第四章：M-absolute 和 M-square 风险度量
思维导图
从第四章开始比较难了
(M^A) 和 (M^2) 控制了组合预期变化的下限
两个重要不等式的推导
首先有
[V_0 = sum_{t=t_1}^{t_n} CF_t e^{-int_0^t f(s)ds}
]
令
[egin{aligned}
V_H &= V_0 e^{int_0^H f(s)ds}\
&= sum_{t=t_1}^{t_n} CF_t e^{-int_0^t f(s)ds} e^{int_0^H f(s)ds}\
&= sum_{t=t_1}^{t_n} CF_t e^{int_t^H f(s)ds}
end{aligned}
]
以及
[V_H^{prime} = sum_{t=t_1}^{t_n} CF_t e^{int_t^H f^{prime}(s)ds}
]
那么
[egin{aligned}
frac{V_H^{prime} - V_H}{V_H} &=
frac{1}{V_0 e^{int_0^H f(s)ds}}
sum_{t=t_1}^{t_n} CF_t (e^{int_t^H f^{prime}(s)ds} - e^{int_t^H f(s)ds})\
&=frac{1}{V_0}sum_{t=t_1}^{t_n} CF_t[e^{int_t^H f(s)ds}(e^{int_t^H Delta f(s)ds}-1)]e^{-int_0^H f(s)ds}\
&=frac{1}{V_0}sum_{t=t_1}^{t_n} CF_te^{-int_0^t f(s)ds}(e^{int_t^H Delta f(s)ds}-1)
end{aligned}
]
记
[h(t) = int_t^H Delta f(s)ds
]
关于 (M^A) 的不等式
(Delta f(t)) 的边界分别是 (K_1) 和 (K_2)，即 (K_1 le Delta f(t) le K_2)
若 (H>t) 时
[h(t) ge K_1(H-t) = K_1 |t-H|
]
若 (H le t) 时
[h(t) ge -K_2(t-H) = -K_2|t-H|
]
于是
[h(t) ge min(K_1, -K_2)|t-H|
]
而
[egin{aligned}
min (K_1, -K_2) &= -max(-K_1, K_2)\
&ge -max(|K_1|, |K_2|)\
&=-K_3
end{aligned}
]
则
[h(t) ge -K_3|t-H|
]
已知
[e^x - 1 ge x
]
那么
[egin{aligned}
frac{V_H^{prime} - V_H}{V_H} & =frac{1}{V_0}sum_{t=t_1}^{t_n} CF_te^{-int_0^t f(s)ds}(e^{int_t^H Delta f(s)ds}-1)\
& ge frac{1}{V_0}sum_{t=t_1}^{t_n} CF_te^{-int_0^t f(s)ds}h(t)\
& ge frac{1}{V_0}sum_{t=t_1}^{t_n} CF_te^{-int_0^t f(s)ds} (-K_3|t-H|)\
& = -K_3M^A
end{aligned}
]
关于 (M^2) 的不等式
记
[frac{dDelta f(t)}{dt} = g(t) le K_4
]
那么
[egin{aligned}
h(t) &= int_t^H Delta f(s)ds\
& = tDelta f(t)|_{t}^{H} - int_t^H s g(s)ds\
& = HDelta f(H) - tDelta f(t) - int_t^H s g(s)ds\
& = (H-t)Delta f(H) + tDelta f(H) - tDelta f(t) - int_t^H s g(s)ds\
& = (H-t)Delta f(H) + tint_t^H g(s)ds - int_t^H s g(s)ds\
& = (H-t)Delta f(H) + int_t^H(t-s) g(s)ds\
end{aligned}
]
若 (H>t) 时
[egin{aligned}
int_t^H (t-s) g(s)ds & ge int_t^H (t-s) K_4ds\
&=-K_4(t-H)^2/2
end{aligned}
]
若 (H le t) 时
[egin{aligned}
int_t^H (t-s) g(s)ds & = -int_H^t (t-s) g(s)ds\
& ge -int_H^t (t-s) K_4ds\
& = -K_4(t-H)^2/2
end{aligned}
]
那么
[h(t) ge (H-t)Delta f(H) -K_4(t-H)^2/2
]
已知
[e^x - 1 ge x
]
那么
[egin{aligned}
frac{V_H^{prime} - V_H}{V_H} & =frac{1}{V_0}sum_{t=t_1}^{t_n} CF_te^{-int_0^t f(s)ds}(e^{int_t^H Delta f(s)ds}-1)\
& ge frac{1}{V_0}sum_{t=t_1}^{t_n} CF_te^{-int_0^t f(s)ds}h(t)\
& ge frac{1}{V_0}sum_{t=t_1}^{t_n} CF_te^{-int_0^t f(s)ds} left((H-t)Delta f(H) -K_4(t-H)^2/2
ight)\
& = (H-D)Delta f(H) -K_4M^2/2
end{aligned}
]
凸性效应（CE）和风险效应（RE）的推导
[egin{aligned}
R(H) &= frac{V_H^{prime} - V_0}{V_0}\
&=frac{sum_{t=t_1}^{t_n} CF_t e^{int_t^H f^{prime}(s)ds}}{sum_{t=t_1}^{t_n} CF_t e^{-int_0^t f(s)ds}} - 1\
&=frac{sum_{t=t_1}^{t_n} CF_t e^{int_0^H f^{prime}(s)ds - int_0^t f^{prime}(s)ds}}{sum_{t=t_1}^{t_n} CF_t e^{-int_0^t f(s)ds}} - 1\
&=frac{e^{int_0^H f^{prime}(s)ds}sum_{t=t_1}^{t_n} CF_t e^{-int_0^t f(s)ds}e^{-int_0^t Delta f(s)ds}}{sum_{t=t_1}^{t_n} CF_t e^{-int_0^t f(s)ds}} - 1\
&=frac{e^{int_0^H f(s) + Delta f(s)ds}sum_{t=t_1}^{t_n} CF_t e^{-int_0^t f(s)ds}e^{-int_0^t Delta f(s)ds}}{sum_{t=t_1}^{t_n} CF_t e^{-int_0^t f(s)ds}} - 1\
&=frac{e^{int_0^H f(s)}sum_{t=t_1}^{t_n} CF_t e^{-int_0^t f(s)ds}e^{int_t^H Delta f(s)ds}}{sum_{t=t_1}^{t_n} CF_t e^{-int_0^t f(s)ds}} - 1\
end{aligned}
]
令
[R_F(H) = e^{int_0^H f(s)ds} - 1
]
记
[k(t) = e^{int_t^H Delta f(s)ds}
]
对 (k(t)) 在 (H) 做 Taylor 展开
[egin{aligned}
k(t) &= e^{int_t^H Delta f(s)ds}\
&= e^{-int_H^t Delta f(s)ds}\
&= k(H) + (t-H)k'(H) + frac{1}{2}(t-H)^2k''(H) + varepsilon\
&= 1 + (t-H)(-Delta f(H)) + frac{1}{2}(t-H)^2(Delta f(H)^2 - frac{d(Delta f(t))}{dt}|_{t=H}) + varepsilon\
&= 1 + (t-H)(-Delta f(H)) + frac{1}{2}(t-H)^2(Delta f(H)^2 - g(H)) + varepsilon\
end{aligned}
]
代入得到
[egin{aligned}
R(H) &= R_F(H) + gamma_1 (D-H) + gamma_2 M^2 + varepsilon\
gamma_1 &= -Delta f(H)(1+R_F(H))\
gamma_2 &= frac{1}{2}(1+R_F(H))(Delta f(H)^2 - g(H))\
gamma_2 &= CE - RE\
CE &= frac{1}{2}(1+R_F(H))Delta f(H)^2\
RE &= frac{1}{2}(1+R_F(H))g(H)
end{aligned}
]