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

返回

Jenny · 2021年04月10日

WCL的计算没懂

  1. FRM II
  2. Credit Risk Measurement And Management

NO.PZ2016082406000083

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.

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？
添加评论
1 个答案

品职答疑小助手雍 · 2021年04月10日

嗨，从没放弃的小努力你好：


这题涉及对概念的理解和运用的，单纯的课上的概念其实都懂的，不过这题理解上涉及WCL的含义，以及月PD和年PD的转换。

首先，月PD和年PD的转换：题目给了年PD是2%，也就是1-0.02，代表一年不违约的概率；一年不违约也就意味着连续12个月不违约，假设月PD是d的话，12个月连续不违约的概率就是（1-d）的12次方。连续12个月不违约的概率和1年不违约的概率是相等的，即1-0.02 = (1-d)的12次方。

然后是WCL的含义：前一步算了月PD是0.168%，也就是那么单个月来说在0.168%概率的情况下，会发生1 million的损失；而1- 0.168% = 99.832%的概率下，是不发生损失的。而题目要求的confidence level是99.9%，超过的99.832%这个数（也可以理解成在被包含在最尾部的那0.168%的概率里，损失是1million），所以WCL=1million。

而Credit var是等于WCL-EL的，所以答案就是C了。


PS：如果题目要求的confidence level是小于99.832%的，比如99%，那WCL就等于0了，此时credit var（按公式是负数，但是var不能为负，否则就是收益了）也就等于0了。

----------------------------------------------
就算太阳没有迎着我们而来，我们正在朝着它而去，加油！

  • 1

    回答

  • 0

    关注

  • 493

    浏览
相关问题

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

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

2020-03-21 16:20 1 · 回答