FM441 Exam 2017 solutions PDF

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FM441 Questions and SolutionsSummer 2017 ExamRohit RahiMarch 19, 2017Instructions to candidatesTime allowed: 2 hours.This paper containsfourquestions, each worth 25 marks. Answerallquestions.The last page contains a list of facts that you may find useful.Calculators are allowed in this examination.1...

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FM441 Questions and Solutions Summer 2017 Exam Rohit Rahi March 19, 2017

Instructions to candidates Time allowed: 2 hours. This paper contains four questions, each worth 25 marks. Answer all questions. The last page contains a list of facts that you may find useful. Calculators are allowed in this examination.

1

1. A stock is currently trading at £100. Its price over the next 6 months evolves as a twostep binomial process. Over each 3-month period, the price can either go up by a factor u = 1.1 or down by a factor d = 1/u. The annual risk-free interest rate is 5% with continuous compounding. Consider a European put on the stock, struck at K = 93 and expiring in 6 months. (a) (5 marks) Draw a tree representing the evolution of the price of the underlying over time, and compute the put payof f at each terminal node.

The binomial tree representing the stock price is: Put payoff . ......... .. ........ . ............ ... ......... . . . . . . . . .......... ... ..... ... ............ .. ...... ...... . . ..... . ...... ...... . . . . . . . . . . ... ... . ... ......... . .. .... ..... ... .. ........ .. .... .... ..... ..... ...... ... ...... ........ ... ....... . . . .. ....... . . . . . ............ ...... ....... ... ...... ..... ..... . . .. ...... ......... . . . ........ . . . . ....... ......... . . . . . . . . .... .. . .......... ..... ... ... ...... .. .... . . ... . .. ........ ... ..... .. . ..... .... . ... .... ....... . ... ..... . ....... .... ... ... ...... ..... . . . ...

110

100

90.91

121

0

100

0

82.64

10.36

(b) (5 marks) Use the risk-neutral pricing approach to determine the initial fair value of the put. At any node, the risk-neutral probability q of the stock price going up subsequently satisfies er∆t = qu + (1 − q)d ⇒ q =

er∆t − d e0.05×0.25 − 1/1.1 = 0.5421. = 1.1 − 1/1.1 u−d

The put is worth zero at node u (Pu = 0). At node d, the put is worth Pd = e−r∆t (1 − q)Pdd = 4.69. At the initial node, the put is worth P0 = e−r∆t (1 − q)Pd = 2.12.

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(c) (10 marks) Now suppose that in 3 months, the stock pays a dividend of £10. On the dividend payment date, the stock price immediately adjusts to its ex-dividend level and then either goes up by a factor u = 1.1 or down by a factor d = 1/u over the subsequent 3 months. Construct a dynamic self-financing strategy that replicates the final payoff of the put. What is the impact of dividend payment on the initial put value?

The tree representing the ex-dividend stock price is:

Put payoff

. ...... ..... . ...... . ..... ... . . . ..... . ....... . ... .... .. ..... ... ..... . . . . ... ... ... .... ...... .. ... ... .. ...... . . . ...... . ...... ..... . .. .... ..... . ....... ... ... ..... .. .... ... . ... ....... . . . . ...... ..... . ...... . . ...... .... .. . . ..... .. ... . .... .. . . ... . ....... . . . ...... ..... . . .. .... ...... .. ...... . . ... .. ... . ... . . .. ... . ..... . . . . ..... .. .... ...... ..... . . .... ... .. .... ....... ... . . . . . ... ... .... ...... ...... . ..... ...... ... .. .... .. ...... ...... ...... ...... . .... ..... . ... ..... .... .. ...... ...... ... ... ...... . .... ... .... .. ..... . . . . .... . .. .... ..... ... ... ...... ....... .... .. ...... ...... ...... . . . . . .... ... . ...... ...... . ...... . ..... .. .... ...... . . ..... ... ... ........ ......... . ..... ..... . ...... . .... .. .... ... ........ . ..... . ....... . ..... ...... .. .... .. ...... ....... .. ... .. ..... ....

110

0

100

100

90.91

89

2.09

4

80.91

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73.55

19.45

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At the interim node i = u, d, we need to hold δi units of the stock and take position βi in the money market account so that δi uSie + er∆t βi = Piu, δi dSie + er∆t βi = Pid , where Sie is the ex-dividend stock price at the intermediate date. At node u, we find δu = −0.1095 and βu = 11.89. The cost of the replicating portfolio is Pu = (−0.1095 × 100) + 11.89 = 0.9452. At node d, we f ind δd = −1 and βd = 91.84. The cost of the replicating portfolio is Pd = −80.91 + 91.84 = 10.93. At the initial node, we need to hold δ0 units of the stock and take position β0 in the money market account so that δ0 uS0 + er∆t β0 = Pu, δ0 dS0 + er∆t β0 = Pd . Note that here we use the cum-dividend price at the intermediate date, since that is the total payof f at the intermediate date of holding one unit of the stock at the initial date. We find δ0 = −0.523 and β0 = 57.75, the initial cost of replication being P0 = (−0.523 × 100) + 57.75 = 5.45. The dividend payment makes the put more expensive.

(d) (5 marks) Check your answer for the put value in part (c) using the risk-neutral valuation method. The risk-neutral probabilities are unchanged, i.e. q = 0.5421. Therefore, Pu = e−r∆t [qPuu + (1 − q)Pud ] = 0.9451, Pd = e−r∆t [qPdu + (1 − q)Pdd ] = 10.9370, P0 = e−r∆t [qPu + (1 − q)Pd ] = 5.45.

2. (a) Consider an economy with two dates (t = 0, 1), S states of the world and N assets. Let D be the S × N matrix of asset payoffs at date 1, and π the asset price vector at date 0. Then the Fundamental Theorem of Asset Pricing (FTAP) states that: (D, π) does not admit arbitrage if and only if there exists a strictly positive vector of state prices ψ % 0 in RS such that π⊤ = ψ⊤ D, i.e. the price of every asset is the state price weighted sum of its payoffs. For the following two questions, assume there is no arbitrage. c LSE ST 2017/FM441 #

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i. (7 marks) Show that if markets are complete, there is a unique state price vector. By the FTAP, there is a vector ψ such that π⊤ = ψ⊤ D. If markets are complete, there are S linearly independent assets. Let the payoff matrix and price vector for these assets be denoted by D ∗ and π∗ , respectively (any collection of S linearly independent assets will do). Note that the S × S matrix D ∗ is nonsingular, hence invertible. Then we must have π⊤ = ψ⊤ D ∗ or D ∗⊤ ψ = π, which has a unique solution: ψ = (D ∗⊤ )−1 π.

ii. (3 marks) Now suppose markets are incomplete. Argue that there are many state price vectors. If markets are incomplete the number of linearly independent assets (say N ∗ ) is less than S. But N ∗ is also the number of independent equations in π⊤ = ψ⊤ D , which is fewer than the number of unknowns S. Hence there are infinitely many solutions for ψ. iii. (5 marks) Show that the state price weighted sum of the payoffs of an attainable claim is the same regardless of the choice of state price vector. Consider an attainable claim y. Then there is a replicating portfolio x such that y = Dx. For any state price vector ψ, we have ψ⊤ y = ψ⊤ (Dx) = (ψ⊤ D)x = π⊤ x which does not depend on the choice of ψ (it is just the cost of the replicating portfolio x).

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(b) (10 marks) Consider the following stock price process: dSt = µt St dt + σt St dWt , where µt and σt are deterministic functions of time. Show that stock returns are lognormal. More precisely, show that the conditional distribution at time t of log(ST /St ) is normal. Derive expressions for the mean and the variance of this distribution. Using Itˆ o’s lemma, we see that ! " 1 1 1 d log St = dSt + − 2 dSt2 2 St St " ! 1 1 1 = (µt St dt + σt St dWt ) + − 2 (µt St dt + σt St dWt )2 St 2 St ! " 1 1 1 = (µt St dt + σt St dWt ) + − 2 (σt2St2dt) 2 St St 2 = (µt − σ t /2)dt + σt dWt Integrating, we obtain log (ST /St ) =

#

t

T

$

µs −

%

σs2/2

ds +

#

T

σs dWs . t

Since µs and σs are non-stochastic, log (ST /St ) is normally distributed, with # T % $ E [log (ST /St )] = µs − σs2/2 ds, #t T Var [log (ST /St )] = σs2 ds. t

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3. For this question, suppose the Black-Scholes assumptions hold. (a) (10 marks) Derive the formula for Γ, the gamma of a call option in the Black-Scholes model (you may use the formula for delta without deriving it). Determine the value of S, denoted by S ∗ , that maximizes Γ. What is the limit of S ∗ as t approaches the maturity date T ? The delta of a call option is Φ(d1 ), where d1 = Therefore, Γ=

log(S/K) + (r + σ 2 /2)(T − t) √ . σ T −t

∂Φ(d1 ) 1 ∂d1 √ φ(d1 ), = = Φ′ (d1 ) ∂S ∂S Sσ T − t

where φ(·) is the standard normal density. Differentiating Γ with respect to S, we get Γ′ (S) = −

1 ∂d1 1 √ √ . φ(d1 ) + φ′ (d1 ) ∂S S 2σ T − t Sσ T − t

Note that 1 2 1 φ(d1 ) = √ e−2 d 1 2π φ′ (d1 ) = −d1 φ(d1 ) ∂d1 1 √ . = ∂S Sσ T − t

Hence, Γ′ (S) = 0 implies

which gives us

1 √ d1 = −1, σ T −t

3 log S ∗ = log K − r(T − t) − σ 2 (T − t) 2 & ! " ' 3 ∗ 2 ⇒ S = K exp − r + σ (T − t) . 2 Therefore, limt→T S ∗ = K. Notice that, prior to maturity, S ∗ < e −r(T −t) K.

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(b) The buyer of an at-the-money forward-start call option, with forward-start date T1 and exercise date T2 > T1 , receives at date T1 a standard European call option struck at K = ST1 and expiring at date T2 . i. (5 marks) What is the price of the forward-start option at date T1 ? The value of the forward-start option when the starting date is reached (i.e. at T1 ) is just the value of a standard ATM European call with time-to-maturity T2 − T1 :

where

CT1 (atm, T2 − T1 ) = C BS (ST1 , ST1 , σ, r, T2 − T1 ) ( ) = ST1 Φ(d1 ) − e−r(T2 −T1 ) Φ(d2 ) , " σ2 * T2 − T1 , r+ 2 * d2 = d1 − σ T2 − T1 . 1 d1 = σ

!

ii. (10 marks) What is the price of the forward-start option at date 0? Relate this to the price of a standard option at date 0. The price of the forward-start option at time 0 is given by ( ) ( ) e−rT1 EQ C BS (ST1 , ST1 , σ, r, T2 − T1 ) = e−rT1 EQ (ST1 ) Φ(d1 ) − e−r(T2 −T1 ) Φ(d2 ) ( ) = S0 Φ(d1 ) − e−r(T2 −T1 ) Φ(d2 ) = C0 (atm, T2 − T1 ). We used the fact that e−rT1 EQ (ST1 ) = S0 . Thus the price of the forward-start ATM call option at date 0 coincides with the price of a standard ATM call option with time-to-expiration T2 − T1 .

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4. (a) The price of a 6-month zero-coupon bond is £0.97943, and that of a 1-year zero-coupon bond is £0.95238. (Both bonds have a face value of £1). i. (4 marks) What is the rate on an FRA for a 6-month loan starting in 6 months, expressed as a simple annualized rate? By trading in zero-coupon bonds today you can lock in a borrowing/savings interest rate over a future period equal to today’s forward interest rate for that period. By entering an FRA, you can lock in the “strike” interest rate defining the payof f of the FRA as a borrowing/savings rate. Therefore, to prevent arbitrage, it must be that the “strike” interest rate in the FRA contract is equal to the forward interest rate at the time that the contract is initiated. Hence, we proceed to compute the forward rate based on the knowledge of zero-coupon bond prices. We know that P (0, 0.5) = 97.943 and P (0, 1) = 95.238. Therefore, expressing rates with semi-annual compounding, 0.97943 = 0.95238 =

1 1+

⇒

y(0,0.5) 2

1

] [1+ y(0,1) 2

2

y(0, 0.5) = 4.2%, y(0, 1) = 4.94%.

⇒

The no-arbitrage level of the rate specified by the FRA is equal to the forward rate F (0, 0.5, 1), which is such that &

1+

y(0, 0.5) 2

' & '& ' y(0, 1) 2 F (0, 0.5, 1) = 1+ 1+ 2 2 ,

which implies that F (0, 0.5, 1) = 1+ 2

1 1+ y(0,0.5) 2 1

[1+ y(0,1) 2 ]

= 2

P (0, 0.5) . P (0, 1)

Therefore, F (0, 0.5, 1) = 5.68%. This must be the rate of the FRA (expressed as a simple rate LK ). Notice that the forward rate can be computed directly from the prices of zero-coupons; it is not necessary to compute zero rates.

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ii. (9 marks) Suppose you are the counterparty in the FRA for a borrower who uses the FRA to hedge the interest rate on a £10m loan. What position in zero-coupon bonds would you use to hedge the risk of the FRA? More precisely, derive a trading strategy which exactly offsets your liability from the FRA transaction and which entails no cash flows, positive or negative, at any other point in time. We are now going to see how one can synthesize, at no cost, the payof f of the FRA by trading in zero-coupon bonds (this is assuming that the fixed rate on the FRA is set at its fair level). In the context of this exercise, as a market maker, you need to pay in a year 1 (L 1 − LK ) × 10, 2 2 where L 1 := L(0.5, 1) is the 6-month rate prevailing in 6 months, LK = 5.68% is 2 the FRA strike, and all amounts are in £m. The (zero-cost) replicating strategy is as follows: • at time 0, buy 6-month zero-coupons with a total face value equal to 10, financing this purchase by selling 1-year zero-coupons. The number of 1-year zeros sold is, therefore, & ' P (0, 0.5) F (0, 0.5, 1) × 10 = 1 + × 10 P (0, 1) 2 & ' LK = 1+ × 10. 2 • after 6 months, you earn 10 and save this amount over the next 6 months at the prevailing rate L 1 . 2

Overall, you break even initially and after six months, and your payof f in a year is exactly + , ' & L1 1 LK 2 = (L 12 − LK ) × 10. − 10 1 + 10 1 + 2 2 2

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(b) (12 marks) Suppose the 1-year and 2-year risk-free rates are currently 2.5% and 2.37%, respectively. In one year’s time, the 1-year risk-free rate will be either 1.5% (with probability 3/4) or 3% (with probability 1/4). Rates are expressed as annual rates with annual compounding. Calculate the current price of a call option on a 2-year zero-coupon bond with face value £100, strike price £97, and expiration in one year. Do all your calculations to 3 decimal places. The current price of the 2-year zero is P (0, 2) =

100 100 = 95.423. = 2 [1.0237]2 [1 + y(0, 2)]

In 1 year’s time, the price of the 2-year zero is P (1, 2) =

100 . 1 + y(1, 2)

Let u denote the “up” state when y(1, 2) = 1.5%, and d the “down” state when y(1, 2) = 3%. Thus we have 100 = 98.522, 1.015 100 = 97.087. Pd (1, 2) = 1.030

Pu(1, 2) =

Let q be the risk-neutral probability of the “up” move. Then the current price of the 2-year zero is given by P (0, 2) = i.e. 95.423 =

qPu(1, 2) + (1 − q)Pd (1, 2) , 1 + y(0, 1) 98.522q + 97.087(1 − q) , 1.025

which gives us q = 0.501. The call option on the 2-year zero is in the money in both states, with Cu = 1.522 and Cd = 0.087. Therefore, the current price of the option is C=

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1 [0.501 × 1.522 + (1 − 0.501) × 0.087] = 0.786. 1.025

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Useful Facts • Black-Scholes formula: The price of a European call option at t < T is given by C = SΦ(d1 ) − e−r(T −t) KΦ(d2 ), where Φ(·) is the standard normal cdf, and d1,2 =

• Standard normal density:

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log(S/K) + (r ± σ 2 /2)(T − t) √ . σ T −t

1 1 2 φ(x) = √ e− 2 x . 2π

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