问题没看懂？不是有9种可能呢，为啥只有Fi的3种？知道先算两个的均值，之后算每个情况的cov再相加，问题如下图：
选项：
A. 
B. 
C. 
解释：
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∑jP(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？
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∑jP(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
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
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
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∑jP(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
