DERIVATIVES: Valuation Methods and Some Extra Stuff Reference: John C. Hull, Options, Futures and Other Derivatives, Prentice Hall

K = Strike price, ST = Price of asset at maturity Payoffs from Options (Terminal Value) European Options, Cost of option is NOT included K = Strike price, ST = Price of asset at maturity Payoff ST K SHORT CALL LONG CALL SHORT PUT LONG PUT 2

Review… Valuation Approaches: Binomial Trees Black-Scholes Monte Carlo Methods Finite Differences

Binomial Trees

Basic Concept: RISK-NEUTRAL VALUATION Assume that the expected return from all traded assets is the risk-free rate Value the payoffs from the derivative by calculating their expected values, and then, use the risk-free rate to discount them

where s is the volatility and Dt is the length of the time step and, In summary… S0u p where s is the volatility and Dt is the length of the time step and, p = ( a - d) / ( u – d ) S0 S0d 1 - p 6 6

The Probability of an Up Move 7 7

Example: Put Option, American Style S0 = 50; K = 50; r =10%; s = 40%; T = 5 months = 0.4167; Dt = 1 month = 0.0833 The parameters imply u = 1.1224; d = 0.8909; a = 1.0084; p = 0.5073 1-p = .4927 8 8

Value of stock Value of option 9 9

[.5073 x 0 + .4927 x 5.45] exp{-.10 x .0833} = 2.66 There is no point in exercising the option at this node (S=K=50), thus early exercise is useless, better to wait, therefore, value is 2.66 10 10

(.5073 x 5.45 + .4927 x 14.64) exp{ -.10 x .083 } = 9.9 However, early exercise, generates 50 – 39.69 = 10.31; thus, the option should be exercised, the correct value for the option at the node is 10.31 (10.31 > 9.9, implies that it is better to exercise NOW!) 11 11

(.5073 x 6.38 + .4927 x 14.64) exp{ -.10 x .083 } = 10.36 NOTE, it is not always better to exercise the option when the option is in the money… 50 - 39.69 = 10.31… better NOT TO exercised at that node 12 12

Estimation of Delta at time Δt, that is, the sensitivity of the option price w/r to S [stock price] 13 13

Black-Scholes

The Black-Scholes Formulas 15 15

The N(x) Function N(x) is the probability that a normally distributed variable with a mean of zero and a standard deviation of 1 is less than x 16 16

Monte Carlos

MONTE CARLO SIMULATION We rely on the risk-neutral approach… We generate a random path and then calculate the payoff of the option at time t=T; we discount the value using the risk-free rate More formally… (i) Generate a random path for S in a risk-neutral world (ii) Determine the payoff at time t = T (iii) Repeat the previous steps many times (iiii) Calculate the mean of the payoffs (from the many paths generated) (iiiii) “Bring the mean” to t= 0 using the risk free rate

ΔS = [Exp Mean S] S Δt + σ S ε sqrt (Δt) Random sample from a distribution N(0, 1) If S denotes the price of a stock that it does not pay a dividend between [0, T], the [Exp Mean S] = risk free rate. Otherwise, see before

In summary… For European options, method is straightforward For American options, we need to take care of an additional complication…

Consider a 3-year American put option (non-dividend-paying stock) than can be exercised at the end of year 1, 2, or 3. Risk-free rate= 6%; S0 = 1.00; K= 1.1 (We also know the volatility). For simplicity, we will assume that we did a “mini Monte Carlo” (only 8 paths)

Cash flows had the option been exercised at t=3 But at the end of year 2, the put option was in the money for paths: 1, 3, 4, 6 and 7. Hence, we need to decide at this points if we exercise or continue… For these paths we assume that V = a + b S + c S2 S, stock value at t = 2; V is the value of continuing discounted to t=2 with the risk free rate

Cash flows had the option been exercised at t=3 V = a + b S + c S2 The values of S (t=2) are: 1.08; 1.07; .97; .77 and .84 The values of V (t=2) are 0.00; 0.07exp(-.06 x 1); 0.18exp(-.06 x 1) 0.20exp(-.06 x 1); 0.09exp(-.06 x 1) S, stock value at t = 2; V is the value of continuing discounted to t=2 with the risk free rate

We determined a, b and c. using a least-squares method V = a + b S + c S2 We determined a, b and c. using a least-squares method MIN sum(1 to 5) { Vi – a – b Si - c Si2 }2 The values of S (t=2) are: 1.08; 1.07; .97; .77 and .84 The values of V (t=2) are 0.00; 0.07exp(-.06 x 1); 0.18exp(-.06 x 1) 0.20exp(-.06 x 1); 0.09exp(-.06 x 1) S, stock value at t = 2; V is the value of continuing discounted to t=2 with the risk free rate

V = a + b S + c S2 We determined a, b and c. using a least-squares method MIN sum(1 to 5) { Vi – a – b Si - c Si2 }2 WE GET IN THIS CASE V = -1.070 + 2.98 S - 1.81 S2 S, stock value at t = 2; V is the value of continuing discounted to t=2 with the risk free rate

V = a + b S + c S2 V = -1.070 + 2.98 S - 1.81 S2 V gives the value, at point t=2, of continuing That is, for paths (1, 3, 4, 6 and 7) we get V = (0.0369 0.0461 0.1176 0.1520 and 0.1565)

That is, for paths (1, 3, 4, 6 and 7) we get V = (0.0369 0.0461 0.1176 0.1520 and 0.1565) But at t=2 the value of exercising is .02 .03 .13 .33 .26 (( 1.1- .84 = .26)) We should only exercise at t=2 for paths 4, 5 and 6

Payoffs assuming that we have exercised at t=3 OR t=2 depending on what is best We repeat the same procedure for time = 1 (when it was worth to exercise at t=1?) and we finally get the full payoff table

Full tableau of cash flows.. Thus, the value of the options is… (1/8) [ 0.07 exp(-0.06 x 3) + 0.17 exp(-0.06 x 1)+ …+ 0.22 exp(-0.06x1) ] = .1144

Finite Differences

Attack the equation directly….

Consider the case of a Put Option, let us assume that the life of the option is T and Smax is a value sufficiently high (put has no value) We want to find f (value of the option) solving the “real thing”

f = f ( t, S) Time discretization: 0, Δt, 2 Δt, ……, T (N intervals, index i) Stock-value discretization: 0, ΔS, 2 ΔS, ……, Smax (M intervals, index j)

Value of the option at time = T is fN,j = MAX (K – j Δ s, 0) Stock Value Index i, time; Index j, S Smax Value of the option at time = T is fN,j = MAX (K – j Δ s, 0) j =1, 2, …, M T Time

Value of the option is K when S = 0 f i,0 = K for i=0,…, N Stock Value Index i, time; Index j, S Smax Value of the option is K when S = 0 f i,0 = K for i=0,…, N T Time

Value of the option is 0 when S = Smax f i, M = 0 for i=0,…, N Stock Value Index i, time; Index j, S Smax Value of the option is 0 when S = Smax f i, M = 0 for i=0,…, N T Time

Stock Value Index i, time; Index j, S Smax We know the value of f (options) along the red lines T Time

Using a finite differences approach (explicit method)

Using a finite differences approach (explicit method)

Using a finite differences approach (explicit method)

Introducing the finite differences approximation into the B-S equation and re-arranging… We get the following approx. to the B-S equation…

With… And…

f i,j+1 f i,j f i+1,j f i,j-1

Stock Value Value of f is known We can set up a linear system of simultaneous eqs by using these points as i,j then we move to the left.. Etc etc Smax T Time Value of f is unknown