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zhouyu · 2019年08月12日

问一道题:NO.PZ2016082405000035

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解释:


为什么是-2.33?


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orange品职答疑助手 · 2019年08月12日

同学你好,根据题目条件,知道违约概率的门槛值是99%对应的分位数,也就是-2.33,所以k和等式左边都是取-2.33的。同学如果有遗忘的话可以看一下基础班的344至345页,上面有相似例题的

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NO.PZ2016082405000035 Suppose a cret position ha correlation to the market factor of 0.5. Whis the realizemarket value this useto compute the probability of reaching a fault thresholthe 99% confinlevel? -0.2500. -0.4356. -0.5825. -0.6243. A fault loss level of 0.01 correspon to -2.33 on the stanrnormstribution. The realizemarket value is computefollows: {l}-2.33=(\frac{-2.33-(0.5)\overline m}{\sqrt{1-0.5^2}})\\\overline m=-0.62430\enarray}\ 这里的correlation默认就是β吗,这两个不是不一样么

2021-05-04 17:45 2 · 回答

NO.PZ2016082405000035 Suppose a cret position ha correlation to the market factor of 0.5. Whis the realizemarket value this useto compute the probability of reaching a fault thresholthe 99% confinlevel? -0.2500. -0.4356. -0.5825. -0.6243. A fault loss level of 0.01 correspon to -2.33 on the stanrnormstribution. The realizemarket value is computefollows: {l}-2.33=(\frac{-2.33-(0.5)\overline m}{\sqrt{1-0.5^2}})\\\overline m=-0.62430\enarray}\ 0.31215=−(0.5)m 这一步是怎么来的

2021-04-05 22:21 1 · 回答

NO.PZ2016082405000035 Suppose a cret position ha correlation to the market factor of 0.5. Whis the realizemarket value this useto compute the probability of reaching a fault thresholthe 99% confinlevel? -0.2500. -0.4356. -0.5825. -0.6243.  A fault loss level of 0.01 correspon to -2.33 on the stanrnormstribution. The realizemarket value is computefollows: l−2.33=−2.33−(0.5)m‾1−0.52−2.33(0.86603)=−2.33−(0.5)m‾0.31215=−(0.5)m‾−0.62430=m‾{l}-2.33=\frac{-2.33-(0.5)\overline m}{\sqrt{1-0.5^2}}\\-2.33{(0.86603)}=-2.33-{(0.5)}\overline m\\0.31215=-{(0.5)}\overline m\\-0.62430=\overline ml−2.33=1−0.52 ​−2.33−(0.5)m​−2.33(0.86603)=−2.33−(0.5)m0.31215=−(0.5)m−0.62430=m 请问解答过程中的l-2.33的l是什么意思?

2021-03-18 10:18 1 · 回答

Suppose a cret position ha correlation to the market factor of 0.5. Whis the realizemarket value this useto compute the probability of reaching a fault thresholthe 99% confinlevel? -0.2500. -0.4356. -0.5825. -0.6243.  A fault loss level of 0.01 correspon to -2.33 on the stanrnormstribution. The realizemarket value is computefollows: l−2.33=−2.33−(0.5)m‾1−0.52−2.33(0.86603)=−2.33−(0.5)m‾0.31215=−(0.5)m‾−0.62430=m‾{l}-2.33=\frac{-2.33-(0.5)\overline m}{\sqrt{1-0.5^2}}\\-2.33{(0.86603)}=-2.33-{(0.5)}\overline m\\0.31215=-{(0.5)}\overline m\\-0.62430=\overline ml−2.33=1−0.52 ​−2.33−(0.5)m​−2.33(0.86603)=−2.33−(0.5)m0.31215=−(0.5)m−0.62430=m 这讲的是哪一个知识点? 在那一页ppt啊?

2020-10-11 13:44 1 · 回答

-0.4356. -0.5825. -0.6243.  A fault loss level of 0.01 correspon to -2.33 on the stanrnormstribution. The realizemarket value is computefollows: l−2.33=−2.33−(0.5)m‾1−0.52−2.33(0.86603)=−2.33−(0.5)m‾0.31215=−(0.5)m‾−0.62430=m‾{l}-2.33=\frac{-2.33-(0.5)\overline m}{\sqrt{1-0.5^2}}\\-2.33{(0.86603)}=-2.33-{(0.5)}\overline m\\0.31215=-{(0.5)}\overline m\\-0.62430=\overline ml−2.33=1−0.52 ​−2.33−(0.5)m​−2.33(0.86603)=−2.33−(0.5)m0.31215=−(0.5)m−0.62430=m k是α的临界分位点,α是服从一般正态分布的,为什么k是-2.33?

2020-09-25 10:01 1 · 回答