ACT4000, MIDTERM #2 ADVANCED ACTUARIAL TOPICS MARCH 16, 2009 HAL W. PEDERSEN

You have 70 minutes to complete this exam. When the invigilator instructs you to stop writing you must do so immediately. If you do not abide by this instruction you will be penalised. All invigilators have full authority to disqualify your paper if, in their judgement, you are found to have violated the code of academic honesty.

Each question is worth 10 points. Provide sufficient reasoning to back up your answer but do not write more than necessary.

This exam consists of 8 questions. Answer each question on a separate page of the exam book. Write your name and student number on each exam book that you use to answer the questions. Good luck!

Question 1. You have written a European call on a share of stock which expires in six months and has a of $55.00. You are given the following information.

• The current price of the stock is S0 = $50.00. • The stock pays no dividends. • The continuous time interest rate r =0.08. • The on the stock is σ =0.25. • The price of the 55-call is $2.37535. • The delta of the 55-strike call you have written is ∆ = 0.41119. • The gamma of the 55-strike call you have written is Γ = 0.04401.

You have decided to delta-gamma hedge your position using a European on a share of stock which expires in six months and has a strike price of $60.00. You are also given the following information. • The price of the 60-put is $8.77541. • The delta of the 60-strike put is ∆ = −0.76322. • The gamma of the 60-strike put is Γ = 0.03491. • The price of a 55-call with remaining maturity of six months less a day when the underlying stock is trading at $51.00 is $2.79428. • The price of a 60-put with remaining maturity of six months less a day when the underlying stock is trading at $51.00 is $8.03167.

(a) [4 points] Compute the position you must take in 60-strike puts to effect the delta-gamma hedge. (b) [6 points] One day after you effect the delta-gamma hedge the stock price has increased by $1.00. What is your profit or less on the hedged portfolio. 1 2 ACT4000 – MIDTERM #2

Question 2. Assume that the Black-Scholes framework holds. The price of a stock that pays no dividends is $30.00. The price of a put option on this stock is $4.00.

You are given: • ∆=−0.28, and • Γ=0.20.

Using the delta-gamma approximation, determine the price of the put option if the stock price changes to $31.50.

Question 3. Consider a model with two stocks. Each stock pays dividends continu- ously at a rate proportional to its price.

Sj(t) denotes the price of one share of stock j at time t.

Consider a claim maturing at time 3. The payoff of the claim is:   Maximum S1(3),S2(3) . You are given the following information

• S1(0) = 100 • S2(0) = 200 • Stock 1 pays dividends of amount (0.05)S1(t) dt between time t and time t+dt. • Stock 2 pays dividends of amount (0.10)S2(t) dt between time t and time t+dt. • The price of a European option to exchange Stock 2 for Stock 1 at time 3 is $10.00. Calculate the price of the claim.

Question 4. Consider a “” which, 1 year from today, will give its owner a 1-year European with a strike price equal to the stock price at that time.

You are given:

• The European call option is on a stock that pays no dividends. • The stock’s volatility is 30%. • The for delivery of 1 share of the stock 1 year from today is $100. • The continuously compounded risk-free interest rate is 8%.

Under the Black-Scholes framework, determine the price today of the forward start option. ACT4000 – MIDTERM #2 3

Question 5. Let S(t) denote the price at time t of a stock that pays dividends continuously at a rate proportional to its price. Consider a European gap option with date T for T>0.

If the stock price at time T is greater than $100, the payoff is S(T ) − 90 otherwise the payoff is 0.

You are given:

(i) S(0) = $80 (ii) The price of a European call option with expiration date T and strike price $100 is $4. (iii) The delta of the call option in (ii) is 0.2.

Calculate the price of the gap option.

Question 6. Explain the notion of a compound call option. Be sure to clearly identify what the underlying is and to indicate what the payoff of the is at expiration.

Question 7. The stochastic process X follows the SDE dXt =0.14 dt + σdWt Xt where W is a standard Brownian motion.

−t 2 Consider the new process Y defined by Yt = f(t, Xt)wheref(t, x)=e x .

Apply Itˆo’s lemma to write an expression for dYt.

Question 8. An asset price follows the diffusion process defined by   St =80· exp 0.08 t +0.25 Wt where W is a standard Brownian motion. The risk-free interest rate is r =0.05. A second asset, denoted X, is available for trade with asset price dynamics dXt = µdt+0.20 dWt Xt and X0 = 20.

Assuming that the market is arbitrage-free, compute µ.

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Question 2

The delta-gamma approximation is merely the Taylor series approximation with up to the quadratic term. In terms of the Greek symbols, the first is Δ, and the second derivative is Γ. The approximation formula is 1 2 P(S + ε) ≈ P(S) + ε Δ + ε Γ. (13.2 & 13.5) 2

With P(30) = 4, Δ = −0.28, Γ = 0.10, and ε = 1.50, we have P(31.5) ≈ 4 + (1.5)(−0.28) + 1 (1.5)2(0.1) 2 = 3.6925 ≈ 3.70.

Question 3

Because of the identity

Maximum( S1(3), S2(3) ) = Maximum( S1(3) – S2(3), 0) + S2(3), the payoff of the claim can be decomposed as the sum of the payoff of the exchange option in statement (v) of the problem and the price of stock 2 at time 3. In a no- arbitrage model, the price of the claim must be equal to the sum of the exchange option price (which is 10) and the prepaid forward price for delivery of stock 2 at −δ2 ×3 time 3 (which is e ×S2(0)). So, the answer is 10 + e−0.1×3×200 = 158.16.

Remark: If one buys e−δ2 ×3 share of stock 2 at time 0 and re-invests all dividends, one will have exactly one share of stock 2 at time 3.

Question 4

This problem is based on 14.21 on page 465 of McDonald (2006).

Let S1 denote the stock price at the end of one year. Apply the Black-Scholes formula to calculate the price of the at-the-money call one year from today, conditioning on S1.

2 2 d1 = [ln (S1/S1) + (r + σ /2)T]/(σ T ) = (r + σ /2)/σ = 0.417, which turns out to be independent of S1.

d2 = d1 − σ T = d1 − σ = 0.117

The value of the forward start option at time 1 is −r C(S1) = S1N(d1) − S1e N(d2) −0.08 = S1[N(0.417) − e N(0.117)] −0.08 ≈ S1[N(0.42) − e N(0.12)] -0.08 = S1[0.6628 − e ×0.5438] = 0.157S1. (Note that, when viewed from time 0, S1 is a random variable.)

Thus, the time-0 price of the forward start option must be 0.157 multiplied by the time-0 price of a security that gives S1 as payoff at time 1, i.e., multiplied by the prepaid forward P price F 1,0 (S) . Hence, the time-0 price of the forward start option is P −0.08 −0.08 0.157× F 1,0 (S) = 0.157×e × F 1,0 (S) = 0.157×e ×100 = 14.5

Remark: A key to pricing the forward start option is that d1 and d2 turn out to be independent of the stock price. This is the case if the strike price of the call option will be set as a fixed percentage of the stock price at the issue date of the call option.

Question 5

In terms of the notation in Section 14.15, K1 = 90 and K2 = 100.

By (12.1), and (12.2a, b), statement (ii) of the problem is −−δTrT 4(0)()=−S e Nd12 Ke Nd () 2, (1) where S ()080,= 1 2 ln(SK (0) /2 )+−+ ( rδσ2 ) T d1 = , σ T and 1 2 ln(SK (0) /2 )+−− ( rδσ2 ) T d2 = d1 − σ T = . σ T Do note that both d1 and d2 depend on K2, but not on K1.

From the last paragraph on page 383 and from statement (iii), we have −δT Δ=eNd()1 = 0.2, and hence equation (1) becomes −rT 4=× 80 0.2 − 100eNd (2 ) , or −rT eNd(2 )=×− (80 0.2 4) /100 = 0.12 .

By (14.15) on page 458, the gap call option price is −−δTrT S(0) e Nd (11 )− Ke Nd ( 2 ) = 80×−× 0.2 90 0.12 = 5.2.

Remark: The payoff of the gap call option is [S(T) – K1]×I(S(T) > K2), where I(S(T) > K2) is the indicator random variable, which takes the value 1 if S(T) > K2 and the value 0 otherwise. Because the payoff can be expressed as S(T)×I(S(T) > K2) – K1×I(S(T) > K2), we can obtain the pricing formula (14.15) by showing that the time-0 price for the time-T payoff S(T)×I(S(T) > K2) is −δT SeNd(0) (1 ) , and the time-0 price for the time-T payoff I(S(T) > K2) is −rT e N(d2). Note that both d1 and d2 are calculated using the strike price K2. We can use risk- neutral pricing to verify these two results: −rT −δT E*[e S(T)×I(S(T) > K2)] = SeNd(0) (1 ) , which is the pricing formula for a European asset-or-nothing (or digital share) call option, and −rT −rT E*[e I(S(T) > K2)] = e N(d2), which is the pricing formula for a European cash-or- nothing (or digital cash) call option. Here, we follow the notation on pages 604 and 605 that the asterisk is used to signify that the expectation is taken with respect to the risk-neutral probability measure. Under the risk-neutral probability measure, the random variable ln[S(T)/S(0)] is normally distributed with mean (r – δ – 1 σ2)T and variance σ2T. 2

The second expectation formula, which can be readily simplified as E*[I(S(T) > K2)] = N(d2), is particularly easy to verify: Because an indicator random variable takes the values 1 and 0 only, we have E*[I(S(T) > K2)] = Prob*[S(T) > K2], which is the same as Prob*(ln[S(T)/S(0)] > ln[K2/S(0)]). To evaluate this probability, we use a standard method, which is also described on pages 590 and 591. We subtract the mean of ln[S(T)/S(0)] from both sides of the inequality and then divide by the standard deviation of ln[S(T)/S(0)]. The left-hand side of the inequality is now a standard normal random variable, Z, and the right-hand side is ln[KS / (0)]−−− ( rδσ2 / 2) T ln[SKr (0) / ]+−− (δσ2 / 2) T 2 = − 2 σ 2T σ T = −d2. Thus, we have E*[I(S(T) > K2)] = Prob*[S(T) > K2], = Prob(Z > −d2) = 1 − N(−d2) = N(d2).

The first expectation formula, −rT −δT E*[e S(T)×I(S(T) > K2)] = SeNd(0) (1 ) , is harder to derive. One method is to use formula (18.29), which is in the syllabus of Exam C, but not in the syllabus of Exam MFE. A more elegant way is the actuarial method of Esscher transforms, which is not part of the syllabus of any actuarial −rT examination. It shows that the expectation of a product, E*[e S(T)×I(S(T) > K2)], can be factorized as a product of expectations, −rT E*[e S(T)] × E**[I(S(T) > K2)], where ** signifies a changed probability measure. It follows from (20.26) and (20.14) that E*[e−rT S(T)] = e−δT S(0). To evaluate the expectation E**[I(S(T) > K2)], which is Prob**[S(T) > K2], one shows that, under the probability measure **, the random variable ln[S(T)/S(0)] is normally distributed with mean (r – δ − 1 σ2)T + σ2T = (r – δ + 1 σ2)T, 2 2 and variance σ2T. Then, with steps identical to those above, we have E**[I(S(T) > K2)] = Prob**[S(T) > K2], = Prob(Z > −d1) = 1 − N(−d1) = N(d1).

Alternative solution: Because the payoff of the gap call option is [S(T) – K1]×I(S(T) > K2) = [S(T) – K2]×I(S(T) > K2) + (K2 – K1)×I(S(T) > K2), the price of the gap call option must be equal to the sum of the price of a European call option with the strike price K2 and the price of (K2 – K1) units of the corresponding cash-or-nothing call option. Thus, with K1 = 90, K2 = 100, and statement (ii), the price of the gap call option is 4 + (100 – 90)×e−rTProb*[S(T) > 100] −rT = 410+ eNd (2 ). On the other hand, from (ii), (iii), and (12.1), it follows that −rT 4=− 80(0.2) 100eNd (2 ). −rT Thus, eNd()2 = 0.12, and the price of the gap call option is 4 + 10×0.12 = 5.2.

A compound option is an option to buy an option. If you think of an ordinary option a!>an asset-analogous to a stock-then a compound option is similar to an ordinary option. Compound options are a little more complicated than ordinary options because there are two strikes and two expirations. one each for the underlying option and for the compound option. Suppose that the current time is 10 and that we have a compound option which at time,) will give us the right to pay x to buy a European call option with strike K. This underlying call will expire at time T > '). Figure 14.2 compares the timing of the exercise decisions for this compound option with the exercise decision for an ordinary call expiring at time T. If we exercise the compound call at time 'I. then the price of the option we receive is C(S. K. T -ld· At time T. this option will have the value max(O. S'{ - K). the same as an ordinary call with strike K. At time 11. when the compound option expires. the value of the compound option is

max[C(SI . K. T - '1) - x. OJ

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The compound option is exercised for 511 > S'. Thus. in order for the compound call to ultimately be valuable. there are two events that must take place. First. at time I) we must have Sf, > S': that is. it must be worthwhile to exercise the compound call. Second. at time T we must have ST > K: that is. it must be profitable to exercise the underlying call. Because two events must occur. the formula for a compound call contains a bivariate cumulative norma] distribution. as opposed to the univariate distribution in the Black-Scholes formula.

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