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

SkipperLin · 2021年04月12日

请问“exp(−10−/β)”里面第二个“-”是什么?谢谢老师

* 问题详情,请 查看题干

NO.PZ202001030300001402

问题如下:

An analyst is using the following exponential function to model corporate default rates:

fY(y)=1βexp(y/β),y0f_Y(y)=\frac{1}{\beta}exp(-y/\beta), y\geq0

where y is the number of years.


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:

FY(10)FY(5)F_Y(10)-F_Y(5)

By the SURVIVAL probability through year five(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)1FY(5)=exp(5/β)exp(10/β)exp(5/β)=1exp(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)

请问“exp(−10−/β)”里面第二个“-”是什么?谢谢老师

1 个答案
已采纳答案

品职答疑小助手雍 · 2021年04月12日

嗨,从没放弃的小努力你好:


读下来,感觉是笔误了多了个负号,我反馈修正下,谢谢指出~

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

  • 1

    回答
  • 0

    关注
  • 545

    浏览
相关问题

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)

2024-04-16 16:26 1 · 回答

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

2024-02-06 15:47 2 · 回答

NO.PZ202001030300001402 请问这个SURVIVprobability的公式应该怎么理解呀?谢谢老师

2021-04-12 14:05 2 · 回答

为什么分子要减去fy(5)

2020-08-02 16:49 1 · 回答