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上小学 · 2023年09月21日

请问此题是什么意思？需要干啥？题目要求是什么意思？

  1. FRM II
  2. Credit Risk Measurement And Management

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=(1−f)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%1−0.40=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=(1−3%)=0.970S_1={(1-3\%)}=0.970, S2=S1(1−3%)=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.

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。

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2 个答案

品职答疑小助手雍 · 2023年09月24日

每年3%违约，下一年活下来97%，这97%里面有3%违约，以此类推，可以算出来累计5年下来，还活下来多少，用1-去活下来的，就是违约的。

这上课讲marginal pd和conditional pd的时候讲过的。


为什么要算五年：因为题目问的就是5年之后，这50个债券的资产池里有百分之多少会违约损失掉（也就是wipe out的意思）。

所以要算5年的累计违约概率。

品职答疑小助手雍 · 2023年09月22日

同学你好，这题给的spread 和recovery rate，可以算出来每年的PD，而每年的PD可以累计算出来5年的cummulative pd。

这样就可以指导5年后，50个债券里违约个数的期望，同时由于有recovery rate，就可以求出整个50个债券里，损失比例的期望值。

题目求的就是我加粗的内容。

这种长的题看不懂的话一般有几个原因：1、英语不太好，读太长的题目记不太住，会忘条件；2、基础知识不牢，对题目给的条件不敏感，或者对题目要求的东西不敏感，不能条件反射性得想到需要使用的知识点。

第一个原因只能多读多练，第二个原因要看看基础班同时多练习题。

上小学 · 2023年09月23日

我计算了PD但是五年累积的PD怎么计算呢？看不懂解释，讲义听了大约会一些直接简单的题目，稍微综合就看不懂题目了。本题不知道需要干啥，这五年是要干啥，为什么用累积概率作为平均违约概率呢

上小学 · 2023年09月23日

另外，8。5被Wiped是什么意思呢？谢谢

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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. 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 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 wipe B.There woullikely 14 faults antranches up to the 8.5% are wipe C.There woullikely 7 faults antranches up to the 3% are wipe There 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. 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）*P 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

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NO.PZ2020033002000078 没看懂这个8.5%是怎么来的

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