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boundary to appear while in the region (D0 /r, ) it is possible for a free boundary to appear. If D0 /r < , then in the entire region (, ), it is possible for a free boundary to appear. Therefore if < max(, D0 /r) = max(1, D0 /r), then
1
Homework Problems (Part I) for Derivative Securities and Dierence Methods
Problems
3
1 Introduction
Problems
1. What is the dierence between taking a long position in a forward contract and in a call option? 2. Suppose the futures price of gold is curre
10 Points Extra Credit: Draw the following graphs:
1) The curve is only increasing and concave up. 2) The curve is only increasing and concave down. 3) The curve is only decreasing and concave up. 4) The curve is only decreasing and concave down.
1
Mathematics 2Y Spring 1995 Probability Theory Contents x1. Basic concepts. Sample space, events, inclusion-exclusion principle,
probabilities. Examples. x2. Independence, conditioning, Baye's formula, law of total probability. Examples. x3. Discrete ran
284
4 Interest Rate Derivative Securities
and using the nal condition, we arrive at b0 (t) = kZ 1 er(T t) , r
which is nonnegative for t T . Dene B c (S, t) = Bc (S, t) b0 (t). Thus Bc (S, t) = B c (S, t) + b0 (t). Substituting this relation into the orig
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4 Interest Rate Derivative Securities
r = (u w). Therefore the equation for any convertible bond is 1 2V 2V 1 2V V + 2 S 2 2 + Sw + w2 2 t 2 S Sr 2 r V V + (u w) rV + kZ = 0. + (rS D(S, t) S r In this equation, only the market price of risk for the i
4 Interest Rate Derivative Securities
261
Then Vf (r, t ) gives the value of the oor and the premium of the oor is given by Vf (r , t ), where r is the spot interest rate at time t . 19. a) S is a random vector and its covariance matrix is B. Let S = AS,
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4 Interest Rate Derivative Securities
T+
V (r, T + ) V (r, T ) =
T
V dt = t
T+
(t T )dt = 1.
T
That is, V (r, T ) = V (r, T + ) + 1 = 1. In the rst problem V (r, T ) = V (r, T + ) is also identically equal to one. Consequently, from T to any t < T ,
4 Interest Rate Derivative Securities
243
with A(T ) = B (T ) = 0 and determine the system of ordinary dierential equations the functions A(t) and B (t) should satisfy. Solution: Substituting V (r, t) = eA(t)rB (t) into the equation yields dA dB r + (a0 +
4 Interest Rate Derivative Securities
Problems
1. a) Suppose the spot interest rate is a known function r(t). Consider a bond with a face value Z and assume that it pays a coupon with a coupon rate k (t), that is, during a time interval [t, t + dt], the c
3 Exotic Options
225
1 2 and being . . . . n b) For the European option with a payo min(S0 , S1 , , Sn ), its price is Vmin (S1 , S2 , , Sn , t) = S0 Nn B10 , B20 , , Bn0 ; 120 , 130 , , (n1)n0
+S1 Nn (B21 , B31 , , B01 ; 231 , 241 , , n01 ) . . . +Sn Nn
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3 Exotic Options
Therefore according to these relations above, for the integral given we have er
= S1 S0 0 S1T
S1T (S1T , S2T ; S1 , S2 , t)dS2T dS1T
1 2 det P1 2 T 1 1 e(1 Pe1 ) P (1 Pe1 )/2 dS2T dS1T S1T S2T 1 = S1 2 1 2 201
= S1
e[1 (z21 )2201 1
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3 Exotic Options
C+ + C = 1, C + C = 0, + +
From the rst two conditions, we have C+ = and W = C+
+
C + + + C = f 2 , f2 f2 C + + C = . + + f2 f2 f2 + , and C = + +
f 1
+ C
f 1
=
+
f 1
+
+
+ +
f 1
.
Subtracting the third condition from the
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3 Exotic Options
c) Let V (S, H ) = SW ( ) and = H/S . Then V S 2V S 2 V H V (S, H ) H Thus 1 2 2 2V V S + (r D0 ) S rV 2 2 S S dW 2 2 d 2 W rW + (r D0 ) W =S 2 d 2 d dW 2 2 d 2 W =S + (D0 r) D 0 W 2 d 2 d and V dW (S, S ) = (1). H d dW H dW 2 = W , d
3 Exotic Options
137
1 + rT A, i.e., (1 + D0 T )S + (1 + rT )A < 0. Thus 1 + D0 T 1 + rT 1 A if < , then when S > , the PDE cannot be used. If 1 + D0 T 1 + rT 1 1 + rT > , then when S > A, the PDE cannot be 1 + D0 T 1 + D0 T 1 + rT A <S< A, we have (1 + D
3 Exotic Options
Problems
1. Consider the following problem: 1 2 2 2V V V t + 2 S S 2 + (r D0 ) S S rV = 0, 0 S, t T, where 1 (S ) and 2 (S ) are continuous functions and 1 (B ) = 2 (B ) may not hold. a) Try to nd such a relation between 1 (S ) and 2 (S )
2 Basic Options
87
P (S, t) and Sf (t) are known functions obtained from S P the solution of the American put option problem. Let P = and P . For them, the formulations are P D0 = D0 1 2 P P 2P P + 2 S 2 + (r D0 )S rP + S 2 2 = 0, t 2 S 2 S S Sf (t) S, 0
Main Subjects of Chapters 3 and 4
1. Closed-form solutions for European barrier options, properties of European and American barrier options, and relations among knock-out, knock-in and vanilla options, Problems 1*1 , 2a,b,d,e, 3* (Chapter 3) 2. Asian opt
1 Introduction
Problems
1. What is the dierence between taking a long position in a forward contract and in a call option? Solution: Taking a long position in a forward contract is an obligation to buy an underlying asset at a xed price. Taking a long pos
2 Basic Options
Problems
1. a) Show
2 1 ex /2 dx = 1. 2
b) Show that
2 2 1 e(xa) /2b dx = 1 b 2
holds for any a and b. (Because this is true and the integrand is always possitive, it can be a probability density function.) c) If the probability densit
2 Basic Options
9
b) Due to the fact that the sum of two normal random variables is a normal variable, it is easy to show that the sum of n normal variables, 1 t + + n t, and the limit of the sum as n are normal variables. Because
n
E [X (t)] = lim E
n i=
16
2 Basic Options
Var [ ] = E 2 E [ ] = = e2b b2 (e 1). a2
2
2
2 1 3b2 1 e 2 e2b a2 a
11. a) Show that if an investment is risk free, then theoretically its return rate must be the spot interest rate. b) Using this fact and Its lemma, derive the Black-Sc
24
2 Basic Options
b) Because V = Erer(T t) , t we have V 1 2V V + 2 S 2 2 + (r D0 )S rV t 2 S t = Erer(T t) rEer(T t) = 0. If E is an amount of money you want to have at time T . Eer(T t) is the amount of money you need to deposit in a bank at time t. 16
2 Basic Options
29
19. Suppose V (S, t) is the solution of the problem 1 2V V V + 2 (S )S 2 2 + (r D0 )S rV = 0 , 0 S, t T, t 2 S S V (S, T ) = V (S ), 0 S.
T
S , = T t and V (S, t) = (S + Pm )V (, ), where Pm is S + Pm a positive constant. a) Show that V
2 Basic Options
33
21. Suppose that S is a random variable which is dened on [0, ) and whose probability density function is G(S ) =
2 2 2 1 e[ln(S/a)+b /2] /2b , 2bS
is the solution of the initial-value problem 2u u , < x < , = x2 u(x, 0) = u (x), < x <
2 Basic Options
39
2 p + 1 2 (t)S 2 p + [r(t) D (t)]S p r(t)p = 0, 0 t 2 2 S S Therefore 0 S, t T, p(S, T ) = max(1 S, 0), 0 S.
p(S, t) = Ee (t) N (d2 ) (S/E )e (t) N (d1 ) = Ee (t) N (d2 ) Se (t) (t) N (d1 ) is the value of a put option with time-depend
48
2 Basic Options
is a linear function of c for c (3.2081, 4.8784). From the data we know (3.2081) = 0.3 and (4.8784) = 0.4. Thus (c) = (3.2081) + (4.8784) (3.2081) (c 3.2081) 4.8784 3.2081 0.1 (c 3.2081). = 0.3 + 4.8784 3.2081
Consequently (4.5) = 0.3 +