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410140980 · 2023年07月20日

这道题的累积违约概率

NO.PZ2020033002000078

问题如下:

In a synthetic CDO, the homogeneous reference portfolio has following characters:

Number of reference entities = 50;

CDS spread, s=180bps=180bp;

Recovery rate f=40%f=40\%.D

Defaults are independent.

The annual default probability on a single name is constant over five years and obeys the relation: s=(1f)PDs={(1-f)}PD.

What is the expected number of defaulting entities over the next five years, and which of the following tranches would lose 100% of the principal invested and hence be entirely wiped out?

选项:

A.

There would likely be 14 defaults and tranches up to the 3% are wiped.

B.

There would likely be 14 defaults and tranches up to the 8.5% are wiped.

C.

There would likely be 7 defaults and tranches up to the 3% are wiped.

D.

There would likely be 7 defaults and tranches up to the 8.5% are wiped.

解释:

D is correct.

考点:CDO

解析:

先算 PD d=1.8%10.40=3.00%d=\frac{1.8\%}{1-0.40}=3.00\%.

5年累积PD d+S1d+S2d+S3d+S4d=3%(1+0.970+0.941+0.913+0.885)=14.1%d+S_1d+S_2d+S_3d+S_4d=3\%(1+0.970+0.941+0.913+0.885)=14.1\%where the survival rates are S1=(13%)=0.970S_1={(1-3\%)}=0.970, S2=S1(13%)=0.941S_2=S_1{(1-3\%)}=0.941, and so on.

The expected number of defaults is therefore 50×14.1%50\times14.1\% = 7.

With a recovery rate of 40%, the expected loss is 8.5% of the notional.

So, all the tranches up to the 8.5% point are wiped out.

老师这道题的题干当中的违约概率是怎么判断的呢?给的公式s=(1-f)*PD 不理解怎么解析算了五年的累积违约概率,这个解题思路是怎样的

2 个答案
已采纳答案

DD仔_品职助教 · 2023年07月23日

嗨,从没放弃的小努力你好:


5年累计的PD是14.1%同时RR是40%,那么损失的期望就是14.1%*60%约等于8.5%

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就算太阳没有迎着我们而来,我们正在朝着它而去,加油!

DD仔_品职助教 · 2023年07月22日

嗨,从没放弃的小努力你好:


同学你好,

根据题目给出的公式:s=(1−f)PD,可以反推出一年的PD=1.8%/(1-0.4)=3%

五年的累计违约概率就等于=第一年违约的概率+第一年没有违约*第二年违约了的概率+前一二年没有违约*第三年违约了的概率+前一二三年没有违约*第四年违约了的概率+前一二三四年没有违约*第五年违约了的概率

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虽然现在很辛苦,但努力过的感觉真的很好,加油!

410140980 · 2023年07月23日

expected loss is 8.5% of the notional. 那这个8.5%怎么算来的

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NO.PZ2020033002000078问题如下 In a synthetic C, the homogeneous referenportfolio hfollowing characters:Number of referenentities = 50; C sprea s=180bps=180bps=180bp; Recovery rate f=40%f=40\%f=40%.efaults are inpennt. The annufault probability on a single name is constant over five years anobeys the relation: s=(1−f)P={(1-f)}P=(1−f)P Whis the expectenumber of faulting entities over the next five years, anwhiof the following tranches woullose 100% of the principinvesteanhenentirely wipeout? A.There woullikely 14 faults antranches up to the 3% are wipeB.There woullikely 14 faults antranches up to the 8.5% are wipeC.There woullikely 7 faults antranches up to the 3% are wipeThere woullikely 7 faults antranches up to the 8.5% are wipe is correct.考点C解析先算 P1.8%1−0.40=3.00%\frac{1.8\%}{1-0.40}=3.00\%1−0.401.8%​=3.00%. 5年累积PS1S2S3S43%(1+0.970+0.941+0.913+0.885)=14.1%S_1S_2S_3S_43\%(1+0.970+0.941+0.913+0.885)=14.1\%S1​S2​S3​S4​3%(1+0.970+0.941+0.913+0.885)=14.1%,where the survivrates are S1=(1−3%)=0.970S_1={(1-3\%)}=0.970S1​=(1−3%)=0.970, S2=S1(1−3%)=0.941S_2=S_1{(1-3\%)}=0.941S2​=S1​(1−3%)=0.941, anso on. The expectenumber of faults is therefore 50×14.1%50\times14.1\%50×14.1% = 7. With a recovery rate of 40%, the expecteloss is 8.5% of the notional. So, all the tranches up to the 8.5% point are wipeout. 烦请详细说下这题在说什么?数字啥意思?问题是啥意思?谢谢,完全读不懂。

2023-09-21 19:43 2 · 回答

NO.PZ2020033002000078问题如下 In a synthetic C, the homogeneous referenportfolio hfollowing characters:Number of referenentities = 50; C sprea s=180bps=180bps=180bp; Recovery rate f=40%f=40\%f=40%.efaults are inpennt. The annufault probability on a single name is constant over five years anobeys the relation: s=(1−f)P={(1-f)}P=(1−f)P Whis the expectenumber of faulting entities over the next five years, anwhiof the following tranches woullose 100% of the principinvesteanhenentirely wipeout? A.There woullikely 14 faults antranches up to the 3% are wipeB.There woullikely 14 faults antranches up to the 8.5% are wipeC.There woullikely 7 faults antranches up to the 3% are wipeThere woullikely 7 faults antranches up to the 8.5% are wipe is correct.考点C解析先算 P1.8%1−0.40=3.00%\frac{1.8\%}{1-0.40}=3.00\%1−0.401.8%​=3.00%. 5年累积PS1S2S3S43%(1+0.970+0.941+0.913+0.885)=14.1%S_1S_2S_3S_43\%(1+0.970+0.941+0.913+0.885)=14.1\%S1​S2​S3​S4​3%(1+0.970+0.941+0.913+0.885)=14.1%,where the survivrates are S1=(1−3%)=0.970S_1={(1-3\%)}=0.970S1​=(1−3%)=0.970, S2=S1(1−3%)=0.941S_2=S_1{(1-3\%)}=0.941S2​=S1​(1−3%)=0.941, anso on. The expectenumber of faults is therefore 50×14.1%50\times14.1\%50×14.1% = 7. With a recovery rate of 40%, the expecteloss is 8.5% of the notional. So, all the tranches up to the 8.5% point are wipeout. 为什么不能用1-e的(-lamb×t)t代入5

2023-07-28 14:36 1 · 回答

NO.PZ2020033002000078 没看懂这个8.5%是怎么来的

2021-09-03 17:15 2 · 回答