phụ lục 2 bảng câu hỏi khảo sát

... symbols References Index 22 5 22 6 22 8 23 1 23 5 23 7 24 0 24 0 24 2 24 4 24 5 24 8 24 9 25 1 25 2 25 3 26 0 26 0 26 3 26 5 26 8 27 1 27 2 27 4 27 5 27 6 27 7 28 2 28 4 28 4 28 6 28 8 29 2 29 3 29 6 29 8 301 303 305 310 315 Foreword ... 156 157 161 1 62 164 165 168 169 1 72 173 174 176 180 1 82 183 188 191 1 92 194 196 197 20 1 20 5 20 9 21 4 21 5 21 7 21 7 21 8 22 0 22 2 x Contents 9.5 Field theory hedging of Treasury Bonds 9.6 Stochastic ... finance 2. 1 Efficient market: random evolution of securities 2. 2 Financial markets 2. 3 Risk and return 2. 4 Time value of money 2. 5 No arbitrage, martingales and risk-neutral measure 2. 6 Hedging 2. 7...

Ngày tải lên: 23/03/2014, 12:20

... option worth? When a stock has a call price of $100 and a value of $ 120 , the option is worth at least $20 ($ 120 - $100 = $20 ) The value of the option clearly depen d s on the value of the stock ... rise to $ 120 by the call date, it would be feasible for this person to exercise his / h e r option, because the shares would still only cost you $100, even though they are worth $ 120 However, ... they should make it less significant (NOVA Online, 20 00) Their solution to this proble m was to become of the most celebrate d discoveries of the 20 th century The solution was rooted in the old...

Ngày tải lên: 08/04/2014, 12:09

... , 2 (x) = x(1−b) /2 ∞ n=0 n − cx2 /2 n j =1 j (2 j + − b) (3. 12) (iii) Special case (a = 0, c = a(b + 1) /2, d = 0): ψ1 (x) = x(b−1) /2 ∞ n=0 n Ax4 /4 , n j =1 j(4 j + b − 1) 2 (x) = x(1−b) /2 ∞ ... (i) and (ii), we obtain y1 (x) = y1 (x) ≤ e−ax /4 ax2 cosh ∀x > 0, (4.13) and therefore, y1 (x) ≤ −ax2 /2 ax2 /4 −ax2 /4 e e +e = 2 + 2e−ax /2 + e−ax ∀x > (4.14) For a > 0, the statement is hence ... ≡ x(1−b) /2 e−ax /4 ψ(x) = x−|m| eλx /4 ψ(x), 2 (2. 2) M Simon and A Ruffing 111 recovering the following Bessel-like equation for the wave function ψ: x2 ψ + xψ + − a2 a(b + 1) (b − 1 )2 x + c− x...

Ngày tải lên: 23/06/2014, 00:20

... R 20 04 20 05 20 06 20 07 0,114861 0,108 927 0,0996 122 0, 024 2895 _ R = i - _ R 0, 028 637 0, 022 703 0,01338 82 -0,0619345 _ ( R i − R) 0,000 820 078 0,00051543 0,00017 924 4 0,008 622 4 0,114861 + 0,108 927 ... năm 20 04 , 20 05 20 06 , 20 07 sau : Lợi nhuận 20 04 20 05 20 06 20 07 Giá trị tổng tài Lợi nhuận / Giá ( VNĐ) 10.7 92. 180.816 11.846. 921 .953 12. 095. 921 .335 3. 327 .045 .21 8 Năm sản (VNĐ) 93.958.889. 121 ... 0,00017 924 4 0,008 622 4 0,114861 + 0,108 927 + 0, 0996 122 + 0, 024 2895 = 0,086 922 43 ⇒ Var = R i _ R =0,086 922 43 *( 0,000 820 078 +0,00051543 + 0,00017 924 4 + 0,008 622 4) ⇒ Var = 0,001783545 Từ đó,sử dụng công...

Ngày tải lên: 09/04/2013, 15:03

... , x + a  : 2 @ p = p = @ p exp , y , x , a 2 @ @ 2 @ y , x , a 2 2 p exp , y , x , a  , @  ay , , a  y2 x , a   x = ,2 + + ,2 p: @ p = p = y , x , a p: x @x @2 p = p = @ y , ... rXt dt + Xt dBt: h y 1 p ; x; y = p exp , 2 log x , r , 2  y 2 It is true but very tedious to verify that p satisfies the KBE p = rxpx + 2x2 pxx: i : 16.5 Connection between stochastic ... , 2 t1 , t0 : The random variable Bt1  , Bt0  has density IP fBt1  , Bt0  dbg = p exp , 2 t b, t  db; 2 t1 , t0 and we are making the change of variable  y = x exp b + r , 2 t1...

Ngày tải lên: 18/10/2013, 03:20

... R0 20 2 R0 303 R0 21 2 R0 313 R1 21 2 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 2 ∂1 α −re 2 sin2 θ ∂1 α −re 2 ∂0 β −re 2 sin2 θ ... the following mess: ds2 = −dt2 + 2GMr 2 dr + 2 d 2 + (r + a2 ) sin2 θ d 2 + (a sin2 θ dφ − dt )2 , ∆ ρ (7.114) where ∆(r) = r − 2GMr + a2 , (7.115) 2 (r, θ) = r + a2 cos2 θ (7.116) and Here ... pµ = m − 2GMr 2 dφ dt 2mGMar + sin2 θ dτ 2 dτ (7.136) THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 21 3 and L = ηµ pµ = − dt m(r + a2 )2 − m∆a2 sin2 θ dφ 2mGMar sin2 θ + sin θ 2 dτ 2 dτ (7.137)...

Ngày tải lên: 23/10/2013, 20:20

... 127 .76 27 .76 n= initial condition 3545.94 4.77 117.74 17.74 20 88.56 22 42. 60 24 13.97 25 96.19 20 .77 4.69 108.51 8.51 923 .13 1087.05 128 7.63 1476.38 12. 81 4.61 100.00 0.00 0.00 344.88 550 .20 722 .67 ... 550 .20 722 .67 6.81 4. 52 92. 16 0.00 -1 0.00 59. 12 177.66 29 3.97 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 3 722 .36 3901.58 4083.63 boundary ... α(δx )2 m ∂ 2u ∂x2 n−1 α u n+1 + u m − 2u n → α(δt )2 m m ∂ 2u ∂t n+1 n−1 um − um → δt ∂u ; ∂t so that the equation for the Dufort–Frankel scheme in Section 8.1(vii) becomes ∂ 2u ∂ 2u ∂u + 2 =...

Ngày tải lên: 25/10/2013, 19:20

... follows that √ d2 − d1 = −σ T d + d1 = and √ σ T ln S0 e−qT X e−r T so that 2 d2 − d1 = (d2 − d1 )(d2 + d1 ) = 2 ln S0 e−qT X e−r T Substituting this into the explicit expression for n(d2 ) in (A) ... n(d2 ) = √ e 2 S0 e−qT X e−r T S0 e−qT n(d1 ) = X e−r T n(d2 ) or √ (C) Differentiating the relationship d1 − d2 = σ T gives   ∂d1 − ∂d2 =    ∂ S0 ∂ S0    ∂d  ∂d2   − =0  ∂r ∂r ∂d2 ... / σ e−kx−k 2 T v(x, T ); x = ln S0 ; T = σ 2T ; k= r − q − 1 2 2 Substituting in the previous equation and doing the algebra reduces the problem to a solution of the equation ∂v ∂ 2v = ∂T ∂x...

Ngày tải lên: 25/10/2013, 19:20

... v EX Theo (2. 5), ta cú 17 E Yt Yt (2) (1) t 3(1 + t ) D Ys (2) Ys (1) t ds 3(1 + T ) D 2 A1sds 3(1 + T ) D A1 t2 t2 = A2 2! Quy np theo k, ta c: E Yt ( k +1) Yt ( k ) A2 k +1 t k +1 ... 2. 2.4 B : Nu thỡ dX t = f1 (t ) + G1 (t )dWt dYt = f (t ) + G2 (t )dWt (2. 15) (2. 16) dX tYt = X t dYt + Yt dX t + G1 (t )G2 (t )dt = ( X t f (t ) + Yt f1 (t ) + G1 (t )G2 (t ) ) dt + ( X t G2 ... ds. 12 ds = t a ds 0 ( ) Thay vo (2. 4), ta c t t X ả E Z Z + tE a ds + E ds E t Xt ữ ữ a2 a2 + a + D2 X ả Xs ( ) s a + D X s ả s nờn X Vỡ 2 a2 + ...

Ngày tải lên: 19/12/2013, 14:05

... nhỏ giá cổ phiếu Delta 0,56 92 nghĩa giá quyền chọn biến động 56, 92% so với thay đổi giá cổ phiếu Ví dụ, giá cổ phiếu $130, tăng $4,0 625 , giá quyền chọn $15,96, tăng $2, 41, khoảng 59% biến động ... dẫn đến giá quyền chọn mua cao Ví dụ: Giả định giá cổ phiếu $130 thay $ 125 ,9375 Nó tạo giá trị N(d1) N(d2) 0,6171 0,51 62, giá trị C $15,96, cao giá trị đạt trước $13,55 CÁC BIẾN SỐ TRONG MÔ ... CÔNG THỨC ĐOẠT GIẢI NOBEL C = S N(d ) − Xe − rc T N(d ) Với ln(S0 /X) + (rc + σ /2) T d1 = σ T d2 = d − σ T N(d1), N(d2) = xác suất phân phối chuẩn tích lũy σ = độ bất ổn hàng năm (độ lệch chuẩn)...

Ngày tải lên: 25/01/2014, 11:20

... 20 1 8.1 Motivation 20 1 8 .2 The Sobolev Space W 1,p (I ) 20 2 1,p 8.3 The Space W0 ... 21 7 8.4 Some Examples of Boundary Value Problems 22 0 8.5 The Maximum Principle 22 9 8.6 Eigenfunctions and Spectral ... = {x = (xn )n≥1 ∈ E; x2n = ∀n ≥ 1} and Y = y = (yn )n≥1 ∈ E; y2n = y2n−1 ∀n ≥ 2n Check that X and Y are closed linear spaces and that X + Y = E Let c ∈ E be defined by 24 The Hahn–Banach Theorems...

Ngày tải lên: 04/02/2014, 11:10

... References 124 Chapter Review 125 Miscellaneous Exercises 125 v vi CHAPTER Contents The Heat Equation 135 2. 1 2. 2 2. 3 2. 4 2. 5 2. 6 2. 7 2. 8 2. 9 2. 10 2. 11 2. 12 2.13 CHAPTER The Wave Equation 21 5 3.1 3 .2 3.3 ... impose the extra requirement that dv1 dv2 u1 + u2 = 0, dt dt ( 12) up = v1 u1 + v2 u2 + v1 u1 + v2 u2 = v1 u1 + v2 u2 , (13) up = v1 u1 + v2 u2 + v1 u1 + v2 u2 , (14) then we find that and the equation ... the dependent and independent variables d2 φ + 2 φ = dx2 d2 φ − 2 φ = dx2 d2 u = dt dT = − 2 kT dt d dw r r dr dr − 2 w = r2 d2 R dR − n(n + 1)R = + 2 dρ dρ In Exercises 7–11, find the general...

Ngày tải lên: 17/02/2014, 14:20

... ∂2F ∂T ∂2F ∂V (17) ∂S = − ∂T = ∂ E ∂S ∂V ∂S ∂V + ∂2E ∂V Next differentiate (16): 1= 0= Thus (17) gives:      ∂2F ∂T =− ∂2F ∂V = ∂2E ∂V ∂2E ∂S ∂2E ∂S ∂2E ∂S − ∂S ∂T ∂S ∂V + ∂2E ∂S∂V −1 ∂2E ... T1 T T2 Now V2 Q+ = (21 ) ΛV (V, T2 )dV V1 Furthermore T2 W= V2 (T ) P dV = Γ T1 V1 (T ) ∂P dV dT ∂T by the Gauss–Green Theorem This identity, (17) and (21 ) imply V2 V1 Let T1 → T2 = T∗ : T2 ΛV ... T2 = T∗ : T2 ΛV (V, T2 )dV = T2 − T1 V2 T2 T1 V2 ΛV (V, T∗ )dV = T∗ V1 V1 V2 (T ) V1 (T ) ∂P dV dT ∂T ∂P (V, T∗ )dV ∂T Divide by V2 − V1 and let V2 → V1 = V∗ , to deduce (22 ) ΛV = T ∂P ∂T (Clapeyron’s...

Ngày tải lên: 17/03/2014, 14:29

... to show jh 1; T 2ij C k 1k2k 2k2 for all 1; 2 S (Rn): We note that h 1; Pt2TPt2 2i ! as t ! and hence it is enough to prove Z d j dt h 1; Pt2TPt2 2idtj C k 1k2k 2k2 for all 1; 2 S (Rn) since P0 ... RtQtTPt2 2i dt j C k 1k2k 2k2 for all 1; 2 S (Rn): t Hence it is enough to show Z j h 1; RtQtTPt2 2i dt j C (k 1k2 + k 2k2) for all 1; 2 S (Rn): 2 t But Z Z i dt j j h 1; RtQtTPt t jhRt 1; QtTPt2 2ij ... 2( x)](x)k2 dt 2t R R Z Rn Z pt(x)j1 ? eihx; ij2dx = ? 2e?j jt: kLt Pt ? Pt 2( x)](x)k2 dt 2t Z = Z 2( 1 ? e?j jt)j'(tj j)j2 dt j ^2( )j2d ^ t Rn Z Z Z ? jj ^ C (1 ? e?j jt) dt + j'(tj j)j2 dt j ^2( )j2d...

Ngày tải lên: 31/03/2014, 15:16

... a2 ϩ jb2 be complex numbers Then the following statements are logically valid: ͉a1 ϩ jb1͉ Ͻ ͉a2 ϩ jb2͉ ↔ a 12 ϩ b 12 Ͻ a 22 ϩ b 22 ͉a1 ϩ jb1͉ Յ ͉a2 ϩ jb2͉ ↔ a 12 ϩ b 12 Յ a 22 ϩ b 22 ͉a1 ϩ jb1͉ Ͼ ͉a2 ... each case, but the groupings differ  32 Chapter One ( (2 ϩ 3)(Ϫ3 Ϫ 1 )2) 2 (5 ϫ (Ϫ4 )2) 2 (5 ϫ 16 )2 8 02 6400 ( (2 ϩ ϫ (Ϫ3) Ϫ 1 )2) 2 ( (2 ϩ (Ϫ9) Ϫ 1 )2) 2 (Ϫ 82) 2 6 42 4096 Inequalities The following general ... Series Fourier Transforms 20 0 20 4 21 8 23 0 23 4 24 0 24 4 24 9 26 1 26 3 27 3 28 6 29 2 306 308 315 318 326 327 335 339 344 347 349 358 376 387 403 410 460 471 478 479 480 4 82 483 485 486 489 490 496 497...

Ngày tải lên: 31/03/2014, 16:19

... (2n (x + y)) t , , 2n n n t t f (2 x) f (2 y) N − T(x), − T(y), , ,N 2n 2n n n n f (2 (x + y)) f (2 x) f (2 y) t N − − , 2n 2n 2n ≥ N T(x + y) − (2: 13) Since, by (2. 8), N f (2n (x + y)) f (2n ... Inequalities and Applications 20 11, 20 11:78 http://www.journalofinequalitiesandapplications.com/content /20 11/1/78 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Hyers, DH: On the stability ... Acad Sci USA 27 , 22 2 22 4 (1941) doi:10.1073/ pnas .27 .4 .22 2 Aoki, T: On the stability of the linear transformationin Banach spaces J Math Soc Japan 2, 64–66 (1950) doi:10 .29 69/ jmsj/0 021 0064 Rassias,...

Ngày tải lên: 20/06/2014, 22:20

... y − y→x y→x 2. 12 Then the mappings t ∈ I → sup{f t, y : x1 t < y < x2 t } and t ∈ I → inf{f t, y : x1 t < y < x2 t } are measurable for each pair x1 , x2 ∈ C I such that x1 t < x2 t for all t ... Society, vol 125 , no 5, pp 1371–1376, 1997 Boundary Value Problems 25 21 P Binding, “The differential equation x f ◦ x,” Journal of Differential Equations, vol 31, no 2, pp ˙ 183–199, 1979 22 E Liz ... t1 , t2 ∈ I such that t1 < t2 , u t1 α t1 and u t

Ngày tải lên: 21/06/2014, 20:20

... with 2n x and 2n y respectively, in 2. 1 , it follows that μ f 2n x 2n y 24 n f 2n x−2n y 24 n −4 f 2n x 2n y 24 n f 2n x−2n y 24 n 24 f 2n y 24 n f 2n x 24 n −3 t f 2n y 24 n 2. 22 ≥ ρ2n x,2n y 24 n t ... respectively in 2. 1 , it follows that μ f 2n x 2n y 23 n f 2n x−2n y 23 n −4 f 2n x 2n y 23 n f 2n x−2n y 23 n 24 f 2n y 23 n f 2n x 23 n −3 t f 2n y 23 n 2. 11 ≥ ρ2n x,2n y 23 n t Taking the limit as n → ∞, ... satisfies 2. 3 Since C 2n x 2. 3 it follows that μC x −C x 2t μC 2n x −C 23 n t 2n x ≥ T μC 2n x −f 2n x 23 n t , μf 2n x −C 2n x ≥ T Ti∞1 ρ0,2i n−1 x 22 i , Ti∞1 ρ0,2i 3n t 23 n t n−1 x 22 i 2. 12 3n t...

Ngày tải lên: 22/06/2014, 02:20

... University, 1 326 Stevenson Center, Nashville, TN 3 724 0, USA Email address: em.diben@vanderbilt.edu Ugo Gianazza: Dipartimento di Matematica “F Casorati”, Universit` di Pavia, Via Ferrata 1, a 27 100 Pavia, ... about the homogeneity of L with respect to the group of dilations λσ1 x1 , ,λσN xN , 2 t (6) with ≤ σ1 ≤ 2 ≤ · · · ≤ σN , which force the coefficients to be polynomials If p, q are polynomials ... otherwise, Δ∞ ur = r u =1 on Br (x) ∩ ∂Ω, in Ω, ur = on ∂Ω\Br (x) (2) He shows existence of a maximal (minimal) solution and proves that u2r ≤ Cur on Ω\B3r (x) (3) He obtains uniqueness of these solutions...

Ngày tải lên: 22/06/2014, 19:20

... Lemma 2. 2 used in the proofs of Theorems 3 .2, 4 .2, and 5 .2 As for generalized versions and applications of Lemma 2. 2, as well as applications of algorithms of the type presented in Remark 2. 3, ... 1-norm of R2 , when h0 (t) = 2t + t + 26 , c0 = and d0 = Thus the results of Theorem 5 .2 hold Also, in this case, the chains needed in the proof of Theorem 5 .2 (cf the proof of Lemma 2. 2) are reduced ... Kluwer Academic, Dordrecht, 20 03, pp 595–615 , Existence results for operator equations in abstract spaces and an application, J Math Anal Appl 29 2 (20 04), no 1, 26 2 27 3 S Heikkil¨ and V Lakshmikantham,...

Ngày tải lên: 23/06/2014, 00:20