Jin WANG · 2020年08月02日
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问题如下：
b. What is the cumulative probability of default within the first ten years given that the company has survived for five years?
选项：
解释：
We need to divide the marginal probability of default between years five and ten:
By the SURVIVAL probability through year five(1-result from Q1).
为什么分子要减去fy(5)
NO.PZ202001030300001402 问题如下 Whis the cumulative probability of fault within the first ten years given ththe company hsurvivefor five years? We neeto vi the marginprobability of fault between years five anten:FY(10)−FY(5)F_Y(10)-F_Y(5)FY(10)−FY(5)the SURVIVprobability through yefive(1-result from Q1).FY(10)−FY(5)=exp(−5/β)−exp(−10/β)F_Y(10)-F_Y(5)=exp(-5/\beta)-exp(-10/ \beta)FY(10)−FY(5)=exp(−5/β)−exp(−10/β)FY(10)−FY(5)1−FY(5)=exp(−5/β)−exp(−10/β)exp(−5/β)=1−exp(−5/β)\frac{F_Y(10)-F_Y(5)}{1-F_Y(5)}=\frac{exp(-5/ \beta)-exp(-10/ \beta)}{exp(-5/ \beta)}=1-exp(-5/ \beta)1−FY(5)FY(10)−FY(5)=exp(−5/β)exp(−5/β)−exp(−10/β)=1−exp(−5/β) 请问为什么是F Y (10)−F Y (5)
NO.PZ202001030300001402 问题如下 Whis the cumulative probability of fault within the first ten years given ththe company hsurvivefor five years? We neeto vi the marginprobability of fault between years five anten:FY(10)−FY(5)F_Y(10)-F_Y(5)FY(10)−FY(5)the SURVIVprobability through yefive(1-result from Q1).FY(10)−FY(5)=exp(−5/β)−exp(−10/β)F_Y(10)-F_Y(5)=exp(-5/\beta)-exp(-10/ \beta)FY(10)−FY(5)=exp(−5/β)−exp(−10/β)FY(10)−FY(5)1−FY(5)=exp(−5/β)−exp(−10/β)exp(−5/β)=1−exp(−5/β)\frac{F_Y(10)-F_Y(5)}{1-F_Y(5)}=\frac{exp(-5/ \beta)-exp(-10/ \beta)}{exp(-5/ \beta)}=1-exp(-5/ \beta)1−FY(5)FY(10)−FY(5)=exp(−5/β)exp(−5/β)−exp(−10/β)=1−exp(−5/β) 如题
NO.PZ202001030300001402 请问这个SURVIVprobability的公式应该怎么理解呀？谢谢老师
NO.PZ202001030300001402 请问“exp(−10−/β)”里面第二个“-”是什么？谢谢老师
