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holicess · 2020年03月21日

问一道题：NO.PZ2016082406000083

  1. FRM II
  2. Credit Risk Measurement And Management

问题如下：

A risk analyst is trying to estimate the credit VAR for a risky bond. The credit VAR is defined as the maximum unexpected loss at a confidence level of 99.9% over a one-month horizon. Assume that the bond is valued at $1,000,000 one month forward, and the one-year cumulative default probability is 2% for this bond. What is the best estimate of the credit VAR for the bond, assuming no recovery?

选项：

A.

$20,000

B.

$1,682

C.

$998,318

D.

$0

解释：

ANSWER: C

First, we have to transform the annual default probability into a monthly probability. Using (1−2%)=(1−d)12{(1-2\%)}={(1-d)}^{12}, we find d=0.00168, which assumes a constant probability of default during the year. Next, we compute the expected credit loss, which is d×$1,000,000=$1,682d\times\$1,000,000=\$1,682. Finally, we calculate the WCL at the 99.9% confidence level, which is the lowest number \(CL_i\)such that P(CL≤CLi)≥99.9%P{(CL\leq CL_i)}\geq99.9\%. We have P(CL=0)=99.83%P{(CL=0)}=99.83\%; P(CL≤1,000,000)=100.00%P{(CL\leq1,000,000)}=100.00\%. Therefore, the WCL is $1,000,000, and the CVAR is $1,000,000−$1,682=$998,318\$1,000,000-\$1,682=\$998,318.

  1. 为什么会有WCL=0 的这个假设
  2. P(CL=0) =99.83%, 所以CL 只有可能等于1682 和0这两个可能性?这个地方没有看懂
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1 个答案

品职答疑小助手雍 · 2020年03月21日

同学你好，先回答第二个问题，因为是一个债券，结果只有违约和不违约两种，所以就有CL=0和CL=1000000两种情况，1682是EL，也就是CL的期望。

第一问那个，其实和WCL的定义有关，记得课上取的那种一竖列的概率对应损失的表么，现在问99.9%的wcl，其实问的就是从0开始累加，累计到超过99.9%的概率时的损失数，就是这个WCL。 而这题，月度违约概率算下来是0.168%。也就是课上那个一竖列的概率和损失情况只有2行，第一行是99.83%对应损失是0, 然后累计100%对应的损失是1000000。所以第一个超过99.9%的数是1000000，也就是wcl是1000000。

像这种分布，如果问的是99%的WCL，那么因为第一个超过99%的数是99.83%，对应损失是0，那么99%的WCL就是0了。

如果我说这些还是看不明白的话，我这打字毕竟没有上课的语言表达的清楚，还是建议把基础班这个WCL的定义部分再听一下了。


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NO.PZ2016082406000083 $1,682 $998,318 $0 ANSWER: C First, we have to transform the annufault probability into a monthly probability. Using (1−2%)=(1−12{(1-2\%)}={(1-}^{12}(1−2%)=(1−12, we fin0.00168, whiassumes a constant probability of fault ring the year. Next, we compute the expectecret loss, whiis $1,000,000=$1,682times\$1,000,000=\$1,682$1,000,000=$1,682. Finally, we calculate the Wthe 99.9% confinlevel, whiis the lowest number \(CL_i\)suthP(CL≤CLi)≥99.9%P{(CL\leq CL_i)}\geq99.9\%P(CL≤CLi​)≥99.9%. We have P(CL=0)=99.83%P{(CL=0)}=99.83\%P(CL=0)=99.83%; P(CL≤1,000,000)=100.00%P{(CL\leq1,000,000)}=100.00\%P(CL≤1,000,000)=100.00%. Therefore, the Wis $1,000,000, anthe CVis $1,000,000−$1,682=$998,318\$1,000,000-\$1,682=\$998,318$1,000,000−$1,682=$998,318.不记得课上有提到过这个计算，可以具体讲一下吗？然后对应讲义具体的哪个部分？

2021-04-10 08:23 1 · 回答

A risk analyst is trying to estimate the cret Vfor a risky bon The cret Vis finethe maximum unexpecteloss a confinlevel of 99.9% over a one-month horizon. Assume ththe bonis value$1,000,000 one month forwar anthe one-yecumulative fault probability is 2% for this bon Whis the best estimate of the cret Vfor the bon assuming no recovery? $20,000 $1,682 $998,318 $0 ANSWER: C First, we have to transform the annufault probability into a monthly probability. Using (1−2%)=(1−12{(1-2\%)}={(1-}^{12}(1−2%)=(1−12, we fin0.00168, whiassumes a constant probability of fault ring the year. Next, we compute the expectecret loss, whiis $1,000,000=$1,682times\$1,000,000=\$1,682$1,000,000=$1,682. Finally, we calculate the Wthe 99.9% confinlevel, whiis the lowest number \(CL_i\)suthP(CL≤CLi)≥99.9%P{(CL\leq CL_i)}\geq99.9\%P(CL≤CLi​)≥99.9%. We have P(CL=0)=99.83%P{(CL=0)}=99.83\%P(CL=0)=99.83%; P(CL≤1,000,000)=100.00%P{(CL\leq1,000,000)}=100.00\%P(CL≤1,000,000)=100.00%. Therefore, the Wis $1,000,000, anthe CVis $1,000,000−$1,682=$998,318\$1,000,000-\$1,682=\$998,318$1,000,000−$1,682=$998,318. 老师如果题目叫你求得CVAR是小于99.83%，P（loss≤0）=99.83%，那么WCL=0，了嘛，CVAR=-EL

2020-10-14 10:23 1 · 回答

$1,682 $998,318 $0 ANSWER: C First, we have to transform the annufault probability into a monthly probability. Using (1−2%)=(1−12{(1-2\%)}={(1-}^{12}(1−2%)=(1−12, we fin0.00168, whiassumes a constant probability of fault ring the year. Next, we compute the expectecret loss, whiis $1,000,000=$1,682times\$1,000,000=\$1,682$1,000,000=$1,682. Finally, we calculate the Wthe 99.9% confinlevel, whiis the lowest number \(CL_i\)suthP(CL≤CLi)≥99.9%P{(CL\leq CL_i)}\geq99.9\%P(CL≤CLi​)≥99.9%. We have P(CL=0)=99.83%P{(CL=0)}=99.83\%P(CL=0)=99.83%; P(CL≤1,000,000)=100.00%P{(CL\leq1,000,000)}=100.00\%P(CL≤1,000,000)=100.00%. Therefore, the Wis $1,000,000, anthe CVis $1,000,000−$1,682=$998,318\$1,000,000-\$1,682=\$998,318$1,000,000−$1,682=$998,318.老师，我看了所有问题的回答还是没没明白99.83%是怎么算出来的还有怎么得出WCL是1,000,000。可以把讲义里这部分的内容帮忙粘贴一下吗？实在是很难和讲义对应上。

2020-09-20 16:51 1 · 回答

A risk analyst is trying to estimate the cret Vfor a risky bon The cret Vis finethe maximum unexpecteloss a confinlevel of 99.9% over a one-month horizon. Assume ththe bonis value$1,000,000 one month forwar anthe one-yecumulative fault probability is 2% for this bon Whis the best estimate of the cret Vfor the bon assuming no recovery? $20,000 $1,682 $998,318 $0 ANSWER: C First, we have to transform the annufault probability into a monthly probability. Using (1−2%)=(1−12{(1-2\%)}={(1-}^{12}(1−2%)=(1−12, we fin0.00168, whiassumes a constant probability of fault ring the year. Next, we compute the expectecret loss, whiis $1,000,000=$1,682times\$1,000,000=\$1,682$1,000,000=$1,682. Finally, we calculate the Wthe 99.9% confinlevel, whiis the lowest number \(CL_i\)suthP(CL≤CLi)≥99.9%P{(CL\leq CL_i)}\geq99.9\%P(CL≤CLi​)≥99.9%. We have P(CL=0)=99.83%P{(CL=0)}=99.83\%P(CL=0)=99.83%; P(CL≤1,000,000)=100.00%P{(CL\leq1,000,000)}=100.00\%P(CL≤1,000,000)=100.00%. Therefore, the Wis $1,000,000, anthe CVis $1,000,000−$1,682=$998,318\$1,000,000-\$1,682=\$998,318$1,000,000−$1,682=$998,318. 老师问个弱弱的问题 这个历史法计算都是假设是贝努力分布嘛 比如三个债券 PA=0.05 PB=0.1 PC=0.2 假设AB违约C不违约的概率0.05*0.1*（1-0.2） 既然是贝努力为啥不是3C2*0.05*0.01*（1-0.2）呢。 我的意思为啥不用C那个公式算？ 什么情况才会用到C那个公式算？

2020-08-17 00:23 2 · 回答