开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)｜3.2.5(安卓)APP下载

返回

为了求职冲呀 · 2019年11月25日

问一道题：NO.PZ2017092702000075

  1. CFA I
  2. Quantitative

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 {$math1}

Cov(RFI)(RDI)=∑i∑jP(RFI,i,RDI,j)(RFI,i−ERFI)(RDI,j−ERDI)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

请问老师在算E(DFI)的时候为什么把25%直接乘了0.25呢？不应该把25%转换成0.25吗？比如：E (DFI)=0.25x0.25+1.5x0.5+1.0x0.25


因为在计算cov的题库里有很多题是直接把百分数转换成小数计算的，什么时候应该直接去掉百分号？什么时候应该转换成小数呢？

添加评论
1 个答案

星星_品职助教 · 2019年11月25日

同学你好，

这种题型里转化和不转化其实都是一致的。答案解析里的计算式E(RFI) = (0.25 × 25) + (0.50 × 15) + (0.25 × 10) = 6.25 + 7.50 + 2.50 = 16.25 。 相当于把百分号当成了一个单位，算出来的结果如果写全了就是 16.25（%）。直接就是百分数的形式。百分号当成单位是外国人的一种习惯，理解成和“元”，“公斤”等一样就行。

转化为小数计算也是可以的，这样算出来的E(RFI) 是0.1625。如果选项里是百分数形式，再转化为16.25%即可。

具体是用小数计算还是把百分号当成单位计算，主要看选项，如果选项都是百分数的形式，那么把百分号当成单位计算可以直接得出答案，加油。

  • 1

    回答

  • 0

    关注

  • 974

    浏览
相关问题

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

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

2020-06-03 23:40 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