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奔跑的咸鱼 · 2021年04月30日

这个题目能不能直接用把全损当做wcl，然后ul=4600-460=4140这么做？

  1. FRM II
  2. Credit Risk Measurement And Management

NO.PZ2016082406000004

问题如下：

A bank has booked a loan with total commitment of $50,000 of which 80% is currently outstanding. The default probability of the loan is assumed to be 2% for the next year and loss given default (LGD) is estimated at 50%. The standard deviation of LGD is 40%. Drawdown on default (i.e., the fraction of the undrawn loan) is assumed to be 60%. The expected and unexpected losses (standard deviation) for the bank are

选项：

A.

Expected loss = $500, unexpected loss = $4,140

B.

Expected loss = $500, unexpected loss = $3,220

C.

Expected loss = $460, unexpected loss = $3,220

D.

Expected loss = $460, unexpected loss = $4,140

解释：

ANSWER: D

First, we compute the exposure at default. This is the drawn amount, or 80%x$50,000=$40,000 plus the drawdown on default, which is 60%x$10,000=$6,000, for a total of CE= $46,000. The expected loss is this amount times   p×E(LGD)=0.02×50%=1%\;p\times E{(LGD)}=0.02\times50\%=1\% ₤or EL = $460. Next, we compute the standard deviation of losses using Equation: σ(CL)=p×σ2(LGD)+p×(1−p)×[E(LGD)]2\sigma{(CL)}=\sqrt{p\times\sigma^2{(LGD)}+p\times{(1-p)}\times{\lbrack E{(LGD)}\rbrack}^2}. The variance is p×σ2(LGD)+p×(1−p)×[E(LGD)]2=0.02(0.4)2+0.02(1−0.02)(0.50)2=0.00810p\times\sigma^2{(LGD)}+p\times{(1-p)}\times{\lbrack E{(LGD)}\rbrack}^2=0.02{(0.4)}^2+0.02{(1-0.02)}{(0.50)}^2=0.00810. Taking the square root gives 0.090. Multiplying by $46,000 gives $4,140. Ignoring σ2(LGD)\sigma^2{(LGD)} gives the incorrect answer of $3,220. Note that the unexpected loss is much greater than the expected loss.

这个题目能不能直接用把全损当做wcl，然后ul=4600-460=4140这么做？

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2 个答案

品职答疑小助手雍 · 2021年05月01日

嗨，努力学习的PZer你好：


这个ul可以认为是credit loss 的标准差（二级里这个概念和计算都不重要），平时说的那个是带置信度的。

如果要二级考这类计算的话不用太看描述，主要看题目给什么条件就好了。

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努力的时光都是限量版，加油！

品职答疑小助手雍 · 2021年04月30日

嗨，爱思考的PZer你好：


不能，给了LGD的波动就直接按公式做了，这题如果LGD的波动不是这个数值，那你不就错了。

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努力的时光都是限量版，加油！

奔跑的咸鱼 · 2021年05月01日

这题的ul跟我说的这种ul最主要区别是什么呀？强化版老师一句带过，我不是很理解

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NO.PZ2016082406000004 A bank hbookea lowith totcommitment of $50,000 of whi80% is currently outstanng. The fault probability of the lois assumeto 2% for the next yeanloss given fault (LG is estimate50%. The stanrviation of LGis 40%. awwn on fault (i.e., the fraction of the unawn loan) is assumeto 60%. The expecteanunexpectelosses (stanrviation) for the bank are Expecteloss = $500, unexpecteloss = $4,140 Expecteloss = $500, unexpecteloss = $3,220 Expecteloss = $460, unexpecteloss = $3,220 Expecteloss = $460, unexpecteloss = $4,140 ANSWER: First, we compute the exposure fault. This is the awn amount, or 80%x$50,000=$40,000 plus the awwn on fault, whiis 60%x$10,000=$6,000, for a totof CE= $46,000. The expecteloss is this amount times   p×E(LG=0.02×50%=1%\;p\times E{(LG}=0.02\times50\%=1\%p×E(LG=0.02×50%=1% ₤or EL = $460. Next, we compute the stanrviation of losses using Equation: σ(CL)=p×σ2(LG+p×(1−p)×[E(LG]2\sigma{(CL)}=\sqrt{p\times\sigma^2{(LG}+p\times{(1-p)}\times{\lbraE{(LG}\rbrack}^2}σ(CL)=p×σ2(LG+p×(1−p)×[E(LG]2 ​. The varianis p×σ2(LG+p×(1−p)×[E(LG]2=0.02(0.4)2+0.02(1−0.02)(0.50)2=0.00810p\times\sigma^2{(LG}+p\times{(1-p)}\times{\lbraE{(LG}\rbrack}^2=0.02{(0.4)}^2+0.02{(1-0.02)}{(0.50)}^2=0.00810p×σ2(LG+p×(1−p)×[E(LG]2=0.02(0.4)2+0.02(1−0.02)(0.50)2=0.00810. Taking the square root gives 0.090. Multiplying $46,000 gives $4,140. Ignoring σ2(LG\sigma^2{(LG}σ2(LG gives the incorreanswer of $3,220. Note ththe unexpecteloss is mugreater ththe expecteloss. 为什么公式里是stanrviation of P没有^2?

2021-05-04 16:50 1 · 回答

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2021-03-26 10:30 1 · 回答

NO.PZ2016082406000004 A bank hbookea lowith totcommitment of $50,000 of whi80% is currently outstanng. The fault probability of the lois assumeto 2% for the next yeanloss given fault (LG is estimate50%. The stanrviation of LGis 40%. awwn on fault (i.e., the fraction of the unawn loan) is assumeto 60%. The expecteanunexpectelosses (stanrviation) for the bank are Expecteloss = $500, unexpecteloss = $4,140 Expecteloss = $500, unexpecteloss = $3,220 Expecteloss = $460, unexpecteloss = $3,220 Expecteloss = $460, unexpecteloss = $4,140 ANSWER: First, we compute the exposure fault. This is the awn amount, or 80%x$50,000=$40,000 plus the awwn on fault, whiis 60%x$10,000=$6,000, for a totof CE= $46,000. The expecteloss is this amount times   p×E(LG=0.02×50%=1%\;p\times E{(LG}=0.02\times50\%=1\%p×E(LG=0.02×50%=1% ₤or EL = $460. Next, we compute the stanrviation of losses using Equation: σ(CL)=p×σ2(LG+p×(1−p)×[E(LG]2\sigma{(CL)}=\sqrt{p\times\sigma^2{(LG}+p\times{(1-p)}\times{\lbraE{(LG}\rbrack}^2}σ(CL)=p×σ2(LG+p×(1−p)×[E(LG]2 ​. The varianis p×σ2(LG+p×(1−p)×[E(LG]2=0.02(0.4)2+0.02(1−0.02)(0.50)2=0.00810p\times\sigma^2{(LG}+p\times{(1-p)}\times{\lbraE{(LG}\rbrack}^2=0.02{(0.4)}^2+0.02{(1-0.02)}{(0.50)}^2=0.00810p×σ2(LG+p×(1−p)×[E(LG]2=0.02(0.4)2+0.02(1−0.02)(0.50)2=0.00810. Taking the square root gives 0.090. Multiplying $46,000 gives $4,140. Ignoring σ2(LG\sigma^2{(LG}σ2(LG gives the incorreanswer of $3,220. Note ththe unexpecteloss is mugreater ththe expecteloss. 为什么求UL的时候要用46000作为AE而不是50000

2021-03-16 11:23 1 · 回答

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