2 Basic Options
79
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 app
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.
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 cal
202
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 dS2
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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 N
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
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)
254
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 identic
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 ran
280
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
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
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)
SCM 370
PRODUCTION AND OPERATIONS MANAGEMENT
(Operations According to Seinfeld)
Fall 2017
Instructor:
Office:
Phone:
E-Mail:
Web Site:
Office Hours:
Dr. Drew Rosen
2054 Computer Information System (CI
184
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
+
182
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
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.
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
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-v
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 (, ),
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 amou
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
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
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 prob
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 bu
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 option
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
126
3 Exotic Options
Consequently, noticing Co (S, tK ; Bl ) = Co (S, tK ; Bl ) = Gc (S ), we obtain Co (S, tk ; Bl )
= max ert
Bl
Co (S , tk+1 ; Bl )G1 (S , tk+1 ; S, tk , Bl )dS ,
Gc (S )
max e