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American College of Gastroenterology - ACCOUNTING - ac550
PROJECT 2Note: Each part of the case is independent of each other. Upload into your Dropbox when you complete the assignment.Part AOn December 31, 2010 Felt Company's inventory burned. Sales and purchases for the year had been $1,400,000 and $980,000, r
Acadia - WOMENS STU - 190
Wang1Hercules,Achilles,Superman.Theseareindividualsidentifiedasheroes.Aside fromthesuperhumanstrength,extraordinarypowers,anddevilishgoodlooks,these heroesallhaveonethingincommon:theyarefictional.Heroes,inthetruestsenseof theword,donotexistinreallifethey
Penn State - CHEM - 466
Ch 6 Entropy and The Boltzmann Law
University of North Carolina, Wilmington - MATH - 6203
2 Basic Options79boundary 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
University of North Carolina, Wilmington - MATH - 6203
602 Basic Options (E ) = Thus we haveln E [ln S + (r D0 ) (T t)] 2 (T t)/2 T t ln E r(T t) [ln S D0 (T t)] 2 (T t)/2 = T t ln S1 ln S0 + 2 (T t)/2 = . T t max(S , E )G(S , T ; S, t)dS = f (S1 , S0 , t; ).er(T t)EConsequently the conclusion is prov
University of North Carolina, Wilmington - MATH - 6203
482 Basic Optionsis 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 3.2081Consequently (4.5) = 0.3 +
University of North Carolina, Wilmington - MATH - 6203
2 Basic Options39 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
University of North Carolina, Wilmington - MATH - 6203
2 Basic Options3321. 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 , 2bSis the solution of the initial-value problem 2u u , < x < , = x2 u(x, 0) = u (x), < x <
University of North Carolina, Wilmington - MATH - 6203
2 Basic Options2919. 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.TS , = T t and V (S, t) = (S + Pm )V (, ), where Pm is S + Pm a positive constant. a) Show that V
University of North Carolina, Wilmington - MATH - 6203
242 Basic Optionsb) 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
University of North Carolina, Wilmington - MATH - 6203
162 Basic OptionsVar [ ] = E 2 E [ ] = = e2b b2 (e 1). a2222 1 3b2 1 e 2 e2b a2 a11. 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
University of North Carolina, Wilmington - MATH - 6203
2 Basic Options9b) 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. BecausenE [X (t)] = lim En i=
University of North Carolina, Wilmington - MATH - 6203
2 Basic OptionsProblems1. a) Show 2 1 ex /2 dx = 1. 2b) Show that 2 2 1 e(xa) /2b dx = 1 b 2holds 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
University of North Carolina, Wilmington - MATH - 6203
1 IntroductionProblems1. 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
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
1Homework Problems (Part I) for Derivative Securities and Dierence MethodsProblems31 IntroductionProblems1. 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
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
Main Subjects of Chapters 3 and 41. 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
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
2 Basic Options87P (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
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
3 Exotic OptionsProblems1. 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 )
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
1263 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 ert Bl Co (S , tk+1 ; Bl )G1 (S , tk+1 ; S, tk , Bl )d
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
3 Exotic Options129c(bE ) , t; r, D0 + 2= er(T t) f =e =cr (T t)= c(S, t; r, D0 ) f(bE )2 (D0 +)(T t) e , 2 (T Eer(T t)2 Bl (t) D0 (T t) Se , 2 (T Eer(T t) t) t)2 Bl (t) , t; r, D0 . SConsequently, we nally have c0 (S, t) = c(S, t; r, D0 ) Bl
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
3 Exotic Options1371 + 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
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
1823 Exotic Optionsc) 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
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
1843 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
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
2023 Exotic OptionsTherefore according to these relations above, for the integral given we have er = S1 S0 0 S1TS1T (S1T , S2T ; S1 , S2 , t)dS2T dS1T1 2 det P1 2 T 1 1 e(1 Pe1 ) P (1 Pe1 )/2 dS2T dS1T S1T S2T 1 = S1 2 1 2 201 = S1e[1 (z21 )2201 1
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
3 Exotic Options2251 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
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
4 Interest Rate Derivative SecuritiesProblems1. 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
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
4 Interest Rate Derivative Securities243with 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 +
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
4 Interest Rate Derivative Securities251 = = =2 + ( + )2 e(T t) ( + ) ( + )e(T t) ( ) 2 ( + )e(T t) ( ) 2 e(T t) ( + )e(T t) ( ) 2 ( + )e(T t) ( ) 2 e( +)(T t)/2 ( + )e(T t) ( )2/ ( + )/ ( )/( + )/2/e( +)(T t)/ ,the two expressions are identical.
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
2544 Interest Rate Derivative SecuritiesT+V (r, T + ) V (r, T ) =TV dt = tT+ (t T )dt = 1.TThat 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 ,
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
4 Interest Rate Derivative Securities261Then 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,
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
2804 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
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
2844 Interest Rate Derivative Securitiesand using the nal condition, we arrive at b0 (t) = kZ 1 er(T t) , rwhich 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
University of North Carolina, Wilmington - MATH 6203 - MATH 6203
1Mathematics 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
Quebec Trois-Rivieres - MA - 141
1. (a) Find parametric equations for the line that passes through the point (2 , 0,1) and is perpendicular to the plane with equation 4 x y 2z = 1. Solution. The direction vector for this line is v = (4,1,2) and it must pass through the point (2, 0,1). Th
Quebec Trois-Rivieres - MA - 141
whose equations are: L1 : 8< : x=t y=2t z=3 , L2 : 8< : x = 1 + 2t y = 3 2t z = 3t. Solution. Since the plane contains the two lines, their direction vectors (1 ,1, 0) and (2,2, 3) are parallel to the plane. Hence their cross product will be a normal vect
Quebec Trois-Rivieres - MA - 141
(1,1, 0) (2,2, 3) = _ ijk 1 1 0 2 2 3 _ = 3i 3j + 0k = (3,3, 0). To get a point in the plane, we can take any point in either line, so just set t = 0 in the equations for L1 to get the point (0, 2, 3). The equation for the plane is
Quebec Trois-Rivieres - MA - 141
z), and then graph the surface z = y2/x for z _ 0. Be sure to label your axes. Solution. Setting z = c 6= 0, we can solve for x to get x = y2/c. These graphs are (sideways) parabolas in the xy-plane, that get less steep as c gets large, and more steep as
Quebec Trois-Rivieres - MA - 141
that these level curves all have a hole where x = 0 since z is not defined there. (See the link to the level curves picture.) In graphing the whole surface, we sketch two cross sections for x = 1 and x = 2. These again are parabolas with equations z = y2
Quebec Trois-Rivieres - MA - 141
3. Calculate the following limits, or show that they do not exist. (a) lim( x,y)! (0,0)e2xyy2 x2 + y2 Solution. As (x, y) ! (0, 0), the numerator approaches e0 = 1, since it is a continuous function of x and y, while the denominator approaches 0 and is
Quebec Trois-Rivieres - MA - 141
(b) lim( x,y)! (0,0)2xy2 x2 + y2 Solution. Convert to polar coordinates to get limr! 0+2r3 cos _ sin2 _ r2 = limr! 0+2r cos _ sin2 _ = (2 cos _ sin2 _) limr! 0+r = 0. (c) lim( x,y)! (0,0)sin(x
Quebec Trois-Rivieres - MA - 141
Solution. W e take the limit as (x, y) approaches (0, 0) along two different lines with equations y = cx. First, let (x, y) approach (0, 0) along the line y = 0. We get lim( x, y) ! (0, 0) y=0sin(x + y) 2x y = limx! 0sin x 2x = limx! 0cos x 2 = 1 2
Quebec Trois-Rivieres - MA - 141
using lHospitals rule for the second equality. If (x, y) approaches (0, 0) along the line y = x instead, we get lim( x, y) ! (0, 0) y=ssin(x + y) 2x y = limx! 0sin(2x) x = limx! 02 cos(2x) 1 = 2, using lHospitals rule for the second equality. Since
Quebec Trois-Rivieres - MA - 141
4. A function y = (y1, y2) is defined by y1 = 3x21 + x22 and y2 = x1x2 1 x1 + 2 . (a) Find the Jacobian matrix _ @yi @xj _ , and say where y is differentiable. Solution. yx = _ @yi @xj _ =@y1 @x1 @y1 @x2 @y2 @x1 @y2 @x2! = _ 6x1 2x22x2 +1 ( x1+2)2 x1 x
Quebec Trois-Rivieres - MA - 141
Approximate y(1.01, 2.02). Solution. Let _y = y(1.01, 2.02) y(1, 2). We know that dy at (1, 2) approximates _y, so we have _y _ dy|(1,2) = yx|(1,2)dx = _ 6 4 5 1 _ 0.01 0.02 _ = _ 0.14 0.07 _ . Here we have taken dx1 = 1.01 1 = 0.01 and dx2 = 2.02 2 = 0.0
Quebec Trois-Rivieres - MA - 141
5. Suppose we have functions z(y1, y2, y3) : R3 ! R2 and y(x1, x2) : R2 cfw_(0, 0) ! R3 given by z= _ z1 = y1y2y3 z2 = y1y2 2 + 2y2y 23and y = 8< : y1 = x2 cos(_x1) y2 = x2 sin(_x1) y3 = ln(x21 + x22 ) (a) Express the Jacobian matrix _ @zi @xj _
Quebec Trois-Rivieres - MA - 141
of the composition z _ y as a product of two matrices (do not evaluate this product). Solution. The chain rule says zx = zyyx. So zx = _ y2y3 y1y3 y1y2 y2 2 2y1y 2 + 2y2 3 4y2y 3 _0 @ _x2 sin(_x1) cos(_x1) _x2 cos(_x1) sin(_x1)2x1 x21 +x22 2x2 x21 +x221
Quebec Trois-Rivieres - MA - 141
Find _ @z2 @x1 _x2(3,1) and simplify your answer. Solution. To get _ @z2 @x1 _x2we must multiply the second row of the first matrix above by the first column of the second matrix, and then we need to evaluate at (x1, x2) = (3,1). We get _ @z2 @x1 _x2
Quebec Trois-Rivieres - MA - 141
Notice that we have y1(3,1) = cos(3_) = 1, y2(3,1) = sin(3_) = 0 and y3(3,1) = ln 10. Thus plugging in x1 = 3 and x2 = 1 in the above, the first and last terms of the sum become 0, and we are left with y2 3_x2 cos(_x1) =
Quebec Trois-Rivieres - MA - 141
(ln 10)2_(1) cos(3_) = _(ln 10)2. 6. Suppose f(x, y) and g(x, y) are differentiable functions, and define h(x, y) = f(x, y)g(x, y).
Quebec Trois-Rivieres - MA - 141
Show that dh = fdg + gdf. Solution. By definition, dh = hxdx + hydy. By the product rule, hx = @ @x (fg) =
Quebec Trois-Rivieres - MA - 141
fxg + fgx and hy = @ @y (fg) = fyg + fg y. Thus, substituting these expressions into the equation for dh and rearranging the terms, we have
Quebec Trois-Rivieres - MA - 141
dh = (fxg + fgx)dx + (fyg + fgy)dy = g(fxdx + fydy) + f(gxdx + gydy) = gdf + fdg. 3
Cal Poly Pomona - DFS - dsfsd
A.P. Biology: Chapter 12 & 13 Test1. The centromere is a region in which a. chromatids are attached to one another. b. metaphase chromosomes become aligned. c. chromosomes are grouped during telophase. d. the nucleus is located prior to mitosis. e. new s
NJIT - MGMT - 691
NEW JERSEY INSTITUTE OF TECHNOLOGY Instructor: Diana Walsh, Esq. Tel.: (973) 596-6418 Fax: (973) 492-5347 E-mail: diana.walsh@njit.edu Webpage: http:/web.njit.edu/~walshdDear Students: As your instructor for MGMT 691 I would like to welcome you to our vi
NJIT - MGMT - 691
a. How should (ethics) we regulate our energy? whom do you think does this well and who benefits?Energy must be controlled by the government.Businesses won't operate without profits motive or without keeping in mind their objective, may be long term. So
NJIT - MGMT - 691
after all the readings of various articles and case law opinions.argue your decisions from a ethical perspective ( motives/consequences of the parties to these cases and is your decision enforcing what is good behavior for the greatest good for the grea
NJIT - MGMT - 691
New meningitis vaccine from India An Indian company called Serum Institute of India Ltd (SIIL) has, in collaboration with the World Health Organization (WHO) and Meningitis Vaccine Project (MVP), developed a new and affordable vaccine for meningococc
NJIT - MGMT - 691
After reading and debating this issue, who do you think is responsible for what goes on in the corporate world? Who, if anyone, should be held accountable and how come? With whom do you agree?Circumstances and pressure have some effect on the decision an
NJIT - MGMT - 691
NEW JERSEY INSTITUTE OF TECHNOLOGY School of ManagementSyllabusDiana Walsh, M.A., J.D (N J Bar) Email: diana.walsh@njit.eduCourse Title: Legal and Ethical Issues Semester: Fall 2007Webpage:http:/web.njit.edu/~walshd/Course No.: Mgmt 691 -851Class T
UC Irvine - CEE - 81B
CEE 81B Lab Report Wells Tsai, 901961031.On the first graph, its easy to see that when the speed is around 40-60 the volume stacks up, and this is for upstream and downstream. This would make sense since there shouldnt be too much of a difference betwee
LSU - HIST - 2055
History 2055 (Spring 2010)Wednesday, January 20 Went over syllabus. o Need to buy books. Assistant: Spencer McBride o Himes 313 o Monday & Wednesday: 12-1:30 o Smcbri4@lsu.eduFriday, January 22Early European Exploration and Settlement and Indians Exp