解体方法

NO.PZ2020010303000012问题如下The monthly return on a hee funportfolio with US1 billion in assets is N(.02, .0003). Whis the stribution of the gain in a month?The funhaccess to a US10 million line of cret thes not count part of its portfolio. Whis the chanththe firm’s loss in a month excee this line of cret?Whwoulthe line of cret neeto to ensure ththe firm’s loss wless ththe line of cret in 99.9% of months (or equivalently, larger ththe LOC in 0.1% of months)?The monthly return is 2%, anthe monthly stanrviation is 1.73%. In US the monthly change in portfolio value ha meof 2% * US1 billion = US20 million ana stanrviation of 1.73% * US1 billion = US17.3 million. The probability ththe portfolio loses more thUS10 million is th(working in millions)Pr(V −10)=Pr(V−2017.3 −10−2017.3)=Pr(Z −1.73)Pr(V -10)=Pr(\frac{V-20}{17.3} \frac{-10-20}{17.3})=Pr(Z -1.73)Pr(V −10)=Pr(17.3V−20​ 17.3−10−20​)=Pr(Z −1.73)Using the normtable, Pr(Z -1.73)=4.18%Here we work in the other rection. First, we finthe quantile where Pr(Z z) = 99.9%, whigives z = -3.09. This is then scaleto the stribution of the change in the value of the portfolio multiply-ing the stanrviation anaing the mean, 17.3 * -3.09 + 20 = -33.46. The funwoulneea line of cret of US33.46 million to have a 99.9% change of having a change above this level.老师好，1、P（Z小于-1.73）的概率，通过网上查到的正态分布表，左列都是从0老师，没有负值，怎么查呢？2、P（Z小于z）的概率是0.01，查表z等于-3.08，这种不是累计概率反函数？具体怎么查表呢？

NO.PZ2020010303000012问题如下The monthly return on a hee funportfolio with US1 billion in assets is N(.02, .0003). Whis the stribution of the gain in a month?The funhaccess to a US10 million line of cret thes not count part of its portfolio. Whis the chanththe firm’s loss in a month excee this line of cret?Whwoulthe line of cret neeto to ensure ththe firm’s loss wless ththe line of cret in 99.9% of months (or equivalently, larger ththe LOC in 0.1% of months)?The monthly return is 2%, anthe monthly stanrviation is 1.73%. In US the monthly change in portfolio value ha meof 2% * US1 billion = US20 million ana stanrviation of 1.73% * US1 billion = US17.3 million. The probability ththe portfolio loses more thUS10 million is th(working in millions)Pr(V −10)=Pr(V−2017.3 −10−2017.3)=Pr(Z −1.73)Pr(V -10)=Pr(\frac{V-20}{17.3} \frac{-10-20}{17.3})=Pr(Z -1.73)Pr(V −10)=Pr(17.3V−20​ 17.3−10−20​)=Pr(Z −1.73)Using the normtable, Pr(Z -1.73)=4.18%Here we work in the other rection. First, we finthe quantile where Pr(Z z) = 99.9%, whigives z = -3.09. This is then scaleto the stribution of the change in the value of the portfolio multiply-ing the stanrviation anaing the mean, 17.3 * -3.09 + 20 = -33.46. The funwoulneea line of cret of US33.46 million to have a 99.9% change of having a change above this level.第一问中，the portfolio loses more thUS10 million，求的是Pr(V −10)；（我理解是亏损大于10m，即return -10)对比着第二问中，or equivalently, larger ththe LOC in 0.1% of months，解答中为什么不是Pr(Z z) = 0.1%，而是Pr(Z z) = 99.9%，我理解只有Pr(Z z) = 0.1%，z = -3.09如果Pr(Z z) = 99.9%，z=3.09此处理解错误在哪里，麻烦老师指正

NO.PZ2020010303000012问题如下The monthly return on a hee funportfolio with US1 billion in assets is N(.02, .0003). Whis the stribution of the gain in a month?The funhaccess to a US10 million line of cret thes not count part of its portfolio. Whis the chanththe firm’s loss in a month excee this line of cret?Whwoulthe line of cret neeto to ensure ththe firm’s loss wless ththe line of cret in 99.9% of months (or equivalently, larger ththe LOC in 0.1% of months)?The monthly return is 2%, anthe monthly stanrviation is 1.73%. In US the monthly change in portfolio value ha meof 2% * US1 billion = US20 million ana stanrviation of 1.73% * US1 billion = US17.3 million. The probability ththe portfolio loses more thUS10 million is th(working in millions)Pr(V −10)=Pr(V−2017.3 −10−2017.3)=Pr(Z −1.73)Pr(V -10)=Pr(\frac{V-20}{17.3} \frac{-10-20}{17.3})=Pr(Z -1.73)Pr(V −10)=Pr(17.3V−20​ 17.3−10−20​)=Pr(Z −1.73)Using the normtable, Pr(Z -1.73)=4.18%Here we work in the other rection. First, we finthe quantile where Pr(Z z) = 99.9%, whigives z = -3.09. This is then scaleto the stribution of the change in the value of the portfolio multiply-ing the stanrviation anaing the mean, 17.3 * -3.09 + 20 = -33.46. The funwoulneea line of cret of US33.46 million to have a 99.9% change of having a change above this level.题目第二问中，Z分布0.1%的分位点是-3.09，是怎么得到的呢？查表吗？我们课上讲的Z分布表，是已知分位点查概率，还可以反过来查吗？用已知的概率查表轴上的分位点？

NO.PZ2020010303000012 问题如下 The monthly return on a hee funportfolio with US1 billion in assets is N(.02, .0003). Whis the stribution of the gain in a month?The funhaccess to a US10 million line of cret thes not count part of its portfolio. Whis the chanththe firm’s loss in a month excee this line of cret?Whwoulthe line of cret neeto to ensure ththe firm’s loss wless ththe line of cret in 99.9% of months (or equivalently, larger ththe LOC in 0.1% of months)? The monthly return is 2%, anthe monthly stanrviation is 1.73%. In US the monthly change in portfolio value ha meof 2% * US1 billion = US20 million ana stanrviation of 1.73% * US1 billion = US17.3 million. The probability ththe portfolio loses more thUS10 million is th(working in millions)Pr(V −10)=Pr(V−2017.3 −10−2017.3)=Pr(Z −1.73)Pr(V -10)=Pr(\frac{V-20}{17.3} \frac{-10-20}{17.3})=Pr(Z -1.73)Pr(V −10)=Pr(17.3V−20​ 17.3−10−20​)=Pr(Z −1.73)Using the normtable, Pr(Z -1.73)=4.18%Here we work in the other rection. First, we finthe quantile where Pr(Z z) = 99.9%, whigives z = -3.09. This is then scaleto the stribution of the change in the value of the portfolio multiply-ing the stanrviation anaing the mean, 17.3 * -3.09 + 20 = -33.46. The funwoulneea line of cret of US33.46 million to have a 99.9% change of having a change above this level. 如题