Brownian motion方面的問題 - 金融分析師
By Harry
at 2010-10-25T00:02
at 2010-10-25T00:02
Table of Contents
ASM P273.
A stock with value S(t) follows an Ito process of the form
dS(t)/dt=0.17dt+0.34dZ
you are given:
(1)The continuously compounded risk-free rate is 0.04.
(2)The stock does not pay any dividends.
以下有三題 前兩題我有算出來
(1)ln(S^2) follows the Ito process
d(lnS^2)=cdt+vdZ
determine c and v.
(2)S^2 follows the Ito process
d(S^2)/S^2=cdt+vdZ
determine c and v.
以上這兩題 我都用Cs=Cs*dS+0.5*Css*((dS)^2)+Ct*dt下去解
但是我是把dS(t)/dt=0.17dt+0.34dZ
看成dS(t)/S=0.17dt+0.34dZ
請問題目給的條件的分母為什麼是dt (if it is dt 這樣dS(t)不就等於零了嗎?)
接下來是我算不出來的問題
(3)A forward F(t) on (S(t))^3 follows the Ito process
dF(t)/dt=cdt+vdZ
Determine c and v.
我們知道以下公式
F(0,T)[S(T)^a]=(S(0)^a)*exp{[a(r-d)+0.5*a*(a-1)*(sigma^2)]*T}...formula(14.7)
其中(0,T)指的是時間 a=3 r=0.04 d=0 sigma=0.34
課本答案寫
A forward has no volatility. See formula(14.7);there is no Z term in the
formula.So v=0. While formula(14.7) started at time 0,the generalization to
F(t,T) would involve replacing T with (T-t):
F(t,T)[S(T)^a]=(S(t)^a)*exp{a(r-d)+0.5*a*(a-1)*sigma^2}(T-t)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^...(***)
F(t,T)[S(T)^a]/F(t,t)[S(t)^a]=exp{a(r-d)+0.5*a*(a-1)*sigma^2}(T-t)
We have used S(t)^a=F(t,t)[S(t)^a]in the second line.(A forward maturing
immediately is equal to the underlying asset.)When we differentiate
with respect to t, a negative sign comes down from the exponent.So c is
negative the exponent. In our case
c=-3(0.04)-0.5(4)(3)(0.34^2)=-0.8136
但是4*3應該是3*2 可能課本寫錯了
以上為解答
但我的想法是我直接算
一樣用Cs=Cs*dS+0.5*Css*((dS)^2)+Ct*dt下去解
即C=F(t,T)[S(T)^a]=(S(t)^a)*exp{a(r-d)+0.5*a*(a-1)*sigma^2}(T-t)
當然dS(t)/dt=0.17dt+0.34dZ中分母的dt我把它看成S
算出來的答案跟課本不一樣
請問為什麼呢?而且我算出來的v也不等於零
這題算了3個多小時
快瘋掉了...
謝謝你看完
--
A stock with value S(t) follows an Ito process of the form
dS(t)/dt=0.17dt+0.34dZ
you are given:
(1)The continuously compounded risk-free rate is 0.04.
(2)The stock does not pay any dividends.
以下有三題 前兩題我有算出來
(1)ln(S^2) follows the Ito process
d(lnS^2)=cdt+vdZ
determine c and v.
(2)S^2 follows the Ito process
d(S^2)/S^2=cdt+vdZ
determine c and v.
以上這兩題 我都用Cs=Cs*dS+0.5*Css*((dS)^2)+Ct*dt下去解
但是我是把dS(t)/dt=0.17dt+0.34dZ
看成dS(t)/S=0.17dt+0.34dZ
請問題目給的條件的分母為什麼是dt (if it is dt 這樣dS(t)不就等於零了嗎?)
接下來是我算不出來的問題
(3)A forward F(t) on (S(t))^3 follows the Ito process
dF(t)/dt=cdt+vdZ
Determine c and v.
我們知道以下公式
F(0,T)[S(T)^a]=(S(0)^a)*exp{[a(r-d)+0.5*a*(a-1)*(sigma^2)]*T}...formula(14.7)
其中(0,T)指的是時間 a=3 r=0.04 d=0 sigma=0.34
課本答案寫
A forward has no volatility. See formula(14.7);there is no Z term in the
formula.So v=0. While formula(14.7) started at time 0,the generalization to
F(t,T) would involve replacing T with (T-t):
F(t,T)[S(T)^a]=(S(t)^a)*exp{a(r-d)+0.5*a*(a-1)*sigma^2}(T-t)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^...(***)
F(t,T)[S(T)^a]/F(t,t)[S(t)^a]=exp{a(r-d)+0.5*a*(a-1)*sigma^2}(T-t)
We have used S(t)^a=F(t,t)[S(t)^a]in the second line.(A forward maturing
immediately is equal to the underlying asset.)When we differentiate
with respect to t, a negative sign comes down from the exponent.So c is
negative the exponent. In our case
c=-3(0.04)-0.5(4)(3)(0.34^2)=-0.8136
但是4*3應該是3*2 可能課本寫錯了
以上為解答
但我的想法是我直接算
一樣用Cs=Cs*dS+0.5*Css*((dS)^2)+Ct*dt下去解
即C=F(t,T)[S(T)^a]=(S(t)^a)*exp{a(r-d)+0.5*a*(a-1)*sigma^2}(T-t)
當然dS(t)/dt=0.17dt+0.34dZ中分母的dt我把它看成S
算出來的答案跟課本不一樣
請問為什麼呢?而且我算出來的v也不等於零
這題算了3個多小時
快瘋掉了...
謝謝你看完
--
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金融分析師
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