[問題] Brownian motion方面的問題

看板CFAiafeFSA (精算師/基金經理人/銀行家)作者howtodowell (well)時間15年前 (2010/10/25 00:02)推噓6(6推 0噓 19→)

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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 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.112.230.16

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