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天蝎独婧 · 2022年04月24日

最后一问根据公式

  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?

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.

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

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DD仔_品职助教 · 2022年04月24日

嗨，努力学习的PZer你好：


这里最后的计算没有问题呀，没有减反呀

前两年的违约概率是 0.02955

前五年的违约概率是 0.09063

这俩都是累计概率，所以2-5年间的违约概率不就是0.09063 − 0.02955 = 0.06107嘛

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虽然现在很辛苦，但努力过的感觉真的很好，加油！

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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？

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