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

claireteng · 2022年07月06日

为什么2-5年违约概率中5年用的是平均hazard rate而不是对应的第5年的hazard rate？

  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.

为什么2-5年违约概率中5年用的是平均hazard rate而不是对应的第5年的hazard rate？

添加评论
1 个答案

李坏_品职助教 · 2022年07月06日

嗨，爱思考的PZer你好：



看一下这个的第二步，我们需要先算出0到5年的违约率，才能知道2道5年的违约率。而0到5年的违约率如果只有第五年的hazard rate显然是高估了实际违约率了。用average更合适。1- exp(-average * 5)就得出了0到5年的违约率了。

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

  • 1

    回答

  • 0

    关注

  • 505

    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 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？

2024-07-29 11:27 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. 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. 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