\alpha^3 C_0  + 3\alpha^2 \beta C_1 + 3 \alpha \beta^2 C_2 + \beta^3 C_3 \quad .
$$

After $n$ days,
$$
\sum_{i=0}^{n} {{n} \choose {i}} \,  \alpha^{n-i} \beta^i  C_i \quad ,
$$
that is easily proved by induction on $n$. Here $C_i$ is the payoff if
the price of the stock, after $n$ days, is $S_0 \, d^i u^{n-i}$.

So far it was abstract gambling. If the seller promises the buyer to sell the
stock, after $n$ days, for $K$ dollars then the value of the promise is
$$
C_i=S_0 d^i u^{n-i} -K \quad, \quad if \quad  S_0 d^i u^{n-i} >K
$$
$$
C_i=0 \quad, \quad if \quad  S_0 d^i u^{n-i} <K \quad .
$$
We just derived the

{\bf Discrete Black-Scholes  Formula (Alias the Cox-Ross-Rubinstein Formula)}

The fair price of the option is
$$
\sum_{i; S_0 d^i u^{n-i} \, > \, K} {{n} \choose {i}} \, \alpha^{n-i} \beta^i  \, (S_0 d^i u^{n-i} -K) 
\quad .
$$
Defining (the {\it partial sum} of the binomial theorem)
$$
B(n,m,p):= \sum_{i=0}^{m}  {{n} \choose {i}} p^i (1-p)^{n-i} \quad ,
$$
this can be expressed, more succinctly as
$$
S_0 \cdot B(n,m, \beta d) - K \cdot B(n,m, \beta)  \quad,
\eqno(CCR)
$$
where $m$ is the solution of $S_0 d^m u^{n-m}=K$, and  $\beta \, = \, \frac{u \, - \, 1}{u-d}$.

{\bf Taking the limit}

Going back to the `continuous', Black-Scholes, scenario, if the expiration date is $T$, then
chop it into $n$ periods, where $n$ is {\bf very very big}. Each of the steps has duration
$$
\Delta t=\frac{T}{n} \quad .
$$

For a stock of `volatility' $\sigma$, it turns out that, for very large $n$,
$$
u=1+\sigma \sqrt{\Delta t} \quad , \quad d=1-\sigma \sqrt{\Delta t} \quad .
$$

Computing $\beta :=  \frac{u-1}{u-d}=\frac{1}{2}$, plugging into $(CCR)$, and taking the limit $n \rightarrow \infty$,
using the  fact, due to de Moivre (see [Z] for an elementary proof) that
$$
B(n,np+ \sqrt{np(1-p)} x) \rightarrow \Phi(x) \quad,
$$
with $p=\beta d=d/2$ for the first term, and with $p=\beta=\frac{1}{2}$ for the second term, of $(CCR)$
one immediately gets (using Maple) the Black-Scholes equation $(BS)$ (with $r=0$) displayed at the beginning of this article.

\bigskip

{\bf References}

[BS] Fischer Black and Myron Scholes, {\it The pricing of options and corporate liabilities},
The journal of political economy (1973), 637-654.

[CCR]  John C. Cox, Stephen A. Ross, and Mark Rubinstein, {\it Option pricing: A simplified approach},
Journal of financial Economics {\bf 7}(3)(1979), 229-263.

[E] Alison Etheridge, ``{\it  A course in Financial Calculus}'', Cambridge University Press, 2002.

[Z] Doron Zeilberger, {\it The Automatic Central Limit Theorems Generator (and Much More!)},
``Advances in Combinatorial Mathematics: Proceedings of the Waterloo Workshop in Computer Algebra 2008 
in honor of Georgy P. Egorychev'', chapter 8, pp. 165-174, (I.Kotsireas, E.Zima, eds. Springer Verlag, 2009.)
\hfill\break
{\tt http://www.math.rutgers.edu/\~{}zeilberg/mamarim/mamarimhtml/georgy.html}

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Doron Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen
Rd., Piscataway, NJ 08854-8019, USA. \hfill\break
Email: {\tt zeilberg at math dot rutgers dot edu}   \quad .
\bigskip
\hrule
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Exclusively published in the Personal Journal of Shalosh B. Ekhad and Doron Zeilberger.
\bigskip
May 14, 2015

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