请问这个题目所用的公式在我们框架图哪里讲到过？谢谢！
问题如下图：
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解释：
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?
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. 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？
NO.PZ2016082406000004 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.1、请问这里的UL和sigma是用什么地方的公式计算的呢？2、为什么最后用sigma乘以EA？这是算的UL吗
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
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. 我有点混淆Cret-Var和Unexpecteloss的概念了。他们都是损失偏离预期损失的部分，请问unexpecteloss是所有偏离expecteloss的平均数而Cvar只是再某个分位点上的数吗？
