Homework Assignments

Homework Assignments Week 1 (p 57) #4.1, 4., 4.3 Week (pp 58-6) #4.5, 4.6, 4.8(a), 4.13, 4.0, 4.6(b), 4.8, 4.31, 4.34 Week 3 (pp 15-19) #1.9, 1.1, 1.13, 1.15, 1.18 (pp 9-31) #.,.6,.9 Week 4 (pp 36-37) #3.1, 3.4, 3.5 Supp. #1, Week 5 (pp 76-79) #5., 5.3, 5.5, 5.6, 5.9, 5.16, 5.18, 5.0 Week 6 (pp 91-9) #6.1, 6., 6.5, 6.6 Week 7 Supp. #3, 4 Week 8 (pp 115-117) # 7., 7.3, 7.4, 7.7, 7.8, 7.10, 7.11, 7.1

Supplement Problems 3. Use a spreadsheet to find the call price for the following barrier option. The stock price at t = 0 is $65, the exercise price is $70, expiration date is 5 periods into the future. Assume the interest rate is 5%, u = 1.1 and d = 0.91. The barrier is set at $80. If the stock rises above $80 at any time at or before the expiration, the call loses all its value. Print your spreadsheet.

Supplement Problems #3 Time t= 0 1 3 4 5 Stock Stock 65 104.683 Strike 70 95.1665 Rate 5% 86.5150 86.6015 Expiratio n 5 78.6500 78.787 # periods 5 71.5000 71.5715 71.6431 Barrier 80 65.0000 65.0650 65.1301 Growth Factor 1.05171 59.1500 59.09 59.684 Discount Factor 0.9519 53.865 53.8803 u Value 1.1 48.981 49.0311 d Value 0.91 44.5737 q Value 0.74353 40.561

Supplement Problems #3 Euro Call 0.0000 0.0000 0.0000 0.0000 0.1383 0.4008 0.446 0.5670 1.6431 0.3114 0.6015 1.161 0.5673 0.819 0.0000 0.5813 0.0000 0.0000 0.0000 0.0000 0.0000

Supplement Problems 4. Suppose two risky assets A and B are trade: 10 140 A 100 B 100 80 40 What is the arbitrage free price of an option (with right but not obligation) which allows the exchange of A for B?

Supplement Problem #4 Solution: Let C t be the value of the option, t = 0, 1. The payoff of the option at t = 1 is (B 1 A 1 ) +. We have the following diagram: 0 C 0 0

Supplement Problem #4 Take α units of A and β units of B to replicate C: 10 140 A 100 B 100 α 80 β 40

Supplement Problem #4 We get the following diagram: 0 = 10α+ 140β C = 100 ( α+β) 0 0 = 80α+ 40β

Supplement Problem #4 0 = 10α+ 140β C = 100 ( α+β) 0 0 = 80α+ 40β Solve for α and β, we get α = -1/8 and β = 1/4 1 1 1 C = 100 100 1.5 0 + = = 8 4 8

Supplement Problem #4 Remark: We could also determine the risk-neutral probability and risk-free rate as follows: = + Q E [A ] (1 r)a 1 0 Q E [B ] = (1 + r)b 1 0 10q + 80(1 q) = (1 + r)100 140q + 40(1 q) = (1 + r)100

Simplify, Supplement Problem #4 Solve for q and r, 40q + 80 = (1 + r)100 100q + 40 = (1 + r)100 q = 3 1 r = 15

Supplement Problem #4 We can also find the price of call option using the risk-neutral probability and risk-free rate as follows: 1 Q 1 1 C = E [C ] = 0 0 1.5 0 1 + = 1+ r 1 3 3 1 + 15

Stock Price in One Step time=0 p time=δ us Stock S 1-p ds

Take u = e α = e σ α σ d = e = e p 1 1 µ = + σ 14

Stock price in n (=4) steps u 3 S u 4 S S us u S uds u ds ud S u 3 ds u d S ds d S ud 3 S d 3 S d 4 S

Section 6. The Multi-period Binomial Model 16

Two-period Binomial Model S uu S u q u Stock S q S ud = S du 1 q q d S d S dd

Two-period Binomial Model 1+ r (1 + r) Bond 1 (1 + r) 1+ r (1 + r)

Two-period Binomial Model No Arbitrage Condition S d < (1+r)S 0 < S u S ud < (1+r)S u < S uu S dd < (1+r)S d < S du

Two-period Binomial Model S uu C uu S u C u Stock q q u Call price S S ud = S du C 0 C ud = C du 1 q C d S d C dd q d S dd

Two-period Binomial Model 1 C = q C + (1 q) C 0 u d 1+ r 1 1 1 = q q C (1 q ) C + (1 q) q C (1 q )C u uu u ud d ud d dd 1+ r 1 r + + + 1+ r 1 = { qq C + q(1 q )C + (1 q)q C + (1 q) (1 q )C u uu u ud d ud d dd} (1 + r) C uu C u Call price C 0 C ud = C du C d C dd

Example Assume the stock price starts at $0 and in each of two time steps may go up by 10% or down by 10%. We suppose that each time step is three months long and the risk-free interest rate is 1% per annum compounded continuously. Let s consider a call option with a strike price of $1.

Example Time t= 0 0.5 0.5 Stock 0 4. Strike 1 Rate 1% 0 19.8 Expiration 0.5 18 # periods 16. Growth Factor 1.0305 Discount Factor 0.9704 probabili ty Path Product Discount u Value 1.1000 3.000 0.455uu 1.3615 d Value 0.9000.056 q Value 0.653 1.8 0.0000 0.4536ud &du 0.0000 0.0000 0.0000 0.109dd 0.0000 1.0000 1.3615 1.8

Multi-period Binomial Model Stock price Call price u 3 S u 4 S u 4 S K S us u S uds u ds ud S u 3 ds u d S u u 3 ds K d S K Finite number of one step models ds d S ud 3 S 0 d 3 S d 4 S 0

Backwards Induction We know the value of the option at the final nodes We work back through the tree using riskneutral valuation to calculate the value of the option at each node, testing for early exercise when appropriate 5

Stock price Call price S 3 us 3 q C ( us3) = max( us3 4 K C 3 ( S 3 ) 1 q ds C ( ) max(, 3 4 ds3 = ds3 K 1 C (S ) = qc (us ) (1 q)c (ds ), 3 3 4 3 4 3 1 r + 1+ r d + q = u d Apply one step pricing formula at each step, and solve backward until initial price is obtained., 0) 0) 1 C (S ) = qc (us ) + (1 q)c (ds ) k 1 k 1 k k 1 k k 1 1+ r

Multi-period Binomial Model Perfect replication is possible Market is complete Real probability is irrelevant Risk neutral probability dominates the pricing formula

Example: American Put Assume the stock price starts at $100 and in each of three time steps may go up by 10% or down by 10%. We suppose that each time step is three time period and the risk-free interest rate is 5% per period compounded continuously. Let s try to price an American put option with a strike price of $100.

Example American Put.74 0 0 0.59 0 0.53 6. 1 10 14. 19 0 0 10.9 7.1

Exotic Option: Knockout Assume the stock price starts at $100 and in each of three time steps may go up by 10% or down by 10%. We suppose that each time step is three time period and the risk-free interest rate is 5% per period compounded continuously. Let s try to price a knockout European call option struck at $105 with the barrier set at $95. If the stock price ever goes below $95, the option is worthless no matter what the stock price is at t = 3.

Example Stock price 11 133.1 100 110 99 108.9 90 89.1 81 7.9

Example Knockout Option 15.85 11.40 0 1.1.81 0 8.1 3.9 0 0

Exotic Option: Look-back Consider a look back option with a three-month expiration. At the end of three months, the buyer is paid the maximum value of the stock over the threemonth period. S = 100 0 T = 3 r = 5% = 0.05 u = 1. d = 0.9

Example r = 5% = 0.05 3 / 1 1 = = 3 1 u = 1. d = 0.9 1 0.05*( ) 1 e 0.9 q = 0.3475 1. 0.9

Example Stock price 144 17.8 100 10 108 19.6 90 97. 81 7.9

Exotic Option: Look-back 1 3 4 5 6 7 8 9 10 11 1 13 A B C D E F G H Lookback Options Stock 100 ProbabiPath Max Val ProbaProduct Rate 0.05 qqq uuu 17.80 0.04 7.4 delta t 0.083 qq(1-q) uud 144.00 0.08 11.33 u 1. qq(1-q) udu 19.60 0.08 10.0 d 0.9 qq(1-q) duu 19.60 0.08 10.0 q 0.347 q(1-q)^dud 108.00 0.15 15.98 q(1-q)^ddu 100.00 0.15 14.80 q(1-q)^udd 10.00 0.15 17.75 (1-q)^3 ddd 100.00 0.8 7.81 1.00 115.31 Option Value = exp(-rt)e^q(payoff) = $113.88

Section 6.3 Proof of the Arbitrage Theorem Omitted References Dynamic Asset Pricing Theory, Darrell Duffie, Princeton University Press (001) Mathematics of Financial Markets, Robert J. Elliott & P. Ekkehard Kopp, Springer (1999) 37

Chapter 7 The Black-Scholes Formula 38

Cox, J.C. Ross, S.A. Rubinstein, M. CRR Model

CRR Model p us 1+r Stock S Bond 1 1-p 1 S u 1+r

CRR Model p us 1+r Stock S Bond 1 1-p 1 S u 1+r q 1 1 (1 + r)s S (1 + r) = u = u 1 1 us S u u u

CRR Model u 3 S u 4 S S us u S S us 1 S u us S q = (1 + r) 1 u u 1 u 1 S u 1 S u 1 3 S u 1 S u 1 4 S u

Binomial Trees Binomial trees are frequently used to approximate the movements in the price of a stock or other asset In each small interval of time the stock price is assumed to move up by a proportional amount u or to move down by a proportional amount d 43

Tree Parameters Parameters p, u, and d are chosen so that the tree gives correct values for the mean & variance of the stock price changes in a risk-neutral world. Mean: Variance: r t e = p u + (1 p)d σ t = pu + (1 p)d e r t A further condition often imposed is u = 1/d 44

Tree Parameters When t is small a solution to the equations is u = e d = e σ σ a p = u a = e t r t t d d 45

Backwards Induction We know the value of the option at the final nodes We work back through the tree using riskneutral valuation to calculate the value of the option at each node, testing for early exercise when appropriate 46

Example: Put Option Consider a five-month European put option on a non-dividend-paying stock when the stock price is $50, the strike price is $50, the risk-free interest rate is 10% per annum, and the volatility is 40% per annum. S = 50 0 K = 50 r = 10% = 0.10 σ= 40% = 0.40 5 T = 5 months = = 0.416667 years 1 Number of period = 5 T 1 t = = = 0.083333 years 5 1 47

Example: Put Option The parameters imply u e e σ t 0.40 0.083333 = = = 1 d = = 0.890947 u 1.1401 Growth Factor a e e r t 0.10(0.083333) = = = 1 Discount Factor = = 0.991701 a 1.008368 48

Example: Put Option The risk neutral probability is q a d = = u d 1.008368-0.890947 = 0.507319 1.1401-0.890947 49

Example: Put Option Time 0.00 0.08 0.17 0.5 0.33 0.4 Stock Price 50 89.07 Strike 50 79.35 Interest Rate 0.1 70.70 70.70 Volatility 0.4 6.99 6.99 Time to Expiration 0.416667 56.1 56.1 56.1 # of steps 5 50.00 50.00 50.00 u Value 1.1401 44.55 44.55 44.55 d Value 0.890947 39.69 39.69 q Value 0.507319 35.36 35.36 Discount factor 0.991701 31.50 Growth factor 1.008368 8.07 50

Example: Put Option Option Tree 0.00 0.00 0.00 0.00 0.64 0.00.11 1.30 0.00 4.3 3.67.66 6.66 6.18 5.45 9.86 9.90 13.81 14.64 18.08 1.93 51

Example: American Put Option Consider a five-month American put option on a non-dividend-paying stock when the stock price is $50, the strike price is $50, the risk-free interest rate is 10% per annum, and the volatility is 40% per annum. 5

Example: American Put Option Time 0 1 3 4 5 Stock Price 50 89.07 Strike 50 79.35 Interest Rate 0.1 70.70 70.70 Volatility 0.4 6.99 6.99 Time to Expiration 0.416667 56.1 56.1 56.1 # of steps 5 50.00 50.00 50.00 u Value 1.1401 44.55 44.55 44.55 d Value 0.890947 39.69 39.69 q Value 0.507319 35.36 35.36 Discount factor 0.991701 31.50 Growth factor 1.008368 8.07 53

Example: American Put Option Option Tree 0.00 0.00 0.00 0.00 0.64 0.00.16 1.30 0.00 4.49 3.77.66 6.96 6.38 5.45 10.36 10.31 14.64 14.64 18.50 1.93 54

Example: American Put Option 89.07 0.00 79.35 0.00 70.70 70.70 0.00 0.00 6.99 6.99 0.64 0.00 56.1 56.1 56.1.16 1.30 0.00 50.00 50.00 50.00 4.49 3.77.66 44.55 44.55 44.55 6.96 6.38 5.45 39.69 39.69 10.36 10.31 35.36 35.36 14.64 14.64 31.50 18.50 8.07 1.93 55

Risk-Neutral Probability = r t a e for a nondividend paying stock = (r r ) t d a e for a stock index where r is the dividend (r r ) t p yield on the index = = a d u d f a e for a currency where r is the foreign risk-free rate a = 1 for a futures contract f d 56

Example: Put Option with Dividend Consider a five-month put option on a stock that is expected to pay a single dividend of $.06 during the life of the option. The initial stock price is $5, the strike price is $50, the risk-free interest rate is 10% per annum, the volatility is 40% per annum, and the ex-dividend date is in 3.5 months. 57

Example: Put Option with Dividend We first construct a tree to model S*, the stock price less the present value of future dividends during the life of the option. 58

Example: Put Option with Dividend Time 0 0.083333 0.166667 0.5 0.333333 0.4167 Stock Price 5 Stock Tree* Strike 50 89.064 Rate 10% 79.35151 volatility 40% 70.698 70.6980 Expiration 0.4167 6.988 6.988 # Steps 5 56.11916 56.119 56.119 Growth Factor 1.0084 49.999 49.999 49.999 Discount Factor 0.9917 44.54666 44.5467 44.5467 u Value 1.14 39.68873 39.68873 d Value 0.8909 35.3606 35.3606 q 0.5073 31.5044 Dividend.0600 8.0688 Ex-Div Date 0.917 59

Example: Put Option with Dividend Adding the present value of the dividend at each node leads to a recombine tree, which is a binomial model for S. Stock Tree 89.064 79.35151 7.7494 70.6980 65.061 6.988 58.13669 58.1706 56.119 5 5.03363 49.999 46.56419 46.5981 44.5467 41.7314 39.68873 37.41 35.3606 31.5044 8.0688 60

Example: Put Option with Dividend Now, use backward induction to price the American Put Option. Am Put 0 0 0 0 0.636065 0.16715 1.30183 0 4.44036 3.771458.664457 6.861097 6.3785 5.4533 10.15907 10.3117 14.45 14.6394 18.4956 1.931 61

Example: Put Option with Dividend 89.0641 0 79.35151 0 7.74945 70.69801 0 0 65.061 6.988 0.636065 0 58.13669 58.1706 56.11916.16715 1.301833 0 5 5.03363 49.999 4.44036 3.771458.664457 46.56419 46.5981 44.54666 6.861097 6.378519 5.453336 41.7314 39.68873 10.15907 10.3117 37.41 35.36056 14.45 14.63944 31.5044 18.4956 8.06876 1.9314 Node Time: 0.0000 0.0833 0.1667 0.500 0.3333 0.4167 6

References Hull, J. (008). Options, Futures, and Other Derivatives, 7th ed. Englewood Cliffs, NJ: Prentice- Hall. Software: DerivaGem for Excel http://www.rotman.utoronto.ca/~hull/software/ 63

Setting up For a fixed time T > 0, positive integer n, let Δ = T/n denote a small increment of time and suppose that, every Δ time units, the price of a security either goes up by the factor u with probability p or goes down by the factor d with probability 1 p. 64

By the CLT and with carefully chosen p, d, and u, the simpler model is going to approach a GBM as we let Δ become smaller and smaller. In order to accomplish this, we need to determine three parameters p, u, and d by matching the mean and variance (from the market). We could take d = 1/u. 65

Let S i = S(iΔ) be the stock price at time iδ for i= 0,1,,,n. Suppose X 1, X,, X n be i.i.d. of modified Bernoulli RVs with parameter p. Let Y n = X 1 + X + + X n. Write u = exp(α), then d = exp(-α). Then X X n n n = n 1 = n 1 = n 1 ( α ) S S u S e S e α X n 66

67 ( ) 1 1 1 1 ( ) ( ) 0 0 n n n n n n n X n n X X n X X n X X X Y S S e S e e S e Se Se α α α α α α + + + + = = = = = =

We need Note that lim αyn = Y N( Tm, Tσ ) n E[ αy ] = αe[ Y ] = αne[ X ] = nα( p 1) For large n, we would like to have n n n Var αy = α Var Y = α nvar X = nα p ( n) ( n) ( n) [1 ( 1) ] nα ( p 1) Tm nα [1 ( p 1) ] Tσ 68

Divide both sides by n, α( p 1) m α [1 ( p 1) ] σ By the 1 st eqn., the nd eqn. is equivalent to α m σ When n, Δ 0. Hence α σ 69

To summarize, we must have α ( p 1) m α σ Solve for α and p, α σ = σ p T n 1 m 1 + σ 70

Take u = e α = e σ α σ d = e = e p 1 m = 1 + σ 71

As n ->, Y S = S e Y N( mt, σ T ) T 0 [ ] 1 ( m+ σ ) T 0 0 P E S = Se = Se T 1 m + σ = µ 1 m = µ σ µ T 7

Hence u = e α = e σ p α σ d = e = e 1 1 µ σ = 1+ σ 73

In CRR Model, we use those parameters to construct the stock tree. The risk neutral probability can be calculated: q = e e σ r t σ t e e t σ t 74

q r t σ t e e = σ t σ t e e 1 [ 1 + + r t o( t) ] 1 σ + t σ + t o( t) = 1 1 1 + σ + t σ + t o( t) 1 + + ( ) σ t σ t o t 1 σ + t r σ + t o( t) = σ t + o( t) 1 r σ 1 + t + o( t) σ = + o( t) 1 1 r σ 1 = + t + o( t) σ 75

76 It is interesting to note that 1 1 1 = + p t σ σ µ 1 1 1 + q t r σ σ

Furthermore, Q Q Q E [ αy ] = αe [ Y ] = αne [ X ] = nα( q 1) n n n 1 r σ nσ t t σ = n t r 1 σ 1 = r σ T 77

And Q Q Q ( α n) = α ( n) = α ( n) Var Y Var Y nvar X = α n[1 (q 1) ] σ ( tn ) 1 = σ T 1 r σ σ t 78

To Summarize, Q S T = 0 1 Y N ( ), P µ σ T σ T 1 Y N ( ), Q r σ T σ T E [ S ] = Se T Se 0 Y rt 79

A Theorem from Lecture 4 Theorem: Suppose Y has a lognormal distribution with parameters μ and σ, and K > 0. Then + µ + σ / µ ln K µ ln K E ( Y K) = e Φ + σ KΦ σ σ Apply the theorem for S T / S 0 with parameters 1 K S µ= (r σ )T, varaice =σ T, and constant = 0

1 K 1 K + 1 ln ln + / r σ T r σ T r T T 0 Q ST K σ σ S K S0 E = e Φ + σ T Φ S0 S0 0 σ T S σ T S0 1 S0 1 ln + r σ T ln + r σ T rt = Φ K K + Φ K e σ T σ T S0 σ T Multiply both sides by S 0 e -rt S0 1 S0 1 ln + r σ T ln + r σ T K K e E S K S T Ke ( ) rt Q + rt = 0Φ + Φ T σ σ T σ T

Let S0 1 S0 1 ln + r σ T ln + r+ σ T K K d + = d1 = ω = + σ T = σ T σ T S0 1 ln + r σ T K d = d = = d1 σ T σ T Use the risk-neutral pricing formula [ ] ( ) rt Q rt Q + T T C0 = e E C = e E S K We get the following Black-Scholes formula for the value of the European call option.

Black-Scholes Formula C = S Φ d Ke Φ d 0 0 ( ) rt ( ) + Here d ± = ln S K 0 1 + r± σ σ T T z - x 1 - Φ(z) = e dx π

Black-Scholes Formula In the Black-Scholes formula, S is the spot price of the stock 0 K is the strike T is the time to maturity r is the risk - free rate σ is the volatility of the stock price

Black-Scholes Formula In the Black-Scholes formula, Ф(d + ) is the probability the option will be exercised under the numeraire measure induced by the asset S. Ф(d - ) is the probability the option will be exercised under the risk neutral measure.

Black-Scholes Formula In a single period model, the Black-Scholes formula can be written as C K = π π 1 + r 0 S ˆ 0 π is the probability the option will be exercised under the risk neutral measure is the probability the option will be exercised under the numeraire measure induced by the asset S. πˆ

Black-Scholes Formula Proof: Let I be the indicator function of the set {S 1 > K}. Then C = ( S K) = ( S K) I + 1 1 1 1 Q C0 = E [ C1] 1 + r 1 Q = E ( 1 ) 1+ S K r 1 Q = E S1 K I 1+ r + [( ) ]

Black-Scholes Formula 1 Q K C0 = E SI 1 E I 1+ r 1+ r [ ] Q[ ] Qˆ SI 1 K Q = SE 0 P ( S1 > K) S1 1+ r Qˆ K = SE [ ] 0 I π 1 + r K = S ˆ 0π π 1 + r

Example 7.1a Suppose that a security is presently selling for a price of 30, the nominal interest rate is 8% (with the unit of time being one year), and the security's volatility is.0. Find the noarbitrage cost of a call option that expires in three months and has a strike price of 34.

Example 7.1a The parameters for B-S formula are 3 T = = 0.5 1 r = 8% = 0.08 σ= 0.0 K = 34 S = 30 0

We get d + 1 Example 7.1a S0 1 30 1 ln + r+ σ T ln + 0.08 + 0. 0.5 K 34 = d 1 = ω = = = 1.10163143 σ T 0. 0.5 d = d = d σ T = 1.10163143-0. 0.5 = 1.0163143 Use Black-Scholes formula 0 0 ( rt + ) ( ) ( ) 0.08(0.5) e ( ) C = S Φ d Ke Φ d = 30Φ 1.00163143 34 Φ 1.10163143 = 0.383 $0.4

At each node: Upper value = Underlying Asset Price Lower value = Option Price Values in red are a result of early exercise. Example 7.1a Strike price = 34 Discount factor per step = 0.9960 37.51738 Time step, dt = 0.0500 years, 18.5 days 3.517376 Growth factor per step, a = 1.0040 35.87651 Probability of up move, p = 0.5336.014 Up step size, u = 1.0457 34.30741 34.30741 Down step size, d = 0.9563 1.145376 0.307413 3.80694 3.80694 0.649091 0.163386 31.3709 31.3709 31.3709 0.36643 0.086838 0 30 30 30 0.06144 0.046153 0 8.6879 8.6879 8.6879 0.0453 0 0 7.433 7.433 0 0 6.334 6.334 0 0 5.08605 0 3.98888 0 Node Time: 0.0000 0.0500 0.1000 0.1500 0.000 0.500

Example 7.1a Stock Price: 30.00 Volatility (% per year): 0.00% Risk-Free Rate (% per year): 8.00% Option Type: Binomial: European Option Data Time to Exercise: 0.500 Exercise Price: 34.00 Tree Steps: 0 Imply Volatility Put Call Price: 0.3537963 Delta (per $): 0.1530867 Gamma (per $ per $): 0.07959319 Vega (per %): 0.048948 Theta (per day): -0.004874 Rho (per %): 0.01108918

Example Find the price of an European contingent claim with the payoff function ϕ (S ) = S T n T

Example Solution: V = e E [S ] rt Q n 0 T n 1 r σ T+σ Tz rt Q = e E Se 0 1 n r σ T+ nσ Tz rt Q n = e E Se 0

Example V e E Se 1 n r σ T+ nσ Tz rt Q n = 0 0 = = = 1 (n 1)rT nσ T n Q nσ Tz 0 Se E e 1 1 (n 1)rT nσ T n σ T n e 0 Se Se n 0 1 (n 1)T r nσ

Example Find the price of an asset-or-nothing European call option with the payoff function ϕ (S ) = S I Where I is the function T T I 1 if S > K T = 0 if S K T

Example Solution: rt Q V = e E SI 0 T = = S Φ 0 ( d ) +

Black-Scholes Formula for European Put Here P = Ke Φ d S Φ d ( ) ( ) rt 0 0 + d ± = ln S K 0 1 + r± σ T σ T z - x 1 - Φ(z) = e dx π

B-S Formula for European Put Proof: z (, ) z x 1-1-Φ(z) = 1 - e dx π - x z x 1-1 - = e dx - e dx π - π - x 1 - = e dx π z 1 = = π Φ(-z) - e x - dx

B-S Formula for European Put By the put-call parity and the Black-Scholes formula for European call, P = C S + Ke 0 0 0 rt = S Φ d Ke Φ d S + Ke ( ) rt ( ) 0 + 0 ( ) rt = S ( ) 0 Φ d+ 1 Ke Φ d 1 ( ) rt = S ( ) 0 1 Φ d+ + Ke 1 Φ d rt = Ke Φ d S Φ d ( ) ( ) 0 + rt