※ 引述《sw0079 (極限操作)》之銘言： : 學校:SFU(溫哥華西岸的學校) : 教師:R.Jones : 科目:Futures, options and other deriatives : 題目:Ito's lemma : 1. Ito's Lemma: : Let the price s(t) of a security follow the Ito process : ds =α s dt +δ s dz : (a) Use Ito's Lemma to determine the process followed by y(t)= ln s(t). S(t) follows the Geometric Brownian motion Since y(t)is a function of S,by Ito's Lemma We can derive the stochastic process followed by y dy = [αS*(1/S)+0+1/2*(δS)^2*(-1/S^2)]dt + δdz dy = (α-1/2*δ^2)dt+δdz----(a) that is,in discrete case ㏑S_t - ㏑S_0 ~ N( (α-1/2*δ^2)T,δ^2*T ) ㏑S_t ~ N( ㏑S_0+(α-1/2*δ^2)T , δ^2*T ) y(3) = ㏑S_3 ~ N( ㏑S_0 + 3*(α-1/2*δ^2) , 3*δ^2 )----(b) (c) Given S_0 ㏑S_t is normally distributed with mean ㏑S_0 +(α-1/2*δ^2)*T and variance δ^2*T X is a random draw from it and has been standardlized,that is ㏑S_t - E(㏑S_t) ㏑S_T - [㏑S_0 + (α-1/2*δ^2)*T] X = ------------------- = ----------------------------------- ~ N(0,1) √Var(㏑S_t) δ√T ㏑S_T = (δ*√T)*X + ㏑S_0 +(α-1/2*δ^2)T S_T = exp[ Xδ√T + ㏑S_0 + (α-1/2*δ^2)T] S_3 = exp[√3Xδ + ㏑S_0 + 3*(α-1/2*δ^2)] 好像是這樣 已經忘的差不多了= = 有錯請指正 : (b) What is the probability distribution of y(3) in terms of y(0), and : (i.e., what is the type of distribution, its mean and its variance) : (c) If you were given s0 = s(0) and a random draw X from the type of : distribution in (b),but it was standardized to have mean 0 and variance 1, : how would you convert it into a random draw of s(3)? : (i.e., of the security price at time 3) : 翻譯: : 讓證卷價錢依照伊藤過程 : a.) 用伊藤過程決定算出y(t)=ln s(t) : b.) y(3)的可能分配是什麼? : (i.e., 怎樣的分配? 平均數跟變異數又是什麼?) : c.) 如果知道s0 = s(0) (s0的0是標在下面的) 而且現在從(b.)的分配裡面隨機抽取X, : 可是X有個mean = 0 and variance = 1, 那當s(3)的時候你的X會變成怎樣? : 我的想法: : 這題其實老師有在黑板寫可是我還是看不太懂 : 我自己用小畫家畫了一下解答 : http://0rz.tw/344Sx : yss=y的開兩次deriative : 請問這就是解答了嗎? 可是總覺得(b)還沒回答完,但是又不知道要怎麼代入 : 另外(c)老師是這樣寫的 : y=ln s --> s= e^y (^是開平方,所以e^2=e的二次方 etc) : convert to S(T) : EXP( ln S(0) + z) --> S(0)*EXP(z) : 這個我也看不懂 伊藤過程好難啊 這些東西到底要怎麼帶入才會算出答案呢?? --

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