1. Present Value and Flat Dollars

Financial Mathematics

April 23, 2007

Notes on Option Valuation - I

1. Present Value and Flat Dollars

When money amounts are expressed (as usual) in dollars, we can define a present value function as follows: For any time t≥ 0, we denote by P(t) the size of a current payment that the market would consider equivalent to a guaranteed payment of $1 at time t. For example, if P(2) = 0.90, then anyone is able to exchange a guaranteed payment of $1 two years from today for a payment of 90 cents today, or vice versa. (We measure t in years.) We assume that the present value function is the same for everyone and is known with certainty by all.

For example, suppose that there is a fixed, universal interest rate r, continuously compounded. Then P(t) = exp(–rt) for every t ≥ 0. In general, P(0) = 1, and normally P is positive and strictly decreasing for t > 0.

We would like in these notes to avoid the complication of multiplying by P(t) or exp(–rt) every time we want to do a calculation involving a future payment. We will therefore adopt the convention that all money amounts mentioned in any context are expressed as present values. For example, if P(2) = 0.90 and a person proposes to make a payment of $1 (actual amount) at time t = 2, we will immediately translate the amount into present value, and for all of our discussions we will refer to the payment as $0.90, understood to be the present value of the payment.

One way to understand this convention is that all amounts are expressed in “flat dollars.” A flat dollar is an imaginary unit of currency with this schedule of exchange rates:

1 actual dollar at time t is equivalent to P(t) flat dollars.

Equivalently,

1 flat dollar at time t is equivalent to 1/P(t) actual dollars.

With this convention, a guaranteed flat dollar at any future time t is worth the same as one flat dollar now. Also, one flat dollar now is worth the same as one actual dollar now, since P(0) = 1.

This convention is equivalent to assuming that there is a universal interest rate r, and that it is equal to zero. That’s obviously a nice simplifying assumption. But we aren’t really making that assumption, since we can always translate between flat dollars and actual dollars using the exchange rate above.

When we derive an important result in flat dollars, usually our last step will be to translate the result into actual dollars, so that we can compare it to similar results in the literature.

Exercise. Let r = 0.05, and assume that P(t) = exp( –0.05t ) for all t ≥ 0. Suppose that a contact says that Alice is to pay Bob $1000 (actual dollars) per year for four years, starting with the first payment one year from now.

(a) What is the present value of the contract? (That is, how much is it worth to Bob at time 0 ?)

(b) How would you restate the contact in flat dollars?

Solution.

(a) This requires computing the sum

V = $1000 exp(–0.05) + $1000 exp(–0.10) + $1000 exp(–0.15) + $1000 exp(–0.20)

= $951.23 + $904.84 + $860.71 + $818.73

= $ 3,535.51.

(b) “Alice is to pay Bob 951.23 flat dollars at time 1, 904.84 flat dollars at time 2, 860.71 at time 3, and 818.73 flat dollars at time 4.”

2. Geometric Brownian Motion (GBM):

We introduce Geometric Brownian Motion (GBM) as a standard model for the price of a share of stock. (In this section it doesn’t matter whether we’re using actual dollars or flat dollars – or Euros or Yen, for that matter.)

The price of a share is constantly changing, and unpredictable. We assume that we know the price S0 at time 0, and that we want a probabilistic model for the price S(t) at time t for t≥0.

Let:

S(t) = price per share at time t

L(t) = ln ( S(t) ) = natural logarithm of the stock price

R(t1, t2) = L(t2) – L(t1) = change in log price from t1 to t2

( Note: R(t1, t2) = ln ( S(t2) / S(t1) ). )

Then the GBM model assumes that R(t1, t2) is normally distributed with these parameters:

mean= t

variance= σ2t

std. dev. =

where t = t2 – t1, and the parameters  and  depend on the particular stock, but not on the time interval.

We call  the drift parameter, and  the volatility parameter. The units of  are inverse time (usually “per year”). The units of  are more complicated: its square 2 has units of inverse time (per year) so the units of  are actually more like “year –1/2 ” or “per year1/2”. People find these units awkward to specify, and so usually characterize values of  as “annual values” or “monthly values” or “daily values” if they are actually year –1/2, month –1/2, or day –1/2 respectively.

A family of random variables S(t) (for all t ≥ 0) is called a Geometric Brownian Motion Process with parameters  and  if the corresponding random variables R(t1, t2) are distributed as above whenever 0 ≤ t1 ≤ t2, and when (watch this extra assumption) the variables R(t1, t2) corresponding to nonoverlapping intervals are independent. In this case we also say that the variables L(t) form a Brownian Motion Process.

Exercise. Suppose that a stock price is modeled by a GBM process with parameters =0.01 and σ=0.20, and suppose that the price at time t = 14 is S(14) = $ 40.0. What is the probability that S(30) > $60.00 ? (All values assume that time is measured in years.)

Solution.

We want:

P ( S(30) > 60 )= P( L(30) > ln 60 )

= P( L(30) – L(14) > ln(60) – L(14) ).

Note that L(14) = ln(S(14)) = ln(40), and L(30)–L(14) is R(14,30).

So,

P ( S(30) > 60 ) = P( R(14,30) > ln(60)-ln(40) ).

Now, R(14,30) is normally distributed with mean

t =  * 16 = 0.16

and standard deviation

= σ * sqrt(16) = σ * 4 = 0.80.

So we can standardize:

P ( S(30) > 60 ) = P( R(14,30) > ln(60)-ln(40) )

= 0.3795.

Notes:

(1) We would have gotten the same answer if we had said that S(0) = $40.00 and asked for P( S(16) > $60 ). All that mattered was the length of the time interval, 16 years. In the GBM model, all time intervals of the same length are equivalent.

(2)  is the cumulative normal distribution function, for which tables are widely available.

(3) It doesn’t matter whether the dollar amounts in this example are actual dollars or flat dollars, as long as they are consistent. We would probably want to use different values of  in the actual-dollar and flat-dollar cases, since stock prices grow more slowly when measured in flat dollars.

3. Expected value of S(t)

Assuming GBM with parameters  and σ, the expected value of S(t) (as viewed from time 0) is given by

E ( S(t) ) = .

This is easily computed if we know the parameters  and  and the initial stock price S(0).

Derivation. To see that this is true, assume that , , and S(0) = S0 are given. Then L0 = ln(S0) is also known, and for any t, we can find the distribution of L(t):

L(t) = L0 + R( 0, t ),

so Δt = t and L(t) is normally distributed with mean L0 + t and standard deviation . Its density function is therefore

S(t) depends on L(t), being given by ex when L(t) = x, so the expected value of S(t) can be computed as the integral of ex times the density function of L(t):

To evaluate this integral, the first step is to combine the two exponents:

Now, rewrite the combined exponent as a single quadratic, and complete the square. We eventually get

The last giant integral is just the integral of a normal distribution with mean and variance --- so the integral is just 1. We finally have

as desired.

The lognormal distribution. Note that S(t) is not normally distributed, but its logarithm L(t) is. In this case we say that S(t) is lognormally distributed or has a lognormal distribution. The mean and standard deviation of L(t) are called the underlying parameters of S(t).

4. The Royalty Story.

Suppose S is lognormal with underlying parameters  and σ, and suppose that the random variable X is given by

X = S if S ≥ K, and

0 if S < K.

Then

E ( X ) = .

This can be proven by evaluating an integral. The integral is complicated, so it is nice that the result can be summarized so simply. We can make the equation more self-contained by giving the random variable X a descriptive name, “S if S ≥ K.” Then the formula becomes

E (S if S ≥ K ) = .

Note that  doesn’t appear in the formula. Note also that  is the underlying standard deviation of S. If we use the notation above and apply this formula with S = S(t), then the underlying standard deviation is and the formula becomes

E (S(t)if S(t) ≥ K ) = .

5. Black-Scholes with a risk-neutrality assumption.

It now matters that we are working in flat dollars. We will be estimating the values of call options, and that is easier if we don’t need to deal with the present value function.

For now, we assume that every participant in the market is “risk-neutral”, which means that they value every asset according to the expected value of its future payoff.

Assume as above that a stock price S(t) follows GBM with parameters  and σ, and assume that the initial price is a known constant,

S(0) = S0.

(Thus, L(0) = ln(S0) is also known.)

The risk-neutrality assumption forces  = – (1/2) σ2. That’s because the time-t payoff from owning a share of stock is, stated as a random variable, just S(t). The expected value of S(t) is

But if everyone values the stock by its expected value, that means that everyone considers one share as equivalent to the right to receive E(S(t)) at time t, which in flat dollars is equivalent to receiving E(S(t)) now. But if one share is equivalent to receiving E(S(t)) now, then that means that S(0) must be equal to E(S(t), which means . That forces +(1/2)σ2 = 0, so  = – (1/2) σ2. (If we were working in non-flat dollars the conclusion would be  = r – (1/2) 2.)

A call option. Assume that a certain call option contract gives the holder the right to buy one share for K at time T, if that is favorable. Thus, the payoff at time T is

X =S(T) – K if S(T) ≥K, and

0otherwise.

The value K is usually stated publicly in actual dollars. We’ll use lower-case k for the flat-dollar equivalent, k = Ke-rT.

The Black-Scholes Formula. With the risk-neutrality assumption, the value of this option is equal to the expected value of its time-T payoff. That’s

V =E (X)

=E ( X1 ) – E ( X2 )

where X1 = S(T) if S(T) ≥k, 0 otherwise

and

X2 = K if S(T) ≥k, 0 otherwise.

(We’re working in flat dollars now, so we’re using k in place of K.)

Now E ( X2 ) is just K P(S(T)≥k), and E(X1) is given by the royalty formula applied to S(T). So:

V= – K P(S(T)≥k)

= – k P(S(T)≥k)

(since with risk-neutrality, S0 must equal E(S(T))).

We calculate:

S(T) ≥k means L(T) ≥ln(k)

means R(0,T) ≥ln(k) – ln(S0)

and R(0,T) is normal with

mean T = -(1/2)σ2T

stdev

P(S(T) ≥k)=P ( R(0,T) ≥ln(k) – ln(S0) )

The other probability is the same, with an extra σ2T term…

= .

So, we have:

V = S0 – K

which simplifies to

V = S0 – k(1)

This is the Black-Scholes formula for the value of a call option, at least in the case of flat dollars.

Note that  doesn’t appear. But that’s just because our risk-neutral assumption tells us that  = -(1/2)σ2, and we have substituted that value for  along the way.

7. Black-Scholes in ordinary dollars.

The dollar amounts in (1) are V, S0, and k. We have assumed that they are given in flat dollars. But outside this class, people will give us S0 and K in ordinary dollars, and they will expect us to compute V in ordinary dollars.

In the case of V and S0, it makes no difference. These quantities are measured at time 0, when flat dollars are the same as ordinary dollars.

But the strike price refers to a payment at time T. If the strike price is K ordinary dollars, then it is equal to k = Ke-rT flat dollars.

So, we can apply equation (1) to determine V, but we must use Ke-rT in place of k wherever it appears. With this change, (1) becomes

V = S0 – K (2)

which is the usual form of the Black-Scholes formula for call options.

Typical test question: Given S0, K, T, σ, and (if we’re not using flat dollars) r, calculateV.

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