solution for stochastic volatility with ρ 0

... Doldrum Downturn Boom $ 100 $85 $ 105 p(1) = 0. 70 p(2) = 0. 20 p(3) = 0. 10 R(1) = 0. 00 R(2) = 0. 15 R(3) = 0. 05 0. 025 0. 06 One can assign probabilities for each scenario, and this in turn gives ... condition: forward rates with stochastic volatility 10. 9 Nonlinear change of numeraire 10. 10 Summary 10. 11 Appendix: Propagator for stochastic volatility 10. 12 Appendix: Effective linear Hamiltonian 10. 13 ... 3.6 Stock price with stochastic volatility 3.7 Merton–Garman equation 3.8 Summary 3.9 Appendix: Solution for stochastic volatility with ρ = Part II 11 13 15 16 18 20 23 25 25 27 30 31 34 38 39...

Ngày tải lên: 23/03/2014, 12:20

... price for an option on a particular stock? In other words, what is the option worth? When a stock has a call price of $ 100 and a value of $1 20, the option is worth at least $ 20 ($1 20 - $ 100 = $ 20) ... $ 100 , even though they are worth $1 20 However, if the value of the stock were $ 80 at the call date, then it would not be feasible to purchase these shares, because you would be paying $ 100 for ... the option to purchase 100 shares a year from now (a date known as the call date) for $ 100 each If the stock were to rise to $1 20 by the call date, it would be feasible for this person to exercise...

Ngày tải lên: 08/04/2014, 12:09

... (i)–(iii)? (i) 0 = 1, αk = for all k ∈ N, and 0 = −B, β1 = d, βk = for all k ∈ N \ {1} (ii) 0 = 1, αk = for all k ∈ N, and 0 = −B, β2 = c, βk = for all k ∈ N \ {2} (iii) 0 = 1, αk = for all k ... equation leaves us with two choices for r, namely, √ r± = ± B ≡ ρ (3.5) The result for F(r + n) where n ∈ N is also the same, since the function F only depends on the parameters 0 and 0 : F(r + n) ... n =0 (ii) Special case (a = d = 0, c = 0) : ψ1 (x) = x(b−1)/2 ∞ n =0 n − cx2 /2 n j =1 j(2 j + b − 1) , ψ2 (x) = x(1−b)/2 ∞ n =0 n − cx2 /2 n j =1 j(2 j + − b) (3.12) (iii) Special case (a = 0, ...

Ngày tải lên: 23/06/2014, 00:20

... 200 4 200 5 200 6 200 7 0, 114861 0, 108 927 0, 0996122 0, 0242895 _ R = i - _ R 0, 028637 0, 022 703 0, 0133882 -0, 0619345 _ ( R i − R) 0, 000 8 200 78 0, 000 51543 0, 000 179244 0, 008 6224 0, 114861 + 0, 108 927 + 0, ... 0, 008 6224 0, 114861 + 0, 108 927 + 0, 09 96122 + 0, 02 42895 = 0, 08692243 ⇒ Var = R i _ R =0, 08692243 *( 0, 000 8 200 78 +0, 000 51543 + 0, 000 179244 + 0, 008 6224) ⇒ Var = 0, 001 783545 Từ đó,sử dụng công thức ... 200 4 , 200 5 200 6 , 200 7 sau : Lợi nhuận 200 4 200 5 200 6 200 7 Giá trị tổng tài Lợi nhuận / Giá ( VNĐ) 10. 792.1 80. 816 11.846.921.953 12 .09 5.921.335 3.327 .04 5.218 Năm sản (VNĐ) 93.958.889.121 108 .734.555.744...

Ngày tải lên: 09/04/2013, 15:03

... normal with mean x + at1 , t0  and variance pt0; t1; x; y = p exp , y , x + at1 , t0  : 2t1 , t0  2t1 , t0 Note that p depends on t0 and t1 only through their difference t1 , t0 This ... by pt0 ; t1; x; y the density (in the y variable) of X t1, conditioned on X t0 = x In other words, IE t ;xhX t1 = The Markov property says that for  t0 IE 0;   Z IR hy pt0; t1; ... and for every ,  Z hX t1 F t0  = IR hy pt0; t1; X t0; y  dy: Example 16.3 (Drifted Brownian motion) Consider the SDE dXt = a dt + dBt: Conditioned on t1 , t0 , i.e., Xt0 ...

Ngày tải lên: 18/10/2013, 03:20

... (7. 50) 176 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 0. 8 Newtonian Gravity massive particles 0. 6 0. 4 L=1 0. 2 0 10 20 30 r 0. 8 Newtonian Gravity massless particles 0. 6 0. 4 L=5 0. 2 0 10 20 r 30 ... tensor: R0 101 R0 202 R0 303 R0 212 R0 313 R1 212 R1 313 R2 323 = = = = = = = = 2 e2(β−α) [ 0 β + ( 0 β)2 − 0 α 0 β] + [∂1 α∂1 β − ∂1 α − (∂1 α)2 ] −re−2β ∂1 α −re−2β sin2 θ ∂1 α −re−2α 0 β −re−2α ... Relativity massive particles 0. 6 0. 4 L=1 0. 2 0 10 20 30 r 0. 8 General Relativity massless particles 0. 6 0. 4 L=5 0. 2 0 10 20 r 30 179 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES limit their radii...

Ngày tải lên: 23/10/2013, 20:20

... 177.66 293.97 3 .00 4.44 84.93 0. 00 -2 0. 00 9.85 42.75 92.85 4.36 78.27 0. 00 -3 0. 00 0 .00 0. 00 0 .00 S Option Payoff m 3722.36 3 901 .58 408 3.63 boundary 30. 10 1. 30 boundary 0. 00 √ (C) The grid spacing ... chapter 97 Numerical Solutions of the Black Scholes Equation Table 8.1 Crank Nicolson example T = 0. 000 0 0. 1667 0. 3333 0. 500 0 t = 0. 000 0 0. 003 3 0. 006 7 0. 0 100 u x = ln S n m f0 4.85 127.76 27.76 ... 17.74 208 8.56 2242. 60 2413.97 2596.19 20. 77 4.69 108 .51 8.51 923.13 108 7 .05 1287.63 1476.38 12.81 4.61 100 .00 0. 00 0 .00 344.88 5 50. 20 722.67 6.81 4.52 92.16 0. 00 -1 0. 00 59.12 177.66 293.97 3 .00 4.44...

Ngày tải lên: 25/10/2013, 19:20

... C0 = (r − q)S0 + σ S0 − rC0 ∂T ∂ S0 ∂ S0 subject to the initial and boundary conditions r C(S0 , 0) = max [0, S0 − X ] r limS 0 C(S0 , T ) → r limS →∞ C(S0 , T ) → S0 e−qT − X e−r T 0 (ii) Let ... values for σ = 20% , T = months are 3.99% and 4 .00 %; for σ = 40% , T = years they are 31 .08 % and 32 .00 % 5.5 ADAPTATION TO DIFFERENT MARKETS (i) The objective of this book is to provide the reader with ... 2.2(i): F0T + P0 er T = X + C0 er T Substitute from the Black Scholes expression for C0 : P0 er T = X {1 − N[d2 ]} − F0T {1 − N[d1 ]} From equation (A1.4), N[d] + N[−d] = 1, so that P0 = e−r T...

Ngày tải lên: 25/10/2013, 19:20

... Yt0 + s b( s )ds + s d (s )dWs t X t = t0 X t0 + s b (s )ds + s 1d ( s )dWs t0 1 t0 t0 t0 t t t t0 = nờn X t = t X t0 + s 1b( s) ds + s 1d ( s)dWs , ú t = exp a ( s)ds Vỡ t0 t0 ... T Xột cỏc trng hp : S0 S > ị ln -0rcT > 0; d1 đ +Ơ - rcT Xe Xe N ( d1 ), N (d ) đ , ú giỏ quyn chn tr thnh: S0 - Xe- rcT Do ú: S0 S - rT Ê ị ln -0rcT Ê 0; d1 đ - Ơ - Nu S0 Ê Xe c ị - rcT Xe ... Exchange, American Journal of Sociology Vol 109 ( 200 3), pp 1 20- 127 [12] K OToole, Stanfords Scholes Wins Nobel for Economics, Stanford News Service (14/ 10/ 1997) [13] A Shah, Black, Merton and Scholes:...

Ngày tải lên: 19/12/2013, 14:05

... 5,69 cổ phiếu • Tuy nhiên, giá quyền chọn giảm khoảng 0, 01 (0, 569) hay 0, 005 69 Vì có 1 .00 0 quyền chọn, quyền chọn tổng cộng giảm 0, 005 69(1 .00 0) = 5,69 • Vì bán quyền chọn, thu 5,69, bù đắp cho khoản ... quyền chọn mua • Ví dụ, delta 0, 5691 nên xây dựng danh mục phòng ngừa delta cách mua 569 cổ phiếu bán 1 .00 0 quyền chọn • Nếu giá cổ phiếu giảm xuống 0, 01, lỗ 0, 01(569) = 5,69 cổ phiếu • Tuy nhiên, ... $125,9375 • Thời gian đến đáo hạn 0, 0959 • Lãi suất phi rủi ro 4,56% • Độ lệch chuẩn = 0, 83 Lãi suất phi rủi ro phải biểu diễn dạng lãi suất ghép lãi liên tục Rc = ln(1 ,04 56) = 4,46 CÔNG THỨC ĐOẠT...

Ngày tải lên: 25/01/2014, 11:20

... Theorem 1.1 with p(x) = g G x • Corollary 1.3 For every x0 ∈ E there exists f0 ∈ E such that f0 = x0 and f0 , x0 = x0 Proof Use Corollary 1.2 with G = Rx0 and g(tx0 ) = t x0 , so that g G = x0 Remark ... But B(x0 , 0 ) ⊂ D(ϕ) for some 0 > (small enough), and thus f, x0 + 0 z ≥ α ∀z ∈ B (0, 1), which implies that f, x0 ≥ α + 0 f On the other hand, we have f, x0 ≤ α, since x0 ∈ D(ψ); therefore ... point y0 ∈ T (U ), so that y0 = T x0 for some x0 ∈ U Let r > be such that B(x0 , r) ⊂ U , i.e., x0 + B (0, r) ⊂ U It follows that y0 + T (B (0, r)) ⊂ T (U ) Using (7) we obtain T (B (0, r)) ⊃ B (0, ...

Ngày tải lên: 04/02/2014, 11:10

... u + 4u = 0, u (0) = 1, (0) = 0; dt dt du d2 u du (0) = 1; b + + 2u = 0, u (0) = 1, dt dt dt du d2 u du (0) = 1; c + + u = 0, u (0) = 1, dt dt dt du d2 u du (0) = d + + 0. 75u = 0, u (0) = 0, dt dt ... = 0, u1 (x) = √ x + 18 dx dx 4x x In Exercises 19–21, use the indicated change of variable to solve the differential equation d dR u (ρ) 19 ρ2 + λ2 ρ R = 0, R (ρ) = dρ dρ ρ 15 20 dφ d ρ dρ dρ ... conditions y (0) = 0. 001 h, y (0) = 24 (Continuation) For an input speed of 25.4 m/s, it is observed that σ ∼ = 600 Hz or 1 200 π radians/s and α = 0. 103 σ Using these values, obtain a graph of the solution...

Ngày tải lên: 17/02/2014, 14:20

... energy and entropy functions for our new model fluid with dissipation As before we start with a form of the First Law: For each cyclic process Γ of our homogeneous fluid body with dissipation, we assume ... , , Xm ) for all W, X1 , , Xm Differentiate with respect to Xk : 0= ∂E ∂E ∂S + ∂S ∂Xk ∂Xk =T =−Pk ✷ We record the definitions (6) by writing m ( 10) dE = T dS − Pk dXk Gibbs’ formula k=1 ... parameters X0 , X1 , , Xm , of which E = X0 , the internal energy, plays a special role Thus we imagine, for instance, a body of fluid, for which there is no temporal or spatial dependence for E,...

Ngày tải lên: 17/03/2014, 14:29

... 11111111111111111111111111 00 000 000 000 000 000 000 000 000 11111111111111111111111111 00 000 000 000 000 000 000 000 000 11111111111111111111111111 00 000 000 000 000 000 000 000 000 11111111111111111111111111 00 000 000 000 000 000 000 000 000 ... 11111111111111111111111111 00 000 000 000 000 000 000 000 000 11111111111111111111111111 00 000 000 000 000 000 000 000 000 11111111111111111111111111 00 000 000 000 000 000 000 000 000 11111111111111111111111111 00 000 000 000 000 000 000 000 000 ... 11111111111111111111111111 00 000 000 000 000 000 000 000 000 11111111111111111111111111 00 000 000 000 000 000 000 000 000 11111111111111111111111111 00 000 000 000 000 000 000 000 000 11111111111111111111111111 00 000 000 000 000 000 000 000 000 ...

Ngày tải lên: 31/03/2014, 15:16

... Fourier Transforms 200 204 218 2 30 234 2 40 244 249 261 263 273 286 292 306 308 315 318 326 327 335 339 344 347 349 358 376 387 403 4 10 4 60 471 478 479 4 80 482 483 485 486 489 4 90 496 497 Contents ... is represented when you write a decimal number For example, 2 704 .53816 ϭ ϫ 103 ϩ ϫ 102 ϩ ϫ 101 ϩ ϫ 100 ϩ ϫ 10 1 ϩ ϫ 10 2 ϩ ϫ 10 3 ϩ ϫ 10 4 ϩ ϫ 10 5 Figure 1.1 The natural numbers can be depicted ... decimals, are: 3/4 ϭ 0. 7 500 00 Ϫ9/8 ϭ Ϫ 1.12 500 00 Examples of the second type of rational number, known as nonterminating, repeating decimals, are: 1/3 ϭ 0. 33333 Ϫ123/999 ϭ 0. 123123123 12 Chapter...

Ngày tải lên: 31/03/2014, 16:19

... 297– 300 (1978) doi: 10. 109 0/S 000 2-9939-1978 -05 07327-1 Bourgin, DG: Classes of transformations and bordering transformations Bull Am Math Soc 57, 223–237 (1951) doi: 10. 109 0/S 000 2-9 904 -1951 -09 511-7 Forti, ... 4455–4472 ( 200 2) doi: 10. 109 0/S 000 2-9947 -02 -03 036-2 Forti, GL: Hyers-Ulam stability of functional equations in several variables Aequ Math 50, 143–1 90 (1995) doi: 10. 100 7/ BF01831117 Forti, GL: ... doi: 10. 109 0/S 000 2-9939-98 -04 6 80- 2 Miheţ, D: The fixed point method for fuzzy stability of the Jensen functional equation Fuzzy Sets Syst 1 60, 1663–1667 ( 200 9) doi: 10. 101 6/j.fss. 200 8 .06 .01 4 Miheţ,...

Ngày tải lên: 20/06/2014, 22:20

... t with g ∈ L1 t0 ε, t0 ≤g t L Hence for s, t ∈ t0 for a a t ∈ t0 ε, t0 L , 5. 10 L , s ≤ t, we have t y∞ t − y∞ s ε, t0 lim n→∞ s yn ≤ t g, 5.11 s and therefore y∞ ∈ AC t0 ε, t0 L Since ε ∈ 0, ... y∞ is a weak lower solution of 1.1 For ε ∈ 0, L and n ∈ N such that tn < t0 ε we have 5. 10 with g ∈ L1 t0 ε, t0 L , hence lim sup yn ∈ L1 t0 ε, t0 L , and for s, t ∈ t0 ε, t0 L , s < t, Fatou’s ... C then there is some t0 ∈ B such that α t0 ⊂ the definitions of B and F yield α t0 > lim sup F t0 , y y → α t0 3.19 − x∗ t0 and α t0 x∗ t0 , but then f t0 , α t 3. 20 Therefore B \ C ⊂ N and thus...

Ngày tải lên: 21/06/2014, 20:20

... t ≤ G t for all t in R The maximal element for Δ in this order is the distribution function 0 given by 0 t 0, if t ≤ 0, 1, if t > 1.5 Definition 1.1 see 20 A mapping T : 0, × 0, → 0, is a continuous ... we get μ3f t ≥ 0 ,y t 2y −24f y 2.4 for all y ∈ X If we replace y in 2.4 by x and divide both sides of 2.4 by 3, we get μf 2x −8f x t ≥ 0 ,x 3t ≥ 0 ,x t 2.5 t ≥ 0 ,x 23 t 2.6 for all x ∈ X and ... ≥ 0 ,2k x 23 k ··· 1/22 −f 2k x /24k t 2k 1 t 2.19 1/2n , by the triangle inequality it n−1 ≥ Tk 0 ,2k x 23 k t 2. 20 Tin 0 ,2i−1 x 23i t for all x ∈ X and t > We replace x with 2m x in 2. 20 to...

Ngày tải lên: 22/06/2014, 02:20

... main results obtained are local uniform estimates for local subsolutions, the Harnack inequality for positive local solutions and the continuity for local solutions The paper is original, the ... boundary Harnack principle for C domains for solutions of the Δ∞ equation, extending results of a previous paper of his, where he had proved the same principle for domains with flat boundaries Then ... elliptic PDEs in divergence form and its applications to free boundary problems They establish for the first time for divergence-form Emmanuele DiBenedetto et al equations with nonsmooth coefficients...

Ngày tải lên: 22/06/2014, 19:20

... increasing with respect to d and f , and decreasing with respect to c Proof Let P be defined by (2.1) with w(t) := d0 + b t c0 + p(s) s a h0 (x)dx ds, t ∈ J, (5.8) where c0 = sup{ c(u) | u ∈ C(J,E)}, d0 ... & Hall/CRC, Florida, 200 0 , Elliptic problems with lack of compactness via a new fixed point theorem, J Differential Equations 186 ( 200 2), no 1, 122–1 40 , Existence results for equations in ordered ... (Catania, 200 0) S Carl and S Heikkil¨ 179 a [6] [7] [8] [9] [ 10] [11] , New algorithms to solve equations and systems in ordered spaces, Neural Parallel Sci Comput ( 200 1), no 3-4, 407 –416 , Existence...

Ngày tải lên: 23/06/2014, 00:20