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quietvivian · 2019年04月14日

问一道题:NO.PZ2016062402000027 [ FRM I ]

有对应的课件页数和截图吗?没找到呀

问题如下图:

选项:

A.

B.

C.

D.

解释:

1 个答案

orange品职答疑助手 · 2019年04月15日

同学你好,这个就是 蒙特卡洛模拟里dS = uSdt+ σSdz这个公式的应用。代入本题数据,得到 dS = 0.14*S*ε*根号下0.01 =0.014*S*ε 。也就是说,
S1 = S0 + dS1 = S0 + 0.014*S0*ε = S0*(1+0.014*0.263),然后S2=S1 + dS2。
本质就是一个递推式。

对应的知识点就是蒙特卡洛模拟里dS = uSdt+ σSdz,对应基础班讲义P200页

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NO.PZ2016062402000027问题如下 Suppose you simulate the pripath of stoHHF using a geometric Brownimotion mol with ift μ = 0, volatility σ = 0.14, antime step Δ = 0.01. Let StS_tSt​ the priof the stotime t. If S0S_0S0​ = 100, anthe first two simulate(ranmly selecte stanrnormvariables are ε1\varepsilon_1ε1​ = 0.263 anε2\varepsilon_2ε2​ = -0.475, whis the simulatestopriafter the seconstep?96.79 99.79 99.97 99.70 The process for the stoprices hmeof zero anvolatility of σ△t=0.14×0.01=0.014\sigma\sqrt{\bigtriangleup t}=0.14\times\sqrt{0.01}=0.014σ△t​=0.14×0.01​=0.014, Henthe first step is S1=S0(1+0.014×0.263)=100.37S_1=S_0{(1+0.014\times0.263)}=100.37S1​=S0​(1+0.014×0.263)=100.37. The seconstep is S2=S1(1+0.014×−0.475)=99.70S_2=S_1{(1+0.014\times-0.475)}=99.70S2​=S1​(1+0.014×−0.475)=99.70请问这道题是哪个知识点

2022-11-18 15:22 4 · 回答

NO.PZ2016062402000027问题如下Suppose you simulate the pripath of stoHHF using a geometric Brownimotion mol with ift μ = 0, volatility σ = 0.14, antime step Δ = 0.01. Let StS_tSt​ the priof the stotime t. If S0S_0S0​ = 100, anthe first two simulate(ranmly selecte stanrnormvariables are ε1\varepsilon_1ε1​ = 0.263 anε2\varepsilon_2ε2​ = -0.475, whis the simulatestopriafter the seconstep? 96.79 99.79 99.97 99.70 The process for the stoprices hmeof zero anvolatility of σ△t=0.14×0.01=0.014\sigma\sqrt{\bigtriangleup t}=0.14\times\sqrt{0.01}=0.014σ△t​=0.14×0.01​=0.014, Henthe first step is S1=S0(1+0.014×0.263)=100.37S_1=S_0{(1+0.014\times0.263)}=100.37S1​=S0​(1+0.014×0.263)=100.37. The seconstep is S2=S1(1+0.014×−0.475)=99.70S_2=S_1{(1+0.014\times-0.475)}=99.70S2​=S1​(1+0.014×−0.475)=99.70 老师您好,如果u不等于0,怎么求?t = u*S*+Starviation * St*

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