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RyanR · 2023年03月19日

为什么损失13也用0.42% 加权

NO.PZ2020011303000054

问题如下:

A one-year project has a 3% chance of losing USD 10million, a 7% chance of losing USD 3 million, and a 90% chance of gaining USD 1 million.

Suppose that there are two independent identical investments with the properties.

What are (a) the VaR and (b) the expected shortfall for a portfolio consisting of the two investments when the confidence level is 95% and the time horizon is one year?

选项:

解释:

有一个项目,3%的概率会损失10m7%损失3m90%概率会获得1m假设这俩投资都是独立相同的,求95%置信区间下1年的VaRES

Losses (USD) of 20, 13, 9, 6, 2, and 2 have probabilities of 0.0009, 0.0042, 0.054, 0.0049, 0.126, and 0.81, respectively.

95%VaR=9

95%ES=[0.0009×20+0.042×13+(0.05-0.0009-0.0042)×9]/0.05=9.534

0.09%的概率损失会超过20,0.42%的概率损失会超过13,5%的概率损失会超过9。

加权平均的话,就是这三种损失的情况总概率是5%

那损失20的概率是0.09%,损失13的概率是0.42%-0.09%=0.33%,损失9的概率是5%-0.42%=4.58%

所以我理解应该用这个概率求平均呀,为啥答案13也是用0.42%加权的。


另外一个问题,我对于-6的prob的是0.49%应该怎么理解有一点不明白,就画了了一个图。请老师帮忙看看这样的理解是否正确。



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已采纳答案

品职答疑小助手雍 · 2023年03月19日

同学你好,正是因为后一个问题你理解的不正确,导致第一个问题也理解的不正确。

这些概率你就想成一层一层叠罗汉(跌满累计概率是100%),但是每一层的厚度不一样,其厚度对应的就是概率,比如损失9的厚度就是5.4%,损失20的厚度就是0.09%。

所以第二个问题的图片里,不应该是-9到-6之间有0.49%。而是-6这个数字占据了一个区间概率为0.49%的区间。

所以回到第一个问题,损失20对应的厚度是0.09%,损失13对应的厚度是0.42%,那么把损失9所占的5.4%里面,分出来5%-0.09%-0.42%=4.49% 就可以联合前两个厚度把尾部的厚度填充到5%。 那么尾部的5%里面就有3个损失数字20,13,9 分别对应的概率权重是0.09%,0.42%,4.49%。

RyanR · 2023年03月19日

明白了!!后面听到后面的课,也看到何老师用这种一个个柱状图的形式解释,。感谢

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