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- たつぞう おとべ
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12 0 T 3.3 0T B 0T E 0T 3.3 T < B T 3.1 (3.1) Pr( < B ) T T *1 * *3 [] *1 Z d dz 9
13 dz = ε d 3. ε dz * ds = µ Sd + σ SdZ 3.3 S d S d S µ S µ σ dz S ds S = µ d + σ dz 3.4 ds S µ d σ d *3= K T [ 0, T ] r C ( S, ) S C ( T ) ( S, ) SN( d ) Ke r N( d ) = d d 1 ln = ( S K ) + ( r + σ )( T ) σ T ( S K ) + ( r σ )( T ) ln = = d1 σ σ T T N(x)
14 N(x) N( x) = 1 π x e y dy 3.6 N (x) 3.4 N(x) x N( ) = 0, N(0) = 1, N( ) = 1 N (x) ( 1 π ) e x x x i (a) (b) (a) 0 1 x (b ( 1 ) e ) N(x) π T ( ) x i x d = µ d + σ dz 3.7 µ σ Z T 1 T 0 T B ( ) 3.8 B B 0 11
15 T T < B T 3.5 T T 3.7 T {( µ σ ) T + dzt } = exp σ 3.9 ln T ( µ σ ) T + dzt = ln σ T ln 0 + ( µ σ )T σ T N(x) µ σ µ Z = σ Z 0 T Expeced Defaul Probabiliy EDP) EDP = Pr = Pr ln = 0 ( T < BT 0 ) ( ln T < ln BT ln 0 ) BT f ( ln ) d ln ln BT = N ln = 1 N T [ ln 0 + ( µ σ ) T ] σ T ( 0 BT ) + ( µ σ ) σ T T T 3.11 [3] f (x) N ( d ) 1
16 N( d ) 1 N( d ) 1 0 T µ B T B 0 0 T (3.11) 0 BT µ σ T T 0 BT µ σ 1T T =1 T B T B T = B 0 = 13
17 S n E0 0 B 0 0 () = B0 + E0 = B0 ns0 4 µ r µ = r 5σ 0 T S ( ) ds = µ S d + σ S dz 3.13 S S µ S σ S n E E = ns
18 { E } 0 T de = µ E d + σ E dz 3.15 E E µ E σ E σ E = σ S µ E = µ S T T T E T = max( B, 0) 3.16 T T B T T ( 0 T ) C (, C ) r( T ) (, ) N( x ) B e N( x ) = T E x 1 ln = ( B ) + ( r + σ )( T ) T σ T x ( B ) + ( r σ )( T ) ln T = = x1 σ σ T T E E = C( ), 3.18 r C 15
19 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1,,,,,, x N re B T x f C C T x f C C x N C C T r T = = = = = = σ σ 3.19 ( ) x f (3.7)(3.18)(*4) ( ) ( ) ( )( ) ( ) ( ) Z x N re r x N Z C C C C E d d d, d, 1 ), ( ), ( d 1 1 σ µ σ σ µ + + = = 3.0 (3.15) ( ) E x N E σ σ 1 = 3.1 = 0 ( ) x N E E σ σ = 3. σ *4 () x x ( ) ( ) Z x x x + =,, σ µ ( ) ( ) Z x x x d, d, d σ µ + =
20 x () G = f ( x, ) G G G 1 G G dg = µ ( x, ) + + { σ ( x, ) } d + σ ( x, ) dz 3.5 x x x 3.. (3.) Hisorical Volailiy 0 n m = σ E 1 n 1 = n 1 n 1 Pi ln Pi i= 1 1 P i σ E n Pi ln Pi i= m
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More information) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4
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More information1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0
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More informationC el = 3 2 Nk B (2.14) c el = 3k B C el = 3 2 Nk B
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