开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)|3.2.5(安卓)APP下载

王佳 · 2024年03月03日

已知正态分布的概率,怎么求分位点?

NO.PZ2020010303000012

问题如下:

The monthly return on a hedge fund portfolio with USD 1 billion in assets is N(.02, .0003). What is the distribution of the gain in a month?

a. The fund has access to a USD 10 million line of credit that does not count as part of its portfolio. What is the chance that the firm’s loss in a month exceeds this line of credit?

b. What would the line of credit need to be to ensure that the firm’s loss was less than the line of credit in 99.9% of months (or equivalently, larger than the LOC in 0.1% of months)?

解释:

a. The monthly return is 2%, and the monthly standard deviation is 1.73%. In USD, the monthly change in portfolio value has a mean of 2% * USD 1 billion = USD 20 million and a standard deviation of 1.73% * USD 1 billion = USD 17.3 million. The probability that the portfolio loses more than USD 10 million is than (working in millions)

Pr(V<10)=Pr(V2017.3<102017.3)=Pr(Z<1.73)Pr(V<-10)=Pr(\frac{V-20}{17.3}<\frac{-10-20}{17.3})=Pr(Z<-1.73)

Using the normal table, Pr(Z<-1.73)=4.18%

b. Here we work in the other direction. First, we find the quantile where Pr(Z < z) = 99.9%, which gives z = -3.09. This is then scaled to the distribution of the change in the value of the portfolio by multiply-ing by the standard deviation and adding the mean, 17.3 * -3.09 + 20 = -33.46. The fund would need a line of credit of USD 33.46 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?

2 个答案

pzqa39 · 2024年03月03日

嗨,努力学习的PZer你好:


FRM题库确实不多,所以把课后题也放在上面让大家进行练习。后续经典题上线之后可以去做一下,经典题的来源主要是PE的汇总整理版本,可能和考试的感觉更接近。

----------------------------------------------
就算太阳没有迎着我们而来,我们正在朝着它而去,加油!

pzqa39 · 2024年03月03日

嗨,爱思考的PZer你好:


是的,通过查表所得,如下图所示,通过0.01去找它对应的分位点。考试的时候很少这样出,因为即使这个表格已经算详细了,依然能看出0.001可以同时对应-3.09、-3.10、-3.08,不容易找出最精准的数值。所以我们大概了解一下可以反过来查表获得就好。

----------------------------------------------
加油吧,让我们一起遇见更好的自己!

王佳 · 2024年03月03日

怎么感觉课后题都是考试不大会出的题。特别是到section 4的时候,课后题很多都是问答,而且考试也没有问答题。目前刚开始复习,所以不确定自己学会了没有,也不太能通过课后题知道自己能不能把学的用到解题中。

  • 2

    回答
  • 0

    关注
  • 251

    浏览
相关问题

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. 是不是只要看到这种不是Z 分布的形式的题目在求解概率的时候都是要先将其化成stanrnormstribution进而求解是吗

2024-07-16 13:05 1 · 回答

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?

2024-05-27 19:23 2 · 回答

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

2024-05-04 15:48 1 · 回答

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

2024-02-06 15:14 1 · 回答