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Pricing Financial Derivatives

Jeremy Questel

University of Toronto

This is a summary of what was presented and discussed
at the May 27th **SIMMER** meeting, along with some problems and
questions to think about.

But what are all these fancy new financial objects, and why are the banks and investment companies hiring hordes of fresh PhD's in math and physics (the so called `rocket scientists')? What are all these analysts doing?

To try to get some idea let's look at a couple of very simple examples. Simple as they are, these models are actually pretty close to the real models that are being used on Bay Street and Wall Street.

You also have the option to put your money in the bank and earn interest. This involves no risk at all and $1 invested becomes $(1+r) at the end of the year. It is natural to assume that d<1+r<u because if 1+r>u there is no point in buying the stock, because you can do better with the bank for sure, and if 1+r< d then there is no point putting your money in the bank, because you can do better with the stock for sure.

Now a guy in a sharkskin suit makes you an offer. Instead of the stock, he will sell you a call option on the stock with strike price K. That means that if after 1 year the stock is worth S_1>K, you can cash it in for S_1-K. Otherwise it is not worth anything. In other words in one year the call option will be worth

V_1=max(S_1 -K, 0).(Note that K should be some number between dS_0 and uS_0 or this is a bit silly.) Now he says look, the probability that the stock will be worth S_1(u)=uS_0 is p, in which case the payoff is V_1(u)=uS_0 -K, and the probability that stock will be worth S_1(d)=dS_0 is 1-p in which case the payoff is V_1(d)=0. So the expected payoff is

p V_1(u) + (1-p)V_1(d) .Of course a dollar a year from now is really worth 1/(1+r) dollars now, so the price now should be

1/(1+r) [p V_1(u) + (1-p)V_1(d)].He offers to sell you the option for one half of that.

It looks like a great deal. Should you take it?

Suppose your initial wealth is $V_0 and you are wondering whether to go for it. You actually have a completely different choice. You could buy Delta_0 shares of the stock and put the rest, (V_0-Delta_0S_0) in the bank. A year later you would have

Delta_0 uS_0 + (1+r) (V_0-Delta_0S_0)if the stock went up, and

Delta_0 dS_0 + (1+r) (V_0-Delta_0S_0)if the stock went down.

On the other hand if you put the V_0 into call options it is worth V_1(u) if the stock goes up, and V_1(d) if the stock goes down.

Let's try something. We have been given S_0, u, d and r and we can choose Delta_0 and V_0. So let's match things after one year.

This is two equations in two unknowns, so we can solve it and the answer isDelta_0 uS_0 + (1+r) (V_0-Delta_0S_0)=V_1(u) Delta_0 dS_0 + (1+r) (V_0-Delta_0S_0)=V_1(d)

Delta_0= (V_1(u)-V_1(d))/(S_1(u) -S_1(d))and

V_0= 1/(1+r)[ q V_1(u) + (1-q) V_1(d) ],where q = (1+r-d)/(u-d).

If the price of the call option was more than V_0, why would you bother? You would do better for sure buying Delta_0 shares and putting the difference in the bank.

On the other hand if the price of the call option was less than V_0 then who would be so stupid as to sell them when they could do better for sure borrowing money and investing in the stock.

So the fair price for the option has to be V_0. You say it is the
*arbitrage price*
because if the price were any different many people in sharkskin suits
(known as
arbitrageurs) would have a field day making money on the difference.

But now we have two different answers for the price of the option, the
expectation price
and the arbitrage price. And the arbitrage price doesn't even depend on
the probabilities
p and 1-p, but uses some new `effective' probabilities q and 1-q
which only depend
on the interest rate and possible price changes.
These are called the *risk neutral probabilities*.

The market forces the arbitrage price to be the true price, so the strange conclusion is that if p is large enough, then the guys offer of half the expectation price is a bad deal!

**Question 2.** A stock price follows the following tree.

PICTURE AVAILABLE IN GRAPHICAL VERSION ONLY

It starts at $20 and after three months either goes up to $22 or down to $18. If it went to $22 then three months later it either goes up to $24.2 or down to $19.8. If it went down to $18 then three months later it either goes up to $19.8 or down to $16.2. The interest rate is still 12% per year. What is the value of an option to buy the stock for $21 in six months?

**Question 3.** A stock price is currently $20 and it is known that at the
end of three months it will be either $22 or $20 or $18.
The interest rate is 12% per annum.
Can you determine
the value today of an option to buy the stock after three months for $21?

The simplest model which is reasonable is called geometric or exponential Brownian motion. Let's try to describe it. The main idea here is that the market is made up of a lot of very little investors, so price movements that we observe are sums of a great number of random small movements. From probability we know that a random quantity which is the sum of a lot of small random quantities always has a Gaussian (or Normal) disribution. But also the stock price grows at some rate µ. All prices also have to be relative to the total price. So in summary we expect

S(t+h) -S(t) =µS(t) h + sigmaS(t)(B(t+h) -B(t))where B(t+h)-B(t) are mean 0 normal random variables. Since the variance of a sum of independent random variables is the sum of the variances, we have to have that the variance of B(t+h)-B(t) is linear in h. sigma plays the role of the constant and it is called the

dS(t) = µS(t) dt + sigmaS(t) dB(t)called a

PICTURE AVAILABLE IN GRAPHICAL VERSION ONLY

Not bad, eh?

Stochastic differential equations are different from ordinary differential equations because the function B(t) turns out to be non-differentiable at every single point with probability one. So the equation is really strange and it took until the 1950's before people even started to make sense of it. In fact these functions are so strange that the ordinary rules of calculus, like the chain rule and the product rule and integration by parts are just not true. But there are analogues. For example the chain rule for Brownian motion turns out to be

df(B(t)) = f'(B(t)) dB(t) + (1/2) f"(B(t)) dt.In normal calculus the last term wouldn't be there. This is called Ito's formula.

You can price a call option with strike price K at time T by analogy with the discrete time case. There is risk neutral probability distribution under which the arbitrage price of the option is

e^(-rT) x Expected value of max( S(T)-K, 0).Using Ito's formula you can check that this can be determined by solving a certain partial differential equation. In this lucky case, one can even write down the solution to this equation. Black, Scholes and Merton did this in the 1970's, and the fact that they finally managed to compute the fair price for an option meant that the whole business could finally get rolling. It's been growing like crazy as more and more fancier types of derivatives are invented, and Merton and Scholes went on to win the Nobel prize in Economics for their work.

**Question 4.** How would you go about constructing a continuous function
which
was nowhere differentiable? (Hint: Think fractals.)

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