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李晨 · 2021年05月04日

计算UL时候，sqr rt后面部分不是LGD^2*standard deviation of PD^2吗？为什么公式里是standard deviation of PD,没有^2?

  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.

为什么公式里是standard deviation of PD,没有^2?

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1 个答案

品职答疑小助手雍 · 2021年05月04日

嗨，努力学习的PZer你好：


用的是PD的方差的，只不过用的是p*（1-p）的形式，下图后面的那个红线。

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