问一道题：NO.PZ2016082406000083

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？

$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。

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？

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