Financial Options Analysis
1 Canonical Example of a Single Period Market Model
There are three tradable securities: a stock S, a call option C and money M. Each security can be bought or sold in unlimited (real) quantities.
• Market for S.
– Current market price S0 = 100.
– Market price at the end of the period, S1, will be either
∗ S1(u) = 125 if the market for S is “up,” or ∗ S1(d) = 80 if the market for S is “down.” Probability that the market for S will be up is 0.80. – A person who buys a unit of S today will pay 100 today, and will receive either 125 or 80 depending on the market outcome. – A person who sells a unit of S today will receive 100 today, but will be obligated to buy back a unit of S at the end-of-period’s prevailing market price (125 or 80).
• Market for C.
– Current market price C0 — to be determined.
– A person who buys a unit of C today will pay C0 today. A person who sells a unit of C today will receive C0 today. – At the end of the period: ∗ A buyer of one unit of C will have the option but not the obligation to buy one unit of S from the seller for the pre-determined, agreed-upon price of 100. ∗ A seller of one unit of C will be obligated to supply a unit of S for the predeter- mined, agreed-upon price of 100 should the buyer wish to exercise the option.
• Market for M.
– Risk-free rate for the period is 10%. – Money can be borrowed or loaned at this rate. Regardless of the market state for S, at the end of the period: ∗ A person who borrowed one unit of M must pay back 1.10. ∗ A person who loaned one unit of M will receive 1.10.
1 • From the perspective of a buyer of C, what are the payoffs at the end of the period?
– If the price of S goes up to 125: ∗ Buyer will exercise the option to acquire S at the below market price of 100. ∗ After buying S for 100 and then selling it for 125, buyer nets 25. – If the price of S goes down to 80: ∗ Buyer will not exercise the option to acquire S at an above market price of 100. ∗ Buyer nets 0. – The payoffs can be expressed mathematically as
C1 = max{S1 − 100, 0}. (1)
– C pays off when S goes up.
• From the perspective of a seller of C, what are the payoffs at the end of the period?
– If the price of S goes up to 125, seller must pay 25. – If the price of S goes down to 80, seller pays nothing.
Questions:
• How much would you pay for a C?
• What is a fair market value for C0?
2 2 Valuation via Discounted Expectation
• What is the future (end-of-period) expected cash flow of C?
– Let p = (0.8, 0.2) denote the objective probability vector. Let C1 denote the random variable whose outcomes are the payoffs of C.
Ep[C1] = 0.8(25) + 0.2(0) = 20. (2)
• What is the present value at the risk-free rate of the expected future cash flow of C?
E [C ] 0.8(25) + 0.2(0) p 1 = = 18.18. (3) 1 + rf 1.10
• Is buying a C for $18 a good deal?
3 Analysis:
• Consider these market actions: 1. Sell 18 units of C. Collect 18(18) = 324. 2. Borrow 676. 3. Use the 1,000 to buy 10 units of S. 4. Buy “beer and pizza.” • What is the portfolio associated with these market actions and what it its cost?
– Portfolio = (10S, -18C, -676M). – Cost of portfolio = (10)(100) + (−18)(18) + (−676)(1) = 0.
• What are the payoffs at the end-of-the period based on these market actions?
Table 1: End-of-period payoffs.
Price of S goes up to 125 Price of S goes down to 80 Cash flow due to the C’s 18(-25) = -450 18(0) = 0 Cash flow due to the loan 1.1(676) = 743.60 1.1(676) = 743.60 Cash flow due to the S’s 10(125) = 1,250 10(80) = 800 Net cash flow 56.40 56.40
• What do you think of the market actions? Beer and Pizza justified? • Can you suggest an even better set of market actions? • Can such a market exist? Is there something wrong here? – NO. Not possible to make unlimited amounts of money risk-free with nothing at stake. This is one form of arbitrage.
– C0 cannot equal 18! – C0 is overpriced: Free money comes from selling it. (Sell an overpriced item!) – Actions of arbitrageurs — the folks who spot such free money opportunities — will drive down the price of C0 until no further free money can be made. • So “discounted expectation” does not yield the correct answer?! – Look at the payoffs associated with a C. Do they look risk-free? – Discounted expectation can work, but you have to use the correct discount rate. (More on this later!)
4 3 Valuation via Replication
• So, what is the correct value for C0?
– The payoffs of the portfolio (10S, −18C) are: ∗ Price of S goes up to 125: (10)(125) + (−18)(25) = 800. ∗ Price of S goes down to 80: (10(80) + (−18)(0) = 800. – Portfolio (10S, −18C) is risk-free. – By the Law of One Price the cost of the portfolio (10S, −18C) = 800/1.1 = 727.27.
– Cost of the portfolio (10S, −18C) = 10(100) + (−18)(C0).
– 727.27 = 1, 000 − 18C0 =⇒ C0 = 15.15.
• Where did the 56.40 come from?
– When C0 = 18, cost of the portfolio (10S, −18C) is only 676. It should have been 727.27, the present value of 800 at risk-free rate 10%.
– To set up a costless portfolio when C0 = 18 only had to borrow 676. Saved 727.27 − 676 = 51.27 = 18(18 − 15.15) risk-free . Future value of 51.27 at 10% = 1.1(51.27) = 56.40.
• Where did the (10S, −18C) come from?
– There are only two market states. – Only two “linearly independent” securities are needed to “span” any payoff vector. – Can form a portfolio of S and M to replicate the payoffs of C. – Basic idea: Find h and m so that (hS, mM) = C. – Portfolio (hS, mM) is called a replicating portfolio.
– Cost of the replicating portfolio is hS0 + m(1) = hS0 + m.
– Law of One Price implies that hS0 + m = C0.
5 • Executing the “game plan:” – State Equations: 125h + 1.1m = 25 (4) 80h + 1.1m = 0. (5) – Solution to the State Equations: 25 − 0 h = = 5/9 (6) 125 − 80 25 − 125(5/9) m = = −40.40. (7) 1.1 – Cost of the replicating portfolio (5/9S, −40.40M):
(5/9)(100) + (−40.40)(1) = 15.15 = C0. (8) – 1C = (5/9S, −40.40M). – 18C = 18(5/9S, −40.40M) = (10S, −727.27M). (Buy 18 of the replicating portfolios.) – 727.27M = (10S, −18C). • What is the correct discount rate? – Expected future value of C = 20.
– Price today is C0 = 15.15. – Discount rate is 20 − 1 = 0.32 or 32%. 15.15 – 20/1.32 = 15.15. • What are the portfolio weights in the replicating portfolio for C? – Value of S = 5/9(100) = 55.55. – Value of M = −40.40. – Value of C = Value of S + Value of M = 55.55 + −40.40 = 15.15. – w = Value of S = 55.55 = 3.6. S Value of C 15.15 w = Value of M = −40.40 = −2.6. M Value of C 15.15 – Sum of portfolio weights = 3.6 + −2.6 = 1, as it should. • What are the expected returns on S and M? 0.8(125)+0.2(80) – E[RS] = 100 − 1 = 0.16 or 16%. – E[RM ] = 0.10 or 10%. • What is the expected return on the replicating portfolio for C?
E[RC ] = wSE[RS] + wM E[RM ] = (3.6)(16%) + (−2.6)(10%) = 32%, as it should! (9)
6 • Any connection to CAPM?
– Sharpe ratio for S:
2 S1 S1 V ar(S1) σS = V ar( − 1) = V ar( ) = 2 . (10) S0 S0 S0 σS1 σS = (11) S0 p(0.8(125)2 + 0.2(80)2) − 1162 = = 0.18. (12) 100 0.16 − 0.10 Sharpe ratio for S = = 1/3. (13) 0.18 – Sharpe ratio for C:
2 C1 C1 V ar(C1) σC = V ar( − 1) = V ar( ) = 2 . (14) C0 C0 C0 σC1 σC = (15) C0 p(0.8(25)2 + 0.2(0)2) − 202 = = 0.66. (16) 15.15 0.32 − 0.10 Sharpe ratio for C = = 1/3! (17) 0.66 • Coincidence?
– Returns on S and C are perfectly positively correlated. – Think about the implications for the Capital Market Line defined by S.
7 4 Valuation via Replication: General Approach
• Given a security V whose payoffs are:
– V1(u) when the market state for S is up.
– V1(d) when the market state for S does down. • Security V is called a derivative security. It “derives” its value from the underlying security S.
• General State Equations to find replicating portfolio for V :
hS1(u) + (1 + rf )m = V1(u). (18)
hS1(d) + (1 + rf )m = V1(d). (19)
• Solution to the General State Equations:
V (u) − V (d) ∆V h = 1 1 = . (20) S1(u) − S1(d) ∆S V (u) − hS (u) m = 1 1 . (21) 1 + rf
• No-Arbitrage Price of V :
V0 = hS0 + m = cost of the replicating portfoio. (22)
8 Example 1 Consider the market for a derivative security P :
• A buyer of one unit of P will have the option but not the obligation to sell one unit of S for the pre-determined, agreed-upon price of 100.
• A seller of one unit of P will be obligated to supply a unit of S for the predetermined, agreed-upon price of 100.
• The payoffs to a buyer of P can be expressed as
P1 = max{100 − S1, 0}. (23)
– If S1 = 125, then P1 = 0.
– If S1 = 80, then P1 = 20. • P pays off when S goes down. (It acts as a hedge.)
• P is called a put option.
What is the no-arbitrage price P0?
9 Analysis:
• What are the State Equations to find the replicating portfolio for P ?
125h + 1.1m = 0 (24) 80h + 1.1m = 20. (25)
• What is the solution to the State Equations? 0 − 20 h = = −4/9 (26) 125 − 80 0 − 125(−4/9) m = = 50.50. (27) 1.1
• What is the cost of the replicating portfolio?
P0 = (−4/9)(100) + 50.50 = 6.06. (28)
• What is the discount rate so that discounted expectation yields the correct answer?
– Expected future value of P = 0.8(0) + 0.2(20) = 4.
– Price today is P0 = 6.06. – Discount rate is 4 − 1 = −0.34 or -34%. 6.06 – 4/0.66 = 6.06.
• What are the portfolio weights in the replicating portfolio for P ?
– Value of S = −4/9(100) = −44.44. – Value of M = 50.50. – Value of C = Value of S + Value of M = −44.44 + −50.50 = 6.06. – w = Value of S = −44.44 = −7.3. S Value of C 6.06 w = Value of M = 50.50 = 8.3. M Value of C 6.06 – Sum of portfolio weights = −7.3 + 8.3 = 1, as it should.
• What are the expected returns on S and M?
0.8(125)+0.2(80) – E[RS] = 100 − 1 = 0.16 or 16%. – E[RM ] = 0.10 or 10%. • What is the expected return on the replicating portfolio for P ?
E[RP ] = wSE[RS]+wM E[RM ] = (−7.3)(16%)+(8.3)(10%) = −34%, as it should! (29)
10 Example 2 The derivative security V has this payoff function: p V1(S1) = S1 − 76 + 2 max{S1 − 115, 0} + 5 max{82 − S1, 0}. (30)
What is the no-arbitrage price V0? Analysis:
• Payoffs = (27, 12).
• h = 1/3. m = −13.3.
• 20.
Example 3 Two “linked” derivatives are C80 and P 45 whose respective payoff vectors are:
C80 payoff vector = (45, 0). (31) P 45 payoff vector = (0, 45). (32)
Analysis:
• The payoff of the portfolio (1S, 1P 45) is (125, 125).
• This portfolio is risk-free. It must earn the risk-free rate of 10%.
• By the Law of One Price:
45 45 45 125/1.1 = S0 + P0 = 100 + P0 =⇒ P0 = 13.63. (33)
• The payoff of the portfolio (1P 45, 1C80) is (45, 45).
• This portfolio is risk-free. It must earn the risk-free rate of 10%.
• By the Law of One Price:
45 80 80 80 45/1.1 = P0 + C0 = 13.63 + C0 =⇒ C0 = 27.27. (34)
• 45 = 125 − 80 = S1(u) − S1(d). This implies h = +/ − 1. • Respective payoffs for C80 and P 45 can be expressed as
80 C1 (S1) = max{S1 − 80, 0}. (35) 45 P1 (S1) = max{125 − S1, 0}. (36)
11 5 Valuation via State Prices
• Reclassify states “up” and “down” as states 1 and 2, respectively.
• Two fundamental securities: E1 and E2.
– E1’s payoff vector is e1 := (1, 0). Its cost is denoted by y1.
– E2’s payoff vector is e2 := (0, 1). Its cost is denoted by y2.
– e1 and e2 are the unit vectors in the plane.
• y1 and y2 are called the state prices. y = (y1, y2) is called the state price vector.
• For a derivative security V whose payoffs are (V1(u),V1(d)): – Observe that
(V1(u),V1(d)) = V1(u)(1, 0) + V1(d)(0, 1) (37)
= V1(u)e1 + V1(d)e2. (38)
– Payoff of portfolio (V1(u)E1,V1(d)E2) is identical to V .
– (V1(u)E1,V1(d)E2) is a replicating portfolio for V . (It is not in “reduced” form in terms of the tradables, S and M.)
– No-arbitrage value V0 must equal cost of portfolio (V1(u)E1,V1(d)E2).
– Cost of replicating portfolio (V1(u)E1,V1(d)E2) is
V1(u) ∗ (cost of E1) + V1(d) ∗ (cost of E2) = V1(u)y1 + V1(d)y2. (39)
• Key Insight: How to find the no-arbitrage value V0 for a derivative security V ? – Pricing via replication: ∗ Find the replicating portfolio and determine its cost. – Pricing via state prices:
∗ Find the values for y1 and y2 first. ∗ Price V via (39)! ∗ No replicating portfolio required! (Though it will be needed to hedge.)
12 Example 4 Let’s return to our canonical example.
• What are the state prices?
– Via replication:
∗ State equations to replicate E1:
125h + 1.1m = 1 (40) 80h + 1.1m = 0. (41)
1 −80/45 ∗ Replicating portfolio is ( 45 S, 1.1 M). 1 80/45 ¯ ∗ Cost of replicating portfolio is 45 (100) − 1.1 = 0.60 = y1. ∗ State equations to replicate E2:
125h + 1.1m = 0 (42) 80h + 1.1m = 1. (43)
−1 125/45 ∗ Replicating portfolio is ( 45 S, 1.1 M). −1 125/45 ¯ ∗ Cost of replicating portfolio is 45 (100) + 1.1 = 0.30 = y2. – Via “linked” derivatives:
∗ Payoff vector of 45 units of E1 is (45, 0). ∗ (45, 0) is the payoff vector for C80 of Example 3. 80 ∗ Cost of 45 units of E1 = 45y1 must equal the cost of C = 27.27. ∗ y1 = 27.27/45 = 0.60. ∗ Payoff vector of 45 units of E2 is (0, 45). ∗ (0, 45) is the payoff vector for P 45 of Example 3. 45 ∗ Cost of 45 units of E2 = 45y2 must equal the cost of P = 13.63. ∗ y2 = 13.63/45 = 0.30. • Using state prices, what is the no-arbitrage value of C?
C0 = 25(0.60) + 0(0.30) = 15.15. It checks! (44)
• Using state prices, what is the no-arbitrage value of the derivative security in Example 2?
V0 = 27(0.60) + 12(0.30) = 20. It checks! (45)
13 • What is the sum of the state prices? [0.90]¯
• What is the reciprocal of the sum of the state prices? [1.1]
• Sum of the state prices always equals the reciprocal of 1 + the risk-free rate!
– The portfolio (1E1, 1E2) pays off 1, risk-free.
– Its cost must equal 1/(1 + rf ).
– Cost of portfolio is y1 + y2.
– Sum of replicating portfolios for E1 and E2: 1 −80/45 −1 125/45 1 ( S, M) + ( S, M) = (0S, M), as it should! (46) 45 1.1 45 1.1 1.1 • Once you know one state price, you can easily find the other state price! WARNING: Assumes you found the first state price correctly!
14 Example 5 Consider this two-state, single-period market model.
• S0 = 36.
• S1(u) = 54 and S1(d) = 24. • Risk-free rate is 5%.
What is the value of the derivative security V whose payoff vector is (111, 61)? Value using the replicating portfolio approach and the state-price approach.
Analysis:
• Valuation via replication.
– What are the values for h and m?[h = 5/3, m = 20]
– What is the value of V0 = hS0 + m? [80] • Valuation via state prices.
– Via replication: 1 −24/30 ∗ Replicating portfolio for E1:( 30 S, 1.1 M). y1 = 0.438122381. −1 54/30 ∗ Replicating portfolio for E2:( 30 S, 1.1 M). y2 = 0.5142585714. ∗ Value of 1/(y1 + y2)? [1.05] ∗ Value of V1(u)y1 + V1(d)y2 =? [80] – Via linked derivatives:
∗ Value of S1(u) − S1(d)? [30]. ∗ Let P 54 denote a security whose payoff vector is (0, 30). ∗ Payoff vector of the portfolio (1S, 1P 54)? [(54, 54)] ∗ No-arbitrage price of P 54? [51.42857143]
∗ Value for y2? [0.5142585714] ∗ Value for 1/(1 + rf )? [0.952380952] ∗ Value for y1? [0.438122381] ∗ Value of V1(u)y1 + V1(d)y2? [80]
15 6 Valuation via Discounted Expectation
• State prices y1 and y2 are the costs of acquiring respective payoff vectors (1, 0) and (0, 1). No arbitrage implies the state prices are positive and unique.
• Let V denote a derivative security whose payoff vector is (V1(u),V1(d)). Define q := y1/(y1 + y2) > 0. We have shown the no-arbitrage value of V is
V0 = y1V1(u) + y2V1(d) (47) y1 y2 = (y1 + y2) V1(u) + V1(d) (48) y1 + y2 y1 + y2 1 = {qV1(u) + (1 − q)V1(d)} (49) 1 + rf 1 = Eq[V1]! (50) 1 + rf • You can use discounted expectation to arrive at the no-arbitrage value for V .
– If you discount by 1 + rf , you must use a special probability vector, (q, 1 − q). – You do not need to know the correct discount rate (as this depends on V ’s payoffs). – You do need the correct probabilities — find these and pricing is easy.
– Keep in mind that (q, 1 − q) = (1 + rf ) ∗ (y1, y2). These probabilities are just a scaled version of the state-prices. Once you know q, it’s easy to find the state prices! • Under q, all securities earn the risk-free rate:
V1 Eq[V1] (1 + rf )V0 Eq[RV ] = Eq[ − 1] = − 1 = − 1 = rf ! (51) V0 V0 V0 • A risk-neutral investor does not care about risk, so will not pay a risk-premium. In a “risk-neutral world” all securities earn the risk-free rate. This is why the correct probabilities are called the risk-neutral probabilities. • Under q, the underlying security S also earns the risk-free rate. This pins down q: 1 S0 = Eq[S1] (52) 1 + rf 1 = [S1(u)q + S1(d)(1 − q)] (53) 1 + rf (1 + rf )S0 = S1(d) + q(S1(u) − S1(d)) (54) (1 + r )S − S (d) q = f 0 1 . (55) S1(u) − S1(d)
• Observe that S0 = S1(u)y1 + S1(d)y2 and y1 + y2 = 1. Solve for y1 and y2!
16 Example 6 Let’s return to the canonical example.
• What are the risk-neutral probabilities?
1.1(100) − 80 q = = 2/3 (56) 125 − 80 = (1 + rf )y1 = 1.1(0.60). (57) 1 − q = 1/3 (58)
= (1 + rf )y2 = 1.1(0.30). (59)
• Using the risk-neutral probabilities, what is the no-arbitrage value for C?
2/3(25) + 1/3(0) C = = 15.15. It checks! (60) 0 1.1
Example 7 Use the risk-neutral probabilities to value the derivative V of Example 5. [q = 0.43.]
17 7 Terminology
A call or put option on an underlying asset S are defined by these characteristics:
• Final payoffs are:
– For a call option: CT = max{ST − K, 0}.
– For a put option: PT = max{K − ST , 0}. • K is called the strike or exercise price.
• T is called the maturity.
• European style option:
– Option can only be exercised at maturity.
• American style option:
– Option can be exercised at any time. – Option holder must decide when to exercise. – Option holder can always sell the option, which is different than exercising it. – Intrinsic value cannot be negative when exercised; otherwise option holder will not exercise. – Option holder receives the option’s intrinsic value when exercised.
∗ For a call option: the payoff at time t is St − K. ∗ For a put option: the payoff at time t is K − St.
18 8 Multi-Period Market Model
We extend the single-period canonical market model to multiple periods.
• In the single-period model, the stock price S either goes up to 125 from 100 or down to 80 from 100.
• As an alternative description, one can say that the stock price S
– goes up by a factor of U = 1.25, or – down by a factor of D = 0.80.
• Binomial Lattice is the most common model of stock price movements::
– Let ∆t denote the length of the time period, measured in years. ∗ ∆t = 0.25 coincides with a 3-month time period. ∗ ∆t = 1/52 coincides with a weekly time period. – Let S(t) denote the stock price at time t. – In each period the stock price S either goes up by a factor of U > 1 or down by a factor of D < 1. ∗ Given S(t), S(t + ∆t) −→ (U ∗ S(t),D ∗ S(t)). – A standard choice for the parameters U and D:
∗ Let σ denote the standard deviation of the ln returns, ln(RS). σ is called the log-volatility. √ √ ∗ Set U = eσ ∆t and D = 1/U = e−σ ∆t.
19 Example 8 For the multi-period canonical example, price a two-period European call option with strike price 100. Determine the self-financed replicating portfolios.
Analysis:
Table 2: Binomial Lattice valuation tree.
0 1 2 S0 = 100 S1(u) = 125 S2(uu) = 156.25 C0 = 20.661 C1(u) = 34.09 C2(uu) = 56.25 RP0 = (0.75S, −55.096M) RP1(u) = (1S, −90.90M)
S1(d) = 80 S2(du) = S2(ud) = 100 C1(d) = 0 C2(ud) = C2(du) = 0 RP1(d) = ∅
S2(dd) = 64 C2(dd) = 0
4/9(56.25) + 4/9(0) + 1/9(0) Eq[C2] Note: 20.661 = = 2 . (61) 1.21 (1 + rf )
Example 9 For the multi-period canonical example, price a two-period European put option with strike price 100. Determine the self-financed replicating portfolios.
Analysis:
Table 3: Binomial Lattice valuation tree.
0 1 2 S0 = 100 S1(u) = 125 S2(uu) = 156.25 P0 = 3.306 P1(u) = 0 P2(uu) = 0 RP0 = (−0.242S, 27.548M) RP1(u) = ∅
S1(d) = 80 S2(du) = S2(ud) = 100 P1(d) = 10.90 P2(ud) = P2(du) = 0 RP1(d) = (−1S, 90.90M)
S2(dd) = 64 P2(dd) = 36
4/9(0) + 4/9(0) + 1/9(36) Eq[P2] Note: 3.306 = = 2 . (62) 1.21 (1 + rf )
20 Example 10 For the multi-period canonical example, price a two-period European call option with strike price 80. Determine the self-financed replicating portfolios.
Analysis:
Table 4: Binomial Lattice valuation tree.
0 1 2 S0 = 100 S1(u) = 125 S2(uu) = 156.25 C0 = 35.35 C1(u) = 52.27 C2(uu) = 76.25 RP0 = (0.8922S, −53.872M) RP1(u) = (1S, −72.72M)
S1(d) = 80 S2(du) = S2(ud) = 100 C1(d) = 12.12 C2(ud) = C2(du) = 20 RP1(d) = (0.5S, −32.32M)
S2(dd) = 64 C2(dd) = 0
4/9(76.25) + 4/9(20) + 1/9(0) Eq[C2] Note: 35.35 = = 2 . (63) 1.21 (1 + rf )
Example 11 For the multi-period canonical example, price a two-period European put option with strike price 80. Determine the self-financed replicating portfolios.
Analysis:
Table 5: Binomial Lattice valuation tree.
0 1 2 S0 = 100 S1(u) = 125 S2(uu) = 156.25 P0 = 1.4692 P1(u) = 0 P2(uu) = 0 RP0 = (−0.1077S, 12.244M) RP1(u) = ∅
S1(d) = 80 S2(du) = S2(ud) = 100 P1(d) = 4.84 P2(ud) = P2(du) = 0 RP1(d) = (−0.4S, 40.40M)
S2(dd) = 64 P2(dd) = 16
4/9(0) + 4/9(0) + 1/9(16) Eq[P2] Note: 1.4692 = = 2 . (64) 1.21 (1 + rf )
21 9 European Put-Call Parity
European Put-Call Parity describes a fundamental relationship between the values of a European call and put options on the same underlying, non-dividend paying stock with the same maturity and same strike price. Knowledge of one price determines the other.
• The portfolio (1S, 1P, −1C) always pays off the strike price K at time T . (Verify this.)
• By the Law of One Price:
S0 + P0 − C0 = Present Value of K. (65)
Example 12 Consider Examples 8 and 9.
• S2 + P2 − C2 = 100 in all states.
• S0 + P0 − C0 = 100 + 3.306 − 20.611 = 82.645 = 100/1.21.
Example 13 Consider Examples 10 and 11.
• S2 + P2 − C2 = 80 in all states.
• S0 + P0 − C0 = 100 + 1.4692 − 35.35 = 66.116 = 80/1.21.
Example 14 Smith knows that the value of a 6-month European call option with strike price K = 24 on a non-dividend paying stock is 6.80. The current value of the stock is 26. He wishes to price a European put option on the same stock with the same strike price and maturity. He knows that the risk-free rate is 2.50% per year, but he is stuck because he does not know the log-volatility σ upon which he would calculate U and D for a binomial lattice calculation. He comes to you for help. What say you? [4.506]
22 10 An American Option
• An American option permits the holder to exercise the option at any date.
• At each time t, the holder of an American option must decide whether to
– exercise the option and receive the option’s intrinsic value, or – do not exercise.
• The holder will choose the action that maximizes the value of the option at that time.
• Keep in mind the holder can always sell the option.
Example 15 Price a two-period American put option on a non-dividend paying stock with these characteristics:
• S0 = 100. • Price path follows a binomial lattice with U = 1.3 and D = 1/1.3.
• Period length = 1 year.
• Risk-free rate is 6% per year.
• Strike price K = 100.
Analysis:
Table 6: Binomial Lattice valuation tree.
0 1 2 S0 = 100 S1(u) = 130 S2(uu) = 169 P0 = 9.845 P1(u) = 0 P2(uu) = 0
S1(d) = 76.923 S2(du) = S2(ud) = 100 P1(d) = max{100 − 76.923, 17.417} = 23.077 P2(ud) = P2(du) = 0 0.5478(0)+0.4522(40.828) 17.417 = 1.06
S2(dd) = 59.172 P2(dd) = 40.828
23 11 Asset Price Dynamics
We price an American put option on a non-dividend paying stock.
• S(0) = 36. σ = 0.30.
• rf = 8%. • Maturity is 5 months.
• K = 40. √ √ • U = e0.3 1/12 = 1.0905 or U = e0.3 1/24 = 1.0632.
Table 7: Valuation using one-month intervals.
0 1 2 3 4 5 36.00 39.26 42.61 46.68 50.90 55.51 33.01 36.00 39.26 42.81 46.68 30.27 33.01 36.00 39.26 Stock prices 27.76 30.27 33.01 25.46 27.76 23.35 0 1 2 3 4 5 4.74 2.69 1.10 0.17 0.00 0.00 6.99 4.43 2.10 0.36 0.00 9.73 6.99 4.00 0.74 Option values 12.24 9.73 6.99 q = 0.5169 14.54 12.24 16.65
Table 8: Valuation using half-month intervals.
0 1 2 3 4 5 6 7 8 9 10 36.00 38.27 40.69 43.26 45.99 48.90 51.98 55.27 58.76 62.47 66.41 33.86 36.00 38.27 40.69 43.26 45.99 48.90 51.98 55.27 58.76 31.85 33.86 36.00 38.27 40.69 43.26 45.99 48.90 51.98 29.96 31.85 33.86 36.00 38.27 40.69 43.26 45.99 28.18 29.96 31.85 33.86 36.00 38.27 40.69 26.50 28.18 29.96 31.85 33.86 36.00 24.93 26.50 28.18 29.96 31.85 Stock prices 23.45 24.93 26.50 28.18 22.06 23.45 24.93 20.75 22.06 19.51 0 1 2 3 4 5 6 7 8 9 10 4.73 3.26 2.03 1.08 0.45 0.11 0.00 0.00 0.00 0.00 0.00 6.30 4.57 3.04 1.76 0.80 0.22 0.00 0.00 0.00 0.00 8.15 6.21 4.40 2.78 1.42 0.46 0.00 0.00 0.00 10.04 8.15 6.14 4.23 2.43 0.95 0.00 0.00 11.82 10.04 8.15 6.14 4.00 1.95 0.00 13.50 11.82 10.04 8.15 6.14 4.00 15.07 13.50 11.82 10.04 8.15 Option values 16.55 15.07 13.50 11.82 q = 0.5119 17.94 16.55 15.07 19.25 17.94 20.49
24 Approximate Value for q:
• We calculate the value of a call option on a non-dividend paying stock by: – Modeling the price process via an N-step binomial lattice. – Computing the risk-neutral probability q. – Computing the discounted expected value of the option’s payoff at time T using q. • Risk-neutral probability q is the unique value that ensures that the discounted expectation of next-period’s stock prices using the risk-free rate of return rf ∆t always equals the current price, namely,
q(USk−1) + (1 − q)(DSk−1) Sk−1 = . (66) 1 + rf ∆t • Unique value for q is (1 + r ∆t)S − DS (1 + r ∆t) − D q = f k−1 k−1 = f . (67) USk−1 − DSk−1 U − D √ √ • Since U = eσ ∆t, D = e−σ ∆t and ∆t ≈ 0, we may use the approximation 1 + x + x2/2 for ex in (67) to obtain √ (1 + r ∆t) − (1 − σ ∆t + σ2∆t/2) q ≈ √ f √ (68) (1 + σ ∆t + σ2∆t/2) − (1 − σ ∆t + σ2∆t/2) √ (r − σ2/2)∆t + σ ∆t = f √ (69) 2σ ∆t 1 ν √ ≈ (1 + ∆t), (70) 2 σ where 2 ν := rf − σ /2. (71) • It can be shown that as the number of steps of the binomial lattice goes to infinity:
S(T ) 2 – The distribution of S(0) is lognormal with parameters ν = (rf − σ /2) and σ: S(T ) ln ∼ N(νT, σ2T ). (72) S(0)
S(T ) µT 2 – Expected growth rate of the stock E[ S(0) ] = e , where µ = ν + σ /2 = rf . Under q the expected growth rate of the stock is the risk-free rate! – Value of a European call option on a non-dividend paying stock is given by Q −rT E [e Max(ST − K, 0)] (73) when the underlying distribution Q is given in (72).
25 Table 9: Expected returns and expected ln returns using half-month intervals.
s10 P {S(10) = s10} s10/S(0) ln s10/S(0) 66.41 0.001235524 1.844802887 0.612372436 58.76 0.011780805 1.632149648 0.489897949 51.98 0.050548837 1.444009273 0.367423461 45.99 0.128529724 1.277556147 0.244948974 40.69 0.214469383 1.130290315 0.122474487 36.00 0.245397552 1.000000000 0.000000000 31.85 0.194990143 0.884728485 -0.122474487 28.18 0.106242502 0.782744492 -0.244948974 24.93 0.037988595 0.692516349 -0.367423461 22.06 0.008049416 0.612688940 -0.489897949 19.51 0.000767517 0.542063358 -0.612372436 Avg = 1.033844090 Avg = 0.014574500
Calculations:
• Theoretical distribution:
– ln S(t)/S(0) = ln S(t) − ln S(0) ∼ N(νt, σ2t). – E[ln S(t)/S(0)] = νt. ν = the expected continuously compounded rate of return. – E[S(t)/S(0)] = e(ν+0.5σ2)t := eµt. µ = the expected growth rate = rf under the risk-neutral probability distribution. – P {S(t) ≤ s} = P { ln S(t)√/S(0)−νt ≤ ln s/S√(0)−νt = Φ( ln s/S√(0)−νt ). σ ∆t σ ∆t σ ∆t 2 2 – ν = rf − 0.5σ = 0.08 − 0.08(0.30 ) = 0.035. – P {S(5/12) ≤ 38} = Φ( ln 38/36−√0.035(5/12) ) = Φ(0.20389) = 0.5808. 0.30 1/24 – E[ln S(5/12)/S(0)] = νt = (0.035)(5/12) = 0.014583333.
• Binomial lattice distribution: ! 10 – P {S(10) = s = S(0)U uD10−u = S(0)U 2u−10} = qu(1 − q)10−u. 10 u √ √ – U = eσ ∆t = e0.3 1/24. √ √ σ ∆t(2u−10) – s10/S(0) = e . ln s10/S(0) = σ ∆t(2u − 10). – Values for q: √ ν 0.035 p ∗ Theoretical value q = 0.5(1 + σ ∆t) = 0.5(1 + 0.30 1/24) = 0.5119. Used in table. (1+rf ∆t)−D ∗ Risk-neutral value q = U−D = 0.5123. – P {S(10) ≤ 38} = 0.5934. – E[ln S(10)/S(0)] = 0.014574500.
26 12 Asset Price Dynamic Examples
• Time is measured in years so that t = 1 corresponds to 1 year from now.
• A stock like Citibank has an annual volatility σ ≈ 28%. A stock like Microsoft has an annual volatility σ ≈ 40%.
• The variance of the natural logarithm of the price process is proportional to time. The standard deviation is proportional to the square-root of time.
Example 16 Suppose the expected proportional rate of return on Citibank or Microsoft stock is µ = 12% per year or 1% per month.
• For t = 1/12 (1 month):
2 2 – ln (St/S0) = Xt ∼ N([0.12 − .28 /2] ∗ (1/12), 0.0808 ).
– A positive 3σ event corresponds to ln (St/S0) = 0.00673 + 3(.0808) or St = 1.283S0.
– A negative 3σ event corresponds to ln (St/S0) = 0.00673 − 3(.0808) or St = 0.790S0. • When t = 3/12 (3 months):
2 2 – ln (St/S0) = Xt ∼ N([0.12 − .28 /2] ∗ (3/12), 0.14 ).
– A positive 3σ event corresponds to ln (St/S0) = 0.0202 + 3(.14) or St = 1.553S0.
– A negative 3σ event corresponds to ln (St/S0) = 0.0202 − 3(.14) or St = 0.670S0. • For Microsoft the corresponding numbers are:
– St = 1.419S0 and St = 0.710S0 for t = 1/12.
– St = 1.847S0 and St = 0.556S0 for t = 3/12.
Example 17 Suppose S0 = 100, and assume µ = 0.12 and σ = 0.40.
• The distribution of ln St is N(ν, σ2), where ν = µ − σ2/2 = 0.12 − (0.40)2/2 = 0.04. S0 • Probability that the stock price will exceed 120 by time t = 1/12 (end of month)?
ln St/S0 − νt ln(1.2) − 0.04/12 P (St > 120) = P ( √ > ) (74) σ t 0.4p1/12 = 1 − Φ(1.55) = 0.0657. (75)
27 13 Black-Scholes European Call Option Pricing Formula
• It is possible to derive a closed-form analytical solution for
Q −rT E [e Max(ST − K, 0)], (76)
the value of a European call option on a non-dividend paying stock.
• It is known as the famous Black-Scholes formula.
• To simplify the derivation to follow we express S(T ) = S(0)eX where X ∼ N(a, b2). Upon substitution, (76) may be expressed via the following equalities:
Z +∞ −rT x 1 − 1 ( x−a )2 = e Max(S0e − K, 0)√ e 2 b dx (77) −∞ 2πb Z +∞ −rT x 1 − 1 ( x−a )2 = e (S0e − K)√ e 2 b dx (78) ln K/S0 2πb Z +∞ Z +∞ −rT 1 x− 1 ( x−a )2 −rT 1 1 (− x−a )2 = S0e √ e 2 b dx − Ke √ e 2 b dx . (79) ln K/S0 2πb ln K/S0 2πb
• By completing the square in the exponent of the exponential, the first integral on the right-hand side of (79) is equivalent to
Z +∞ 2 −rT a+b2/2 1 − 1 ( x−(a+b ) )2 S0e e √ e 2 b dx, (80) ln K/S0 2πb which may be further simplified via the following chain of equalities:
+∞ 2 Z 1 1 2 −rT a+b /2 − 2 y = S0e e 2 √ e dy (81) ln K/S0−(a+b ) b 2π 2 2 ln K/S − (a + b ) = S e−rT ea+b /2 [1 − Φ( 0 )] (82) 0 b 2 ln S0/K + (r + σ /2)T = S0Φ[ √ ] (83) σ T := S0Φ(d1), (84)
after substituting the definitions for a = (r − σ2/2)T and b2 = σ2T .
• The second integral on the right-hand side of (79) is equivalent to
Z +∞ −rT 1 − 1 y2 Ke √ e 2 dy, (85) ln K/S0−a b 2π
28 which may be further simplified via the following chain of equalities:
ln K/S − a = Ke−rT [1 − Φ( 0 )] (86) b ln S /K + a = Ke−rT [Φ( 0 )] (87) b ln S /K + (r − σ2/2)T = Ke−rT [Φ( 0 √ )] (88) σ T −rT := Ke Φ(d2), (89)
after substituting once again for the definitions for a and b.
• Note that √ d2 = d1 − σ T. (90)
• Let C/S denote the value of the call option as a percentage of the current stock price, let
κ := S/Ke−rT , (91) 1 √ κ2 := σ T. (92)
Note that C hln κ κ i 1 hln κ κ i = Φ 1 + 2 − Φ 1 − 2 , (93) S κ2 2 κ1 κ2 2
which shows that one only needs the values for κ1 and κ2 to compute the value of the call option.
29 14 Call Option Pricing Examples
• The famous Black-Scholes European call option formula is given by
−rT C(K,T ) = SΦ(d1) − Ke Φ(d2) (94)
where recall that Φ(·) denotes the cumulative standard normal density, and
ln (S/K) + (r + σ2/2)T d1 = √ , (95) σ T ln (S/K) + (r − σ2/2)T √ d2 = √ = d1 − σ T. (96) σ T
Example 18 Suppose S0 = 100, σ = 0.28 and rf = 5%. As a function of the K and T :
ln (100/K) + 0.0892T d1(K,T ) = √ , 0.28 T √ d2(K,T ) = d1(K,T ) − 0.28 T, −0.05T C(K,T ) = 100Φ(d1(K,T )) − Ke Φ(d2(K,T )).
Now consider the following options.
• Cost of a 1-month call option with strike price = 100?
– d1 = 0.0920, d2 = 0.0111, Φ(d1) = 0.5367, Φ(d2) = 0.5044. – C(100, 1/12) = 100(0.5367) − 99.5842(0.5044) = 3.44.
• Cost of a 3-month call option with strike price = 100?
– d1 = 0.1593, d2 = 0.0193, Φ(d1) = 0.5632, Φ(d2) = 0.5076. – C(100, 3/12) = 100(0.5632) − 98.75778(0.5076) = 6.19.
• Cost of a 1-month call option with strike price = 105?
– d1 = −0.5117, d2 = −0.5925, Φ(d1) = 0.3043, Φ(d2) = 0.2768. – C(105, 1/12) = 100(0.3043) − 104.5634(0.2768) = 1.49.
• Cost of a 3-month call option with strike price = 105?
– d1 = −0.1892, d2 = −0.3292, Φ(d1) = 0.4250, Φ(d2) = 0.3710. – C(105, 3/12) = 100(0.4250) − 103.6957(0.3710) = 4.03.
30 Example 19 Consider the previous example when σ = 40%.
• Cost of a 1-month call option with strike price = 100?
– d1 = 0.0938, d2 = −0.0217, Φ(d1) = 0.5375, Φ(d2) = 0.4916. – C(100, 1/12) = 100(0.5375) − 99.5842(0.4916) = 4.79.
• Cost of a 3-month call option with strike price = 100?
– d1 = 0.1625, d2 = −0.0375, Φ(d1) = 0.5646, Φ(d2) = 0.4850. – C(100, 3/12) = 100(0.5646) − 98.75778(0.4850) = 8.56.
• Cost of a 1-month call option with strike price = 105?
– d1 = −0.3287, d2 = −0.4441, Φ(d1) = 0.3712, Φ(d2) = 0.3284. – C(105, 1/12) = 100(0.3712) − 104.5634(0.3284) = 2.78.
• Cost of a 3-month call option with strike price = 105?
– d1 = −0.0815, d2 = −0.2815, Φ(d1) = 0.4625, Φ(d2) = 0.3891. – C(105, 3/12) = 100(0.4625) − 103.6957(0.3891) = 5.90.
Observations:
• Call option values increase as the time to expiration increases.
• Call option values decrease as the strike price increases.
• Both observations are true in general.
31 15 Chooser Option
Consider a non-dividend paying stock whose initial stock price S(0) = 62. The price process of S has a log-volatility σ − 0.20. Risk-free rate is 2.5% continuously compounded.
Value a 5-month option with strike price K = 60 in which after exactly 3 months the holder must declare this option to be either a European call or European put option.
Table 10: Chooser option valuation.
0 1 2 3 4 5 62.00 65.68 69.59 73.72 78.11 82.75 58.52 62.00 65.68 69.59 73.72 55.24 58.52 62.00 65.68 52.14 55.24 58.52 Stock prices 49.21 52.14 46.45 0 1 2 3 4 5 4.6686 7.0172 10.1423 13.9743 18.2315 22.7488 2.3054 3.876 6.2971 9.7137 13.7248 0.7216 1.4359 2.8571 5.6849 0 0 0 Call option values 0 0 0 0 1 2 3 4 5 2.0469 0.8344 0.1797 0 0 0 3.2858 1.5022 0.3627 0 0 5.1091 2.6646 0.7322 0 7.6107 4.6364 1.4782 Put option values 10.6604 7.8602 13.5462 0 1 2 3 4 5 6.0483 7.3187 10.1423 13.9743 4.7847 4.4847 6.2971 Chooser option values 5.1091 2.6646 7.6107
Observations:
• 6.0483 > max{4.6686, 2.0469}.
• 6.0483 < 4.6686 + 2.0469.
32 16 Asian Option
Consider a non-dividend paying stock S whose price process follows a binomial lattice with U = 2 and D = 0.5. R = 1.25 and S0 = 4. Define
t X Yt := Sk, t = 0, 1, 2, 3 k=0 to be the sum of the stock prices between times zero and t.
Consider a (European) Asian call option that expires at time three and has a strike price K = 4; that is, its payoff at time t = 3 is
nY o max 3 − 4, 0 . 4 An Asian call option is like a European call option, except the payoff of the option is based on the average stock price rather than the final stock price.
Let Vt(s, y) denote the price of this option at time t if St = s and Yt = y. In particular, ny o V (s, y) = max − 4, 0 . 3 4
Questions:
1. Develop an algorithm for computing Vt recursively. In particular, write a formula for Vt in terms of Vt+1.
2. Apply the algorithm to compute V0(4, 4), the price of the Asian option at time zero.
3. Provide a formula for δt(s, y), the number of shares of stock that should be held by the replicating portfolio at time t if St = s and Yt = y. 4. What is the value of a (European) Asian put option that expires at time t = 3 and has a strike price K = 4; that is, its payoff at time three is
n Y o max 4 − 3 , 0 . 4
33 Analysis:
1. Risk-neutral q = 0.5. Vt(s, y) = (1/1.25) 0.5Vt+1(us, y + us) + 0.5Vt+1(ds, y + ds) .
2. V3(32, 60) = 11; V3(8, 36) = 5; V3(8, 24) = 2; V3(2, 18) = 0.5; V3(8, 18) = 0.5; V3(2, 12) = V3(2, 9) = V3(0.5, 7.5) = 0.
V2(16, 28) = (1/1.25)[0.5(11) + 0.5(5)] = 6.40.
V2(4, 16) = (1/1.25)[0.5(2) + 0.5(0.5)] = 1.0.
V2(4, 10) = (1/1.25)[0.5(0.5) + 0.5(0)] = 0.20.
V2(1, 7) = (1/1.25)[0.5(0) + 0.5(0)] = 0.
V1(8, 12) = (1/1.25)[0.5(6.4) + 0.5(1.0)] = 2.96.
V1(2, 6) = (1/1.25)[0.5(0.2) + 0.5(0)] = 0.08.
V0(4, 4) = (1/1.25)[0.5(2.96) + 0.5(0.08)] = 1.216.
3. V (us, y + us) − V (ds, y + ds) δ (s, y) = t+1 t+1 . t (u − d)S
Remark. The value of this Asian option, 1.216, equals the discounted expectation of the final payoffs using the risk-free rate and the risk-neutral probability, i.e., 1 11 + 5 + 2 + 0.5 + 0.5 + 0 + 0 + 0 1.216 = . (1.25)3 8
Simulation is an especially useful computational approach for valuing path-dependent, European-style derivative securities, since their value can be obtained as a (discounted) sample average of the value along a “sample path.”
34 17 Stochastic Interest Rates
We consider a two-period, stochastic volatility, random interest rate model. The stock prices and interest rates are provided in Table 11. Value the European option whose final payoffs are V2 = max{S2 − 7, 0}.
Table 11: A stochastic volatility, random interest rate model
t = 0 t = 1 t = 2 S0 = 4, r0 = 25% S1(U) = 8, r1(U) = 25% S2(UU) = 12 S1(D) = 2, r1(D) = 50% S2(UD) = S2(DU) = 8 S2(DD) = 2
Analysis:
The risk-neutral q changes along the tree since the risk-free rate is stochastic. Otherwise, all the calculations are the same, since at each node the replicating portfolio and corresponding risk-neutral discounted expectation ideas still apply. The solution is provided in Table 12.
Table 12: Solution to the stochastic volatility, random interest rate model
t = 0 t = 1 t = 2 ¯ 1 ¯ 1 1.004 = 1.25 [0.5(2.4) + 0.5(0.1)] 2.4 = 1.25 [0.5(5) + 0.5(1)] 5 ¯ 1 0.1 = 1.5 [(1/6)(1) + (5/6)(0)] 1 0
35