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

返回

okbella · 2024年07月29日

5年的dp

  1. FRM I
  2. Valuation And Risk Models

NO.PZ2020011303000101

问题如下：

If the hazard rate is 1.5% per year for the first three years and 2.5% per year for the next three years, what is the probability of default during the first two years? What is the average hazard rate for the first five years? What is the probability of default between years two and five?

选项：

解释：

The probability of default during the first two years is 1-exp(−0.015 × 2) = 0.02955. The average hazardrate during the first five years is (1.5 × 3 + 2.5 × 2)/5 = 1.9%. The probability of default during the first five years is 1-exp(−0.019 × 5) = 0.09063. The probability of default between years two and five is 0.09063 − 0.02955 = 0.06107.

题目问：如果前三年的hazard rate为每年 1.5%，后三年为每年 2.5%，那么前两年的违约概率是多少？前五年的平均hazard rate是多少？第二年和第五年之间违约的概率是多少？

PD(0-2)=1-e^(-h*t)=1-e^(-1.5%*2)=0.02955

平均h for 0-5 year=(1.5%*3+2.5%*2)/5=1.9%

PD(0-5)=1-e^(-1.9%*5)=0.09063

PD(2-5)=PD(0-5) - PD(0-2)=0.09063-0.02955=0.06107

老师您好，我计算b问题的时候直接这么算的：

1 - （e^ -0.015*3)*(e^ -0.025*2) = 0.090627

这样是不是也可以的？

添加评论
1 个答案

李坏_品职助教 · 2024年07月29日

嗨，努力学习的PZer你好：


1 - （e^ -0.015*3)*(e^ -0.025*2) = 0.090627这个也是计算在五年内违约的概率。可以的。

----------------------------------------------
努力的时光都是限量版，加油！

  • 1

    回答

  • 0

    关注

  • 58

    浏览
相关问题

NO.PZ2020011303000101问题如下If the hazarrate is 1.5% per yefor the first three years an2.5% per yefor the next three years, whis the probability of fault ring the first two years? Whis the average hazarrate for the first five years? Whis the probability of fault between years two anfive? The probability of fault ring the first two years is 1-exp(−0.015 × 2) = 0.02955. The average hazarate ring the first five years is (1.5 × 3 + 2.5 × 2)/5 = 1.9%. The probability of fault ring the first five years is 1-exp(−0.019 × 5) = 0.09063. The probability of fault between years two anfive is 0.09063 − 0.02955 = 0.06107. 题目问如果前三年的hazarrate为每年 1.5%，后三年为每年 2.5%，那么前两年的违约概率是多少？前五年的平均hazarrate是多少？第二年和第五年之间违约的概率是多少？P0-2)=1-e^(-h*t)=1-e^(-1.5%*2)=0.02955平均h for 0-5 year=(1.5%*3+2.5%*2)/5=1.9%P0-5)=1-e^(-1.9%*5)=0.09063P2-5)=P0-5) - P0-2)=0.09063-0.02955=0.06107 如果0-5年可以用算数平均算h，那最后一问为什么不用2-5年算数平均h来计算违约概率呢？

2023-06-23 23:24 1 · 回答

NO.PZ2020011303000101 问题如下 If the hazarrate is 1.5% per yefor the first three years an2.5% per yefor the next three years, whis the probability of fault ring the first two years? Whis the average hazarrate for the first five years? Whis the probability of fault between years two anfive? The probability of fault ring the first two years is 1-exp(−0.015 × 2) = 0.02955. The average hazarate ring the first five years is (1.5 × 3 + 2.5 × 2)/5 = 1.9%. The probability of fault ring the first five years is 1-exp(−0.019 × 5) = 0.09063. The probability of fault between years two anfive is 0.09063 − 0.02955 = 0.06107. 为什么2-5年违约概率中5年用的是平均hazarrate而不是对应的第5年的hazarrate？

2022-07-06 20:53 1 · 回答

NO.PZ2020011303000101问题如下If the hazarrate is 1.5% per yefor the first three years an2.5% per yefor the next three years, whis the probability of fault ring the first two years? Whis the average hazarrate for the first five years? Whis the probability of fault between years two anfive? The probability of fault ring the first two years is 1-exp(−0.015 × 2) = 0.02955. The average hazarate ring the first five years is (1.5 × 3 + 2.5 × 2)/5 = 1.9%. The probability of fault ring the first five years is 1-exp(−0.019 × 5) = 0.09063. The probability of fault between years two anfive is 0.09063 − 0.02955 = 0.06107. t1 t2是对应2年和5年吗？概率不能为负，所以减反了？还是我那里想错了

2022-04-24 18:31 1 · 回答

NO.PZ2020011303000101 能列一下式子吗

2021-10-12 09:01 1 · 回答