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张大龙 · 2025年02月24日

想问下求ES的时候，损失是9的权重算法

  1. FRM I
  2. Valuation And Risk Models

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：

A.

VaR = USD 6 million, ES = USD 9.534 million

B.

VaR = USD 9 million, ES = USD 9.534 million

C.

VaR = USD 9 million, ES = USD 13 million

D.

VaR = USD 6 million, ES = USD 13 million

解释：

有一个项目，3%的概率会损失10m，7%损失3m，90%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VaR和ES？

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

10-1 和-1 10 概率一共是3%*90%*2=0.54。

我理解答案用 5%-0.0009-0.0042 是要在VAR线以上算剩余的概率，相当于线性插值法么？

但是为什么这个是配给-9呢？而不是 用5%-0.54-0.0042配给-13，或者5%-0.54-0.0009配给-20。

此外，这也带来我另一个问题，VAR 95%的话，题目中不是正好95%，而是差一些，这时候的处理方式是什么？

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1 个答案

李坏_品职助教 · 2025年02月24日

嗨，爱思考的PZer你好：


这三个概率累计起来 = 0.0591， 大于5%了。


也就是说，按照9这个loss，累计的概率，已经超过了题目要求的5%的尾部概率。

但是如果只考虑20和13这两个loss，累计概率是0.51%，还远远达不到5%.


可以这么理解：现在有三个碗（三个Loss）去盛水，需要盛5%的水。现在水已经装满了20和13这两个碗，而9这个碗却没有盛满。9这个碗一共盛了5% - 0.42% - 0.09%这么多水，所以是(0.05-0.0009-0.0042)×9

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