vivian_zm · 2020年08月11日
问题如下：
Suppose a credit position has a correlation to the market factor of 0.5. What is the realized market value that is used to compute the probability of reaching a default threshold at the 99% confidence level?
选项： -0.2500.
-0.4356.
C.-0.5825.
D.-0.6243.
解释：
D A default loss level of 0.01 corresponds to -2.33 on the standard normal distribution. The
realized market value is computed as follows:
请问这里题目中给出的不是correlation=0.5吗，为什么直接代入到beta=0.5呢？
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
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
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？
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？
-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？
