ACT4000, MIDTERM #2ADVANCED ACTUARIAL TOPICS

MARCH 16, 2009HAL W. PEDERSEN

You have 70 minutes to complete this exam. When the invigilator instructs you tostop writing you must do so immediately. If you do not abide by this instruction youwill be penalised. All invigilators have full authority to disqualify your paper if, intheir 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 answerbut do not write more than necessary.

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

Question 1. You have written a European call option on a share of stock whichexpires in six months and has a strike price of $55.00. You are given the followinginformation.

• The current price of the stock is S0 = $50.00.• The stock pays no dividends.• The continuous time interest rate r = 0.08.• The volatility 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 put optionon a share of stock which expires in six months and has a strike price of $60.00. Youare 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 thedelta-gamma hedge.

(b) [6 points] One day after you effect the delta-gamma hedge the stock price hasincreased 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 stockthat 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 thestock 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 “forward start option” which, 1 year from today, will giveits owner a 1-year European call option with a strike price equal to the stock price atthat time.

You are given:

• The European call option is on a stock that pays no dividends.• The stock’s volatility is 30%.• The forward price 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 startoption.

ACT4000 – MIDTERM #2 3

Question 5. Let S(t) denote the price at time t of a stock that pays dividendscontinuously at a rate proportional to its price. Consider a European gap optionwith expiration 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 identifywhat the underlying is and to indicate what the payoff of the compound option is atexpiration.

Question 7. The stochastic process X follows the SDE

dXt

Xt

= 0.14dt + σ dWt

where W is a standard Brownian motion.

Consider the new process Y defined by Yt = f(t, Xt) where f(t, x) = e−tx2.

Apply Ito’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 + 0.25Wt

)

where W is a standard Brownian motion. The risk-free interest rate is r = 0.05. Asecond asset, denoted X, is available for trade with asset price dynamics

dXt

Xt= µdt + 0.20dWt

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 derivative is Δ, and the second derivative is Γ. The approximation formula is P(S + ε) ≈ P(S) + ε Δ + 1

2 ε2 Γ. (13.2 & 13.5)

With P(30) = 4, Δ = −0.28, Γ = 0.10, and ε = 1.50, we have

P(31.5) ≈ 4 + (1.5)(−0.28) + 12 (1.5)2(0.1)

= 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 time 3 (which is 2 3e−δ × ×S2(0)). So, the answer is

10 + e−0.1×3×200 = 158.16. Remark: If one buys 2 3e−δ × 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 Exercise 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.

d1 = [ln (S1/S1) + (r + σ2/2)T]/( Tσ ) = (r + σ2/2)/σ = 0.417, which turns out to be independent of S1.

d2 = d1 − Tσ = d1 − σ = 0.117 The value of the forward start option at time 1 is

C(S1) = S1N(d1) − S1e−r

N(d2)

= S1[N(0.417) − e−0.08 N(0.117)]

≈ S1[N(0.42) − e−0.08 N(0.12)]

= S1[0.6628 − e-0.08 ×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

price )(1,0 SFP

. Hence, the time-0 price of the forward start option is

0.157× )(1,0 SFP

= 0.157×e−0.08× )(1,0 SF = 0.157×e−0.08×100 = 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 1 2 24 (0) ( ) ( )T rTS e N d K e N dδ− −= − , (1) where ( )0 80,=S

212 2

1ln( (0) / ) ( )S K r T

dT

δ σσ+ − +

= ,

and

d2 = d1 − σ T = 21

2 2ln( (0) / ) ( )S K r TT

δ σσ+ − −

.

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

1( )Te N dδ−Δ = = 0.2, and hence equation (1) becomes 24 80 0.2 100 ( )rTe N d−= × − , or

2( ) (80 0.2 4) /100 0.12rTe N d− = × − = .

By (14.15) on page 458, the gap call option price is 1 1 2(0) ( ) ( )T rTS e N d K e N dδ− −− = 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 1(0) ( )TS e N dδ− , and the time-0 price for the time-T payoff I(S(T) > K2) is e−rT 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: E*[e−rT S(T)×I(S(T) > K2)] = 1(0) ( )TS e N dδ− , which is the pricing formula for a European asset-or-nothing (or digital share) call option, and E*[e−rT I(S(T) > K2)] = e−rT 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

2 σ2)T and variance σ2T.

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

2

22

ln[ / (0)] ( / 2)K S r T

T

δ σ

σ

− − − = 2

2ln[ (0) / ] ( / 2)S K r TT

δ σσ+ − −

−

= −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, E*[e−rT S(T)×I(S(T) > K2)] = 1(0) ( )TS e N dδ− , 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 examination. It shows that the expectation of a product, E*[e−rT S(T)×I(S(T) > K2)], can be factorized as a product of expectations, E*[e−rT 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 − 12 σ2)T + σ2T = (r – δ + 1

2 σ2)T,

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] = 24 10 ( )rTe N d−+ .

On the other hand, from (ii), (iii), and (12.1), it follows that 24 80(0.2) 100 ( ).rTe N d−= −

Thus, 2( )rTe N d− = 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 ordinaryoption.

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 thetiming of the exercise decisions for this compound option with the exercise decision foran ordinary call expiring at time T.

If we exercise the compound call at time 'I. then the price of the option we receiveis 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. thevalue of the compound option is

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

:e=..- ." ..

;,ff.f:IGURE 14.2" Ordinary call

The timing of exercisedecisions for a

compound call optionon a call compared withan ordinary call option.

to

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10

f

T

Option expiration

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T

Buy compound option Oecision to exercisecompound option

Expiration ofunderlying option(if compound optionwa, exercised)

We only exercise the compound option if the stock price at time I, is sufficiently great

that the value of the call exceeds the compound option strike price. x. Let S' be the

critical stock price above which the compound option is exercised. By definition. S'satisfies

C(5'.K.T-I,)=x (14.111

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 worthwhileto 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 tothe univariate distribution in the Black-Scholes formula.

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