尾部平均是简单平均法何时又变成加权平均了？

NO.PZ2020011303000054 问题如下 A one-yeprojeha 3% chanof losing US10million, a 7% chanof losing US3 million, ana 90% chanof gaining US1 million.Suppose ththere are two inpennt inticinvestments with the properties. Whare (the Van(the expecteshortfall for a portfolio consisting of the two investments when the confinlevel is 95% anthe time horizon is one year? 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 (US of 20, 13, 9, 6, 2, an−2 have probabilities of 0.0009, 0.0042, 0.054, 0.0049, 0.126, an0.81, respectively. 95%VaR=995%ES=[0.0009×20+0.042×13+(0.05-0.0009-0.0042)×9]/0.05=9.534 看了也不知道再算什么？

NO.PZ2020011303000054问题如下 A one-yeprojeha 3% chanof losing US10million, a 7% chanof losing US3 million, ana 90% chanof gaining US1 million.Suppose ththere are two inpennt inticinvestments with the properties. Whare (the Van(the expecteshortfall for a portfolio consisting of the two investments when the confinlevel is 95% anthe time horizon is one year? 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 (US of 20, 13, 9, 6, 2, an−2 have probabilities of 0.0009, 0.0042, 0.054, 0.0049, 0.126, an0.81, respectively. 95%VaR=995%ES=[0.0009×20+0.042×13+(0.05-0.0009-0.0042)×9]/0.05=9.534 老师您好，关于这个一年期项目，为什么损失会有顺序？又不是三年期项目，第一年，第二年的情况因为时间先后会有顺序。不太明白

NO.PZ2020011303000054问题如下 A one-yeprojeha 3% chanof losing US10million, a 7% chanof losing US3 million, ana 90% chanof gaining US1 million.Suppose ththere are two inpennt inticinvestments with the properties. Whare (the Van(the expecteshortfall for a portfolio consisting of the two investments when the confinlevel is 95% anthe time horizon is one year? 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 (US of 20, 13, 9, 6, 2, an−2 have probabilities of 0.0009, 0.0042, 0.054, 0.0049, 0.126, an0.81, respectively. 95%VaR=995%ES=[0.0009×20+0.042×13+(0.05-0.0009-0.0042)×9]/0.05=9.534 老师这个题能画个图吗

NO.PZ2020011303000054问题如下 A one-yeprojeha 3% chanof losing US10million, a 7% chanof losing US3 million, ana 90% chanof gaining US1 million.Suppose ththere are two inpennt inticinvestments with the properties. Whare (the Van(the expecteshortfall for a portfolio consisting of the two investments when the confinlevel is 95% anthe time horizon is one year? 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 (US of 20, 13, 9, 6, 2, an−2 have probabilities of 0.0009, 0.0042, 0.054, 0.0049, 0.126, an0.81, respectively. 95%VaR=995%ES=[0.0009×20+0.042×13+(0.05-0.0009-0.0042)×9]/0.05=9.534 请问为何10+3的损失下概率是2*10和3的各自概率？而10+10的损失，或者3+3的损失(相同数)下概率是各自概率相乘，而不用乘以2呢？谢谢！

NO.PZ2020011303000054 问题如下 A one-yeprojeha 3% chanof losing US10million, a 7% chanof losing US3 million, ana 90% chanof gaining US1 million.Suppose ththere are two inpennt inticinvestments with the properties. Whare (the Van(the expecteshortfall for a portfolio consisting of the two investments when the confinlevel is 95% anthe time horizon is one year? 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 (US of 20, 13, 9, 6, 2, an−2 have probabilities of 0.0009, 0.0042, 0.054, 0.0049, 0.126, an0.81, respectively. 95%VaR=995%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。