最后一个小问题怎么求出来的？

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？

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？

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？

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