Financial Engineering 6

References: Villalobos, Luenberger, CBOE

Lecture 18

Introduction to Option Pricing

Lecture Topics • Quick Review • Option Pricing • Simple Lattice • Example • Financial Engineering

Definition of an Option • Option is a privilege sold by one party to another that offers the

buyer the right, but not the obligation, to buy (call) or sell (put) a security at an agreed-upon price during a certain period of time or on a specific date.

• Call Option is an option contract giving the owner the right, but not the obligation, to buy a specified amount of an underlying security at a specified price within a specified time.

• Put Option is an option contract giving the owner the right, but not the obligation, to sell a specified amount of an underlying security at a specified price within a specified time.

Definition of an Option • An Option is:

– A type of derivative. – A security, just like a stock, bond, or index. – A binding contract with strictly defined terms

and properties. – Insurance. – Hedge!

Option Pricing • From a previous lecture on forward contracts suppose that you

sell a forward short position on a commodity at certain price F0. • One way to minimize your risk is to buy the commodity at the

spot price S0 and store it to meet your obligation at time t. • Since you are achieving a perfect hedge, there is no risk and

the price of the contract should be just the price of carrying the commodity to time t (interest rate and storage).

• If we assume that the price of storage is negligible then the price of selling the contract should be S0R, where R is a risk free rate.

• Thus, pricing a forward contract is simple.

• The question we should ask is: can the same procedure be applied to pricing options?

Option Pricing • Since the option gives the holder the right, but not the

obligation to exercise, we can’t use the previous methodology. • For example, suppose that you sell a European call option for

one share of a stock with a strike price of $100 and a maturity date of one month from your selling date.

– Let’s assume that the spot price of the stock at the time we sell the option is $100.

– In this example, you could buy a share of the underlying stock just in case that the option holder decides to exercise the option.

• Now assume that at expiration the price of the underlying stock is $100.01 or higher, what was the cost for you?

• What if it is 99.99 or less?

Option Pricing • One of the main questions asked in Financial Engineering is the

“fair” or market price for an option. • One way to answer this question is by using Binomial Lattices. • First we observe than in a short period of time the underlying

stock can behave in only two ways: – Its price goes up. – Its price goes down.

• Since the price of the option is tied to the price of the underlying stock its value is determined (up to a point) by the movement of the underlying stock.

• We also observe that for an option to be attractive, its price must also have the same expected value (risk-neutral) that other investments have such as the stock itself or a risk-free investment.

• For example, consider three investments: A stock, a call option, and a risk-free investment.

Example • Consider the following set of transactions:

– Sell three call options at a price of C dollars each. – Buy two shares of the underlying stock at $100 each with a current

spot price of the stock at $100. – Borrow $163.64 at 10% until the maturity date of one period.

• The total cash flow at time zero is: 3C – 200 + 163.64 = 3C - 36.36

• The following diagram represent what can happen to the underlying stock and the call options

• Suppose that there are only two possible scenarios: – The price of the stock goes up to $120. – Or, it goes down to $90. – What is the fair price of the option under these scenarios?

S

Su

Sd

C

Cup

Cdown

Example • If the price of the stock goes up to $120, the total cash flow will be:

+ Selling the two shares of stock at 2 x 120 = 240. − Paying the price of the three options at 3 x (120 − 100) = 60. − Repaying the loan at 163.64 x 1.1) = 180. For a total of 240 − 60 − 180 = $0 total cash flow at the settlement.

• If the price of the stock goes down to $90, the total cash flow will be: + Selling the two shares of stock (2*90 = 180) - Paying the price of the three options (3*0=0) - Repaying the loan (163.64*1.1) = 180 For a total of 180 − 0 − 180 = $0 total cash flow at the settlement. Note, our total cost is $20 + 16.36 = 36.36.

• Under both scenarios the total cash flow is zero, so there is no risk associated with issuing the stock; only cost is the interest rate for the loan.

• Thus price of the option is C = 36.36 / 3 = 12.12 • We have determined the fair price of an option (not only European) with one

period before maturity.

Option Pricing Procedure

• In the previous example the only information that we needed to know was:

– The amount that the underlying asset must move up. – The amount that the underlying asset must move down. – The risk-free rate of return for the loan.

• In general the procedure to find the price of the option consists of: – Selling one call option. – Buying x units of the underlying asset. – Borrowing an amount b at the risk-free rate r.

Option Pricing • Lattices for a share of stock, an American call option, and a risk-free

investment:

• If the price of the stock goes up with a probability of p, then: – The final price of the stock is S multiplied by u or Su. – The final price of the call option is Max(Su - K, 0) where K is the

strike price. – The final price of the risk-free investment is R = (1 + r), where r is

the risk-free rate.

• If the price of the stock goes down with a probability of 1 – p, then: – The final price of the stock is Sd. – The final price of the call option is Max(Sd - K, 0). – The final price of the risk-free investment is R = (1 + r).

S

Su

Sd

p

1-p C

Max(Su-K,0)

Max(Sd-K,0)

p

1-p 1

R

R

p

1-p

Option Pricing • If we assume that the investor has the alternative to invest in a

combination of: – x units of the underlying stock. – b units of the risk-free investment. – Then for a risk-neutral investor to consider an option, the

return given by an option should match the combined return of the above investment.

• Based on this observation the following relationships follow:

Solving this series of linear equations we get the value of the equivalent call option as:

RbdxC RbuxC

d

u

+= +=

  

  

− −

+ − −

==+ du Cdu Ru

C du dR

R Cbx

1

Option Pricing • Expressed in a different form:

• Under the assumption that u > R > d, we can think of q as the probability of the underlying stock going up in the reference period, and result in the risk-neutral price for the call option.

– Note, q is not really a probability, but it takes on a value between 0 and 1.

• These formulas can be used in combination of binomial lattices to find the price of call options.

( )( )1 1

where

u dC qC q CR R d

q u d

= + −

− =

−

Example • Using the previous formula and a binomial lattice, let’s find the

price of option used in the prior example.

S = 100

Su = 120

Sd = 90

C

Cup = Max(Su - K, 0) = Max(20,0) = 20

Cdown = Max(Sd - K, 0) = Max(-10, 0) = 0

1

1 1.1 0.9 1.2 1.1 20 0

1.1 1.2 0.9 1.2 0.9 12.12

u d R d u R

C C C R u d u d

− −  = + − − 

− −  = + − −  =

K = 100

So, u = 1.2 and d = 0.9 R = 1 + 0.1 = 1.1

Parameters • Before we can use the previous formula, we need to determine

the parameters u and d.

• In order to determine the value of these parameters, we will assume that the stock prices follow the lognormal distribution.

• This allows us, among other things, to use the binomial lattice’s multiplicative behavior of total returns.

• So, let’s have a quick lognormal refresher.

• The Lognormal model for stock price assumes that in a small period of time, ∆t, the stock price changes by an amount that is normally distributed with parameters:

• Where S is the current stock price and µ is the instantaneous rate of return.

• In a small time ∆t, the natural logarithm ln(S) of the current stock price will change (Ito’s Lemma) by an amount that is normally distributed with parameters:

Lognormal Stock Prices • If: ( ) ( ) ( ) ( )2 21 1 0

0

ln Nor , ln ln Nor , s

s s s

µ σ µ σ  

≈ → ≈ +   

tS Deviation Standard ∆=σtS∆= µMean

( ) t∆+= 25.Mean σµ tDeviation Standard ∆= σ

The Lognormal Model • Let St = stock price at time t, then ln(St) is normally distributed

with parameters:

( )tS 20 5.lnMean σµ ++= tσ=Deviation Standard

1

ln −t

t

S S

( )25. σµ −

1

ln −t

t

S S

• If we average the values of we get an estimate of

• If we take the standard deviation of we get an estimate of σ.

Parameter Estimation • If a stock follows a lognormal distribution

– with parameters µ and σ – Current price of the stock is S – Price of the stock at any time t is St

• Then the mean and variance of St is given by:

[ ] ( )1222 −= ttt eeSSVar σµ[ ] tt SeSE µ=

Parameter Estimation • If the prices follow a lognormal distribution

– And period ∆t is chosen such that it is small with respect to 1 – Then the parameters of the binomial lattice can be selected

such that:

t v

p ∆  

  

+= σ2

1 2 1 teu ∆= σ ted ∆−= σ

 

  

  

  

 =

0

ln S S

Var Tσ 

  

  

  

 =

0

ln S S

Ev T

Expected growth rate (typically converted to annual)

Volatility (typically converted to annual)

Probability of the stock going up

Stock price up multiple

Stock price down multiple

Luenberger Example • Consider a stock:

– With volatility of its logarithm of σ = 0.20. – The current price of the stock is $62. – The stock pays no dividends.

• A certain call option on this stock has an expiration date 5 months from now and a strike price of $60.

• The current rate of interest is 10% compounded monthly.

• Let’s determine the value of the option assuming the option expires in one month.

C

Cu = Max(Su - K,0)

Cd = Max(Sd-K,0)= 0

Luenberger Example • ∆t = 1 month = 1/12

1.05943 .1

1 1.00833 12

tu e

R

σ ∆= =

= + =

0.9439

0.5577

td e R d

q u d

σ− ∆= = −

= = −

( )( ) ( )( )( ) 144.30684.55577.0 00833.1

1 1

1 =+=−+= du CqqCR

C

= Max((62)(1.05943) - 60, 0) = 5.684

=Max((62)(0.9439) - 60, 0) = 0

Multi-period Options • Usually the options involve more than one period to expiration

– The more periods used, the closer the estimation!

• If more than one period is involved we need to extend the lattice to include the extra periods.

• This involves first calculating the stock prices and then using these prices to calculate the prices of the options.

Suu

Sud

Sdd

S

Su

Sd

Cuu

C Cud

Cdd

Cu

Cd

Dell Example • Call Option for Dell Computers with an Expiration date of May

22, 2004. • Information on value of Option and stock taken on April 18,2004

from Yahoo.

• Let’s assume we have the following data:

Position Num OptSym Expire Days Strike Type IV Vol OI Buy 1 DLQEZ MAY04 35 32.5 Call 17.10% 272 12612 @ 3.1

Dell Computer Corp. Option Trade with 1X Entry

Debit Profit Max Profit Max Risk Delta (Shares) Gamma Vega Theta $310.00

$-10.00 Unlimited $-310.00 1.0 0.0494 $1.02 $-0.21

35.60 35.60 Unlimited% Unlimited% Downside Breakeven Upside Breakeven Max Profit/Max Risk Max Profit/Debit

S = 35.36 K = 32.5

0.3828ln 0

= 

  

  

  

 =

S S

Ev T 0

ln 0.7011T S

Var S

σ   

= =     

Example • Consider the previous option, and assume the expiration date

is in two weeks, S = 35.36, K = 32.50, and r = 4.32% per year.

• Choosing ∆T as one week (1/52) we have

• Now let’s find the value of a call option due in two weeks.

0.7011 1/52e e 1.10211tu σ ∆= = = .0432

1 1.000832 52

R = + =

0.7011 1/52e e 0.90735td σ− ∆ −= = = 0.47998 R d

q u d −

= = −

1 1 1 1 0.3828 1 / 52 0.53786

2 2 2 2 0.7011 v

p t σ    

= + ∆ = + =       

0.3828ln 0

= 

  

  

  

 =

S S

Ev T 0

ln 0.7011T S

Var S

σ   

= =     

Example

( )( ) ( )( )( )1 11 0.47998 10.45 (1 0.47998)(2.86) 6.50 1.000832u uu ud

C qC q C R

= + − = + − =

Cuu=Max(42.95-32.5,0) =10.45

Cud=2.86

Cdd=0

Cu=6.50

Cd=1.37

C=3.83

Suu=42.95

S=35.36 Sud=35.36

Sdd=29.11

Su=38.97

Sd=32.08

35.36 1.10211Su = × 35.36 1.10211 1.10211Suu = × ×35.36 0.90735Sd = ×

• See Excel file “Lecture 18 Example (Excel)”.

Example • Excel lattice vs drawn lattice

Cuu=Max(42.95-32.5,0) =10.45

Cud=2.86

Cdd=0

Cu=6.50

Cd=1.37

C=3.83

Suu=42.95

S=35.36 Sud=35.36

Sdd=29.11

Su=38.97

Sd=32.08

Week 0 1 2 Stock Value 35.36 38.97 42.95Up

32.08 35.36 29.11Down

Week 0 1 2 Option Value 3.83 6.50 10.45Up

1.37 2.86 0Down

Assignments • Finish reading chapters 13 and 14 in the 2nd edition.

• Finish reading chapters 11 and 12 in the 1st edition.