ID 3 3 2 2 2 1 1 2 3 2 3 4 5-1 -
- 2 -
1-3 -
1 (1) (2) (3) (4) (1) 2 (2) (3) - 4 -
(4) (1) (2) 52 52 CAPM r r = α + β ( r r ) + ε t ft Mt ft t r t r Mt r ft ε t β β σ σ M = M σ M σ M - 5 -
r t r r Mt ft ft r α Jensen Jensen s ndex CAPM α = 0 α 0 CAPM α 20 r = α + β r + ε t Mt t 20 1 2000 10 2005 10 TOPIX Yahoo! 1 2 ˆR - 6 -
2000 10 2005 10 t ****** 1%5%10% - 7 -
2 60% 80% 90% 100% 75% 20% 3 26 41% 45% 100% 30% 16 11 16 40% 70% HDDFDB LSI 80% - 8 -
DG 30% 16 100% 80% NEOMAX 60% 40%Nd-Fe-B 55% 35% 90% 50,000 ISO9001ISO14001 30 20 Markowtz - 9 -
subject to { } n n 2 p = j j = 1 j= 1 mn σ ωωσ ω r p n = ωr n = 1 = 1 ω = 1 ω 0.02 σ 2 p r p ω 10 500 2% 3 2000 10 2005 10 3 3 5.4981 2 13 ω = 0.02 2 1 10 2% 10-10 -
2 1985 STOCK - 11 -
nvestment bankng 1994 21 2001 1997 1996 2 2002!! 2005 2002 2002-12 -