问一道题：NO.PZ2020011303000145 [ FRM I ]

问题如下：

The 5 and 95 percentiles of a lognormal loss distribution are 50 and 200. What is the 99.7 percentile of the distribution?

解释：

The 5 and 95-percentiles of the logarithm of the loss are ln(50) and ln(200), (i.e., 3.912 and 5.298). The logarithm of the loss has a mean of (3.912 + 5.298)/2 = 4.605 and a standard deviation of (5.298 – 4.605)/1.645 = 0.421. The 99.7-percentile of the logarithm of the loss is 4.605 + N^-1(0.997) × 0.421 = 5.763. The 99.7-percentile of the loss is therefore e^5.763 =318.3

老师你好，请问这一步标准差是怎么得来？

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小刘_品职助教 · 2020年03月21日

同学你好，

这个先知道了95%分位点的数=mean+1.645*standard deviation

（1.645这个数字是查表得到的，记住就可以）

反求出standard deviation

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NO.PZ2020011303000145 问题如下 The 5 an95 percentiles of a lognormloss stribution are 50 an200. Whis the 99.7 percentile of the stribution? The 5 an95-percentiles of the logarithm of the loss are ln(50) anln(200), (i.e., 3.912 an5.298). The logarithm of the loss ha meof (3.912 + 5.298)/2 = 4.605 ana stanrviation of (5.298 – 4.605)/1.645 = 0.421. The 99.7-percentile of the logarithm of the loss is 4.605 + N-1(0.997) × 0.421 = 5.763. The 99.7-percentile of the loss is therefore e5.763 =318.3 题目问损失的对数正态分布5%和95%的分位点对应的是50和200，求99.7%的损失是多少？对于对数正态分布，5%和95%的损失为ln(50)和ln(200)，也就是3.912和5.298，均值为(3.912+5.298)/2=4.605，标准差为 (5.298 – 4.605)/1.645= 0.421，ps.1.645是95%的z值。99.7% 分位点的值是4.605+ N-1(0.997) *0.421=5.763，损失=e^5.763=318.3。 1.stanrviation 为什么这么算？ (5.298 – 4.605)/1.645 = 0.421我用计算器算出来的是0.693147，为什么不这么计算标准差？2.1.645是怎么得出的？如果是查表，具体怎么查？谢谢。

2023-04-12 16:10 1 · 回答

请问可以解析一下这道题的答案嘛？不记得上课哪里说过了

2020-07-30 08:47 1 · 回答

问一道题：NO.PZ2020011303000145 [ FRM I ]

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