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三言午寺 · 2020年06月03日

问一道题:NO.PZ2017092702000075

问题如下:

An analyst produces the following joint probability function for a foreign index (FI) and a domestic index (DI).

The covariance of returns on the foreign index and the returns on the domestic index is closest to:

选项:

A.

26.39%².

B.

26.56%².

C.

28.12%².

解释:

B is correct.

The covariance is 26.56, calculated as follows. First, expected returns are

E(RFI) = (0.25 × 25) + (0.50 × 15) + (0.25 × 10) = 6.25 + 7.50 + 2.50 = 16.25 and E(RDI) = (0.25 × 30) + (0.50 × 25) + (0.25 × 15) = 7.50 + 12.50 + 3.75 = 23.75. Covariance is

Cov(RFI)(RDI)=ijP(RFI,i,RDI,j)(RFI,iERFI)(RDI,jERDI)Cov{(R_{FI})}{(R_{DI})}=\sum_i\sum_jP{(R_{FI,i},R_{DI,j})}{(R_{FI,i}-ER_{FI})}{(R_{DI,j}-ER_{DI})} = 0.25[(25 – 16.25)(30 – 23.75)] + 0.50[(15 – 16.25)(25 – 23.75)] + 0.25[(10 – 16.25) (15 – 23.75)] = 13.67 + (–0.78) + 13.67 = 26.56

0.25是两件事情的联合发生概率,而在用E(RFI) = (0.25 × 25) + (0.50 × 15) + (0.25 × 10) = 6.25 + 7.50 + 2.50 = 16.25这个式子计算的时候,应该用的是每种情况下RFI单独发生的概率才对吧,为什么可以用联合发生概率算呢

1 个答案

星星_品职助教 · 2020年06月04日

同学你好,

这么理解比较抽象,把这个表格类比成当经济好的时候RFI概率是25%,值为0.25;经济中等时候RFI概率是15%,值为0.5;经济差的时候RFI概率是10%,值为0.25,。

然后加权平均求RFI的平均概率就可以了。

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NO.PZ2017092702000075问题如下analyst proces the following joint probability function for a foreign inx (FI) ana mestic inx ().The covarianof returns on the foreign inx anthe returns on the mestic inx is closest to:A.26.39%². B.26.56%². C.28.12%².B is correct.The covarianis 26.56, calculatefollows. First, expectereturns areE(RFI) = (0.25 × 25) + (0.50 × 15) + (0.25 × 10) = 6.25 + 7.50 + 2.50 = 16.25 anE(R) = (0.25 × 30) + (0.50 × 25) + (0.25 × 15) = 7.50 + 12.50 + 3.75 = 23.75. Covarianis Cov(RFI)(R)=∑i∑jP(RFI,i,R,j)(RFI,i−ERFI)(R,j−ER)Cov{(R_{FI})}{(R_{})}=\sum_i\sum_jP{(R_{FI,i},R_{,j})}{(R_{FI,i}-ER_{FI})}{(R_{,j}-ER_{})}Cov(RFI​)(R​)=∑i​∑j​P(RFI,i​,R,j​)(RFI,i​−ERFI​)(R,j​−ER​) = 0.25[(25 – 16.25)(30 – 23.75)] + 0.50[(15 – 16.25)(25 – 23.75)] + 0.25[(10 – 16.25) (15 – 23.75)] = 13.67 + (–0.78) + 13.67 = 26.56请问老师,组合的方差为等于零,是代表什么深层含义?

2024-05-26 23:06 2 · 回答

NO.PZ2017092702000075 问题如下 analyst proces the following joint probability function for a foreign inx (FI) ana mestic inx ().The covarianof returns on the foreign inx anthe returns on the mestic inx is closest to: A.26.39%². B.26.56%². C.28.12%². B is correct.The covarianis 26.56, calculatefollows. First, expectereturns areE(RFI) = (0.25 × 25) + (0.50 × 15) + (0.25 × 10) = 6.25 + 7.50 + 2.50 = 16.25 anE(R) = (0.25 × 30) + (0.50 × 25) + (0.25 × 15) = 7.50 + 12.50 + 3.75 = 23.75. Covarianis Cov(RFI)(R)=∑i∑jP(RFI,i,R,j)(RFI,i−ERFI)(R,j−ER)Cov{(R_{FI})}{(R_{})}=\sum_i\sum_jP{(R_{FI,i},R_{,j})}{(R_{FI,i}-ER_{FI})}{(R_{,j}-ER_{})}Cov(RFI​)(R​)=∑i​∑j​P(RFI,i​,R,j​)(RFI,i​−ERFI​)(R,j​−ER​) = 0.25[(25 – 16.25)(30 – 23.75)] + 0.50[(15 – 16.25)(25 – 23.75)] + 0.25[(10 – 16.25) (15 – 23.75)] = 13.67 + (–0.78) + 13.67 = 26.56 請問是否應該根據這個公式,最後還要除以3? 另外,0.25,0.5,0.25在表格中代表甚麽?謝謝

2022-10-20 01:35 2 · 回答

NO.PZ2017092702000075 所以这一题在按计算器也有陷阱嘛 我是这么按的 25 - 16.25 = x ( 30 - 23.75 ) / 4 + ( 15 - 16.25 ) * (23.75-25)/2+(10-16.25)*(15-23.75)*0.25

2021-12-18 19:50 1 · 回答

26.56%². 28.12%². B is correct. The covarianis 26.56, calculatefollows. First, expectereturns are E(RFI) = (0.25 × 25) + (0.50 × 15) + (0.25 × 10) = 6.25 + 7.50 + 2.50 = 16.25 anE(R) = (0.25 × 30) + (0.50 × 25) + (0.25 × 15) = 7.50 + 12.50 + 3.75 = 23.75. Covarianis Cov(RFI)(R)=∑i∑jP(RFI,i,R,j)(RFI,i−ERFI)(R,j−ER)Cov{(R_{FI})}{(R_{})}=\sum_i\sum_jP{(R_{FI,i},R_{,j})}{(R_{FI,i}-ER_{FI})}{(R_{,j}-ER_{})}Cov(RFI​)(R​)=∑i​∑j​P(RFI,i​,R,j​)(RFI,i​−ERFI​)(R,j​−ER​) = 0.25[(25 – 16.25)(30 – 23.75)] + 0.50[(15 – 16.25)(25 – 23.75)] + 0.25[(10 – 16.25) (15 – 23.75)] = 13.67 + (–0.78) + 13.67 = 26.56是不是又简便算法 手算20分钟

2020-04-07 10:02 2 · 回答