Introduction

I will introduce a `binomial' model for stock market since it is mathematically simple (but elegant) but still retains the basic things of this topic. (derivative pricing)

  

Figure: The `one-step' binomial model for stock market

In figure , we assume that the transition probability of the stock from s₀ to s_up is m while from s₀ to s_down is (obviously) 1-m. We also assume that the money put in a bank acts predictably with interest rate r. Put the money deposited in a bank in a fancier way: define b₀ as the value of a bond at box zero (i.e time zero) so selling a bond means borrowing money while buying a bond means depositing money. Therefore, the two financial instruments we can handle are stocks and bonds. Now the only question we are interested is:

Given a derivative that has a payout⁷ (at time T) of f(2) dollars when the stock goes up and f(1) dollars when the stock goes down, how much⁸ does it worth?⁹

First I want to give you the result and then proceed to the argument!

1.

Let $(\phi,\psi)$¹⁰ be the unit of stocks and bonds one holds respectively. These two values are the so-called contents of a portfolio.

2.

The cost of this portfolio at time zero is $v=\phi s_{0}+\psi b_{0}$.

3.

Let $(\phi_{0},\psi_{0})$ be the solution of the set of the simultaneous equations¹¹

  

$$\phi_{0} s_{up}+\psi_{0} b_{0}\exp(rT)$$ = f(2) (2)

$$\phi_{0} s_{down}+\psi_{0} b_{0}\exp(rT)$$ = f(1) (3)

4.

Upon solving and , we would get

$$\phi_{0} \frac{f(2)-f(1)}{s_{up}-s_{down}}$$ = (4)

$$\psi_{0} \frac{\exp(-rT)}{b_{0}}(f(2)-\frac{(f(2)-f(1))s_{up}}{s_{up}-s_{down}})$$ = (5)

5.

Then the price, p, of this derivative must be

$$\large {p=\phi_{0} s_{0} + \psi_{0} b_{0}}$$ (6)

In other words,

$$\large {p=s_{0}({{f(2)-f(1)}\over{s_{up}-s_{down}}})+e^{-rT}(f(2)-{{s_{up}(f(2)-f(1))}\over {s_{up}-s_{down}}})}$$ (7)