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ActivePortfolioManagement/P193 命题3/命题3.md
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组合S定义

总收益定义为

[ R_P = 1 + i_F + r_P ]
其中 ( i_F ) 是无风险利率,( r_P ) 是超额收益。
无风险资产的总收益是

[ R_F = 1 + i_F ]
组合 ( S ) 是如下问题的最优解:

[ \min E[R_P^2] ]
其中 ( P ) 可包含风险资产与无风险资产。


构造组合 ( P(w) )

取任意组合 ( P ),构造

[ P(w) : \quad R_{P(w)} = R_S + w (R_P - R_S) ]
定义

[ g_P(w) = E[R_{P(w)}^2] ]
代入得:

[ g_P(w) = E[R_S^2] + 2w E[R_S(R_P - R_S)] + w^2 E[(R_P - R_S)^2] ]


一阶条件

因为 ( S ) 是最小值点,考虑 ( w=0 ) 对应组合 ( S ),应有

[ g_P'(0) = 0 ]
计算导数:

[ g_P'(w) = 2 E[R_S(R_P - R_S)] + 2w E[(R_P - R_S)^2] ]
在 ( w=0 ) 时:

[ g_P'(0) = 2 E[R_S(R_P - R_S)] = 0 ]
所以

[ E[R_S(R_P - R_S)] = 0 \quad \text{对任意组合 } P ]

[ E[R_S R_P] = E[R_S^2] \tag{1} ]


展开协方差形式

[ E[R_S R_P] = \text{Cov}(R_S, R_P) + E[R_S] E[R_P] ]
由 (1) 得:

[ \text{Cov}(R_S, R_P) + E[R_S] E[R_P] = E[R_S^2] \tag{2} ]


对无风险组合 ( F ) 应用 (2)

对 ( P = F )(无风险组合),( R_F = 1 + i_F ) 是常数,所以

[ \text{Cov}(R_S, R_F) = 0 ]
于是 (2) 给出:

[ 0 + E[R_S] R_F = E[R_S^2] ]

[ E[R_S^2] = E[R_S] R_F \tag{3} ]


回到一般组合 ( P )

由 (2)

[ \text{Cov}(R_S, R_P) + E[R_S] E[R_P] = E[R_S^2] ]
用 (3) 替换 ( E[R_S^2] )

[ \text{Cov}(R_S, R_P) + E[R_S] E[R_P] = E[R_S] R_F ]
所以

[ \text{Cov}(R_S, R_P) = E[R_S] (R_F - E[R_P]) \tag{4} ]


用超额收益表示

超额收益 ( r_P = R_P - R_F ),所以

[ E[R_P] = R_F + E[r_P] ]
代入 (4) 右端:

[ \text{Cov}(R_S, R_P) = E[R_S] (R_F - (R_F + E[r_P])) = - E[R_S] E[r_P] \tag{5} ]

[ R_S = R_F + r_S, \quad R_P = R_F + r_P ]
所以

[ \text{Cov}(R_S, R_P) = \text{Cov}(R_F + r_S, R_F + r_P) = \text{Cov}(r_S, r_P) ]
因为 ( R_F ) 是常数。

于是 (5) 变为:

[ \text{Cov}(r_S, r_P) = - E[R_S] E[r_P] \tag{6} ]


解出 ( E[r_P] )

由 (6)

[ E[r_P] = \frac{-1}{E[R_S]} \text{Cov}(r_P, r_S) ]

[ \phi = \frac{-1}{E[R_S]} = \frac{-1}{E[1 + i_F + r_S]} ]

[ E[r_P] = \phi , \text{Cov}(r_P, r_S) \tag{8A-28} ]


证明 (8A-29) 第二个等式

由 (3) 得

[ E[R_S] R_F = E[R_S^2] = \sigma_S^2 + (E[R_S])^2 ]
所以

[ \sigma_S^2 = E[R_S] R_F - (E[R_S])^2 = E[R_S] (R_F - E[R_S]) ]
定义 ( f_S = E[r_S] = E[R_S] - R_F ),则

[ R_F - E[R_S] = - f_S ]
于是

[ \sigma_S^2 = E[R_S] (-f_S) ]

[ E[R_S] = - \frac{\sigma_S^2}{f_S} ]
因此

[ \phi = \frac{-1}{E[R_S]} = \frac{-1}{-\sigma_S^2 / f_S} = \frac{f_S}{\sigma_S^2} ]
得证。