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

粗眉毛辣椒油 · 2024年01月15日

麻烦老师解释一下

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)?

选项:

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.

The probability that the portfolio loses more than USD 10 million is than (working in millions),为什么要统一单位为million?做题思路是什么?

第二题看不懂,能翻译一下以及解释下考点吗?

3 个答案

pzqa27 · 2024年01月19日

嗨,爱思考的PZer你好:


1麻烦解释一下问题a和解析中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. 的意思?

意思就是说,有一个组合,这个组合每个月变化的均值是20m,每个月价值变化的标准差是1.73%,如果这个标准化写成美金的形式,就是17.3m.

2.“这个题题干说有一个对冲基金有1个billion的资产,并且这个资产的每月回报服从正态分布,这个正态分布的均值是0.02,方差是0.0003。”第二问计算分位点,不是用这里的均值和标准差呢?

这里请参考下有关分布标准化的课程,也就是这个课程。

----------------------------------------------
虽然现在很辛苦,但努力过的感觉真的很好,加油!

pzqa27 · 2024年01月18日

嗨,爱思考的PZer你好:


1. Pr(Z<-1.73)=4.18%这个课件的表中是不是没有

这个课件是没有纳入Z表,不过网上Z表很多,同学可以百度一下。

 这里如何运用置信区间知识点啊,完全不懂。

这里跟置信区间没关系,这里只是使用到了置信区间对应分为点的公式而已。

比如这个μ±1.65σ,它指的是取2个分位点,这俩分位点中间的面积占比是90%,那么两端的占比就是10%,由于正态分布对称,所以左边尾巴的占比是5%。

这里我们通过查表(查表的方法在早读课里有讲解),可以得知0.01%累积概率对应的Z值是-3.09,那么直接套用公式μ-Zσ即可

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

pzqa27 · 2024年01月15日

嗨,从没放弃的小努力你好:


这个题题干说有一个对冲基金有1个billion的资产,并且这个资产的每月回报服从正态分布,这个正态分布的均值是0.02,方差是0.0003。

第二题问的是,公司的损失小于99%的分位点是多少?

那么先找出99%分为点对应的Z值,即-3.09,然后带入下图的公式即可,把其中的1.65换成-3.09即可

----------------------------------------------
虽然现在很辛苦,但努力过的感觉真的很好,加油!

粗眉毛辣椒油 · 2024年01月17日

1. Pr(Z<-1.73)=4.18%这个课件的表中是不是没有? 2.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. 3.公司的损失小于99%的分位点是多少? 那么先找出99%分为点对应的Z值,即-3.09,然后带入下图的公式即可,把其中的1.65换成-3.09即可。这里如何运用置信区间知识点啊,完全不懂。

粗眉毛辣椒油 · 2024年01月18日

1麻烦解释一下问题a和解析中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. 的意思? 2.“这个题题干说有一个对冲基金有1个billion的资产,并且这个资产的每月回报服从正态分布,这个正态分布的均值是0.02,方差是0.0003。”第二问计算分位点,不是用这里的均值和标准差呢?

  • 3

    回答
  • 0

    关注
  • 247

    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.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?

2024-03-03 01:47 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. 如题

2024-02-06 15:14 1 · 回答