这里的volatility为什么要乘以根号下0.01？
问题如下图：
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解释：
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
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*
NO.PZ2016062402000027 ift term的计算不懂，题目中给的0，为啥计算机过程中显示1？
请问，本题中的波动率是什么时间跨度？t=0.01是什么时间跨度？
有对应的课件页数和截图吗？没找到呀
