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Churning · 2020年03月22日

问一道题：NO.PZ2017092702000051 [ CFA I ]

  1. CFA I
  2. Quantitative

问题如下：

Annual returns and summary statistics for three funds are listed in the following table:

The fund that shows the highest dispersion is:

选项：

A.

Fund PQR if the measure of dispersion is the range.

B.

Fund XYZ if the measure of dispersion is the variance.

C.

Fund ABC if the measure of dispersion is the mean absolute deviation.

解释：

C is correct.

The mean absolute deviation (MAD) of Fund ABC’s returns is greater than the MAD of both of the other funds.

MAD=∑in∣Xi−X‾∣nMAD=\frac{\displaystyle\sum_i^n{\vert Xi-\overline X\vert}}n

where \(\overline X\) is the arithmetic mean of the series.

MAD for Fund ABC =

[−20−(−4)]+[23−(−4)]+[−14−(−4)]+[5−(−4)]+[−14−(−4)]5=14.4%\frac{{\lbrack-20-{(-4)}\rbrack}+{\lbrack23-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}+{\lbrack5-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}}5=14.4\%

MAD for Fund XYZ =

[−33−(−10.8)]+[-12−(−10.8)]+[−12−(−10.8)]+[-8−(−10.8)]+[11−(−10.8)]5=9.8%\frac{{\lbrack-33-{(-10.8)}\rbrack}+{\lbrack\text{-12}-{(-10.8)}\rbrack}+{\lbrack-\text{12}-{(-10.8)}\rbrack}+{\lbrack\text{-8}-{(-10.8)}\rbrack}+{\lbrack\text{11}-{(-10.8)}\rbrack}}5=\text{9}\text{.8}\%

MAD for Fund PQR =

[−14−(−5)]+[-18−(−5)]+[6−(−5)]+[-2−(−5)]+[3−(−5)]5=8.8%\frac{{\lbrack-\text{14}-{(-\text{5})}\rbrack}+{\lbrack\text{-18}-{(-\text{5})}\rbrack}+{\lbrack\text{6}-{(-\text{5})}\rbrack}+{\lbrack\text{-2}-{(-\text{5})}\rbrack}+{\lbrack\text{3}-{(-\text{5})}\rbrack}}5=\text{8}\text{.8}\%

A and B are incorrect because the range and variance of the three funds are as follows:

The numbers shown for variance are understood to be in "percent squared" terms so that when taking the square root, the result is standard deviation in percentage terms. Alternatively, by expressing standard deviation and variance in decimal form, one can avoid the issue of units; in decimal form, the variances for Fund ABC, Fund XYZ, and Fund PQR are 0.0317, 0.0243, and 0.0110, respectively.

老师问下为什么不选B呢
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1 个答案

星星_品职助教 · 2020年03月22日

同学你好，

B选项直接看表格里的standard deviation就可以，Fund ABC最大，而不是XYZ

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FunXYZ if the measure of spersion is the variance. FunAif the measure of spersion is the meabsolute viation. C is correct. The meabsolute viation (MA of FunABC’s returns is greater ththe Mof both of the other fun. MA∑in∣Xi−X‾∣nMA\frac{\splaystyle\sum_i^n{\vert Xi-\overline X\vert}}nMAni∑n​∣Xi−X∣​ where \(\overline X\) is the arithmetic meof the series. Mfor FunA= [−20−(−4)]+[23−(−4)]+[−14−(−4)]+[5−(−4)]+[−14−(−4)]5=14.4%\frac{{\lbrack-20-{(-4)}\rbrack}+{\lbrack23-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}+{\lbrack5-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}}5=14.4\%5[−20−(−4)]+[23−(−4)]+[−14−(−4)]+[5−(−4)]+[−14−(−4)]​=14.4% Mfor FunXYZ = [−33−(−10.8)]+[-12−(−10.8)]+[−12−(−10.8)]+[-8−(−10.8)]+[11−(−10.8)]5=9.8%\frac{{\lbrack-33-{(-10.8)}\rbrack}+{\lbrack\text{-12}-{(-10.8)}\rbrack}+{\lbrack-\text{12}-{(-10.8)}\rbrack}+{\lbrack\text{-8}-{(-10.8)}\rbrack}+{\lbrack\text{11}-{(-10.8)}\rbrack}}5=\text{9}\text{.8}\%5[−33−(−10.8)]+[-12−(−10.8)]+[−12−(−10.8)]+[-8−(−10.8)]+[11−(−10.8)]​=9.8% Mfor FunPQR = [−14−(−5)]+[-18−(−5)]+[6−(−5)]+[-2−(−5)]+[3−(−5)]5=8.8%\frac{{\lbrack-\text{14}-{(-\text{5})}\rbrack}+{\lbrack\text{-18}-{(-\text{5})}\rbrack}+{\lbrack\text{6}-{(-\text{5})}\rbrack}+{\lbrack\text{-2}-{(-\text{5})}\rbrack}+{\lbrack\text{3}-{(-\text{5})}\rbrack}}5=\text{8}\text{.8}\%5[−14−(−5)]+[-18−(−5)]+[6−(−5)]+[-2−(−5)]+[3−(−5)]​=8.8% A anB are incorrebecause the range anvarianof the three fun are follows: The numbers shown for varianare unrstooto in \"percent square" terms so thwhen taking the square root, the result is stanrviation in percentage terms. Alternatively, expressing stanrviation anvarianin cimform, one cavoithe issue of units; in cimform, the variances for FunABFunXYZ, anFunPQR are 0.0317, 0.0243, an0.0110, respectively. 老师，我用计算器78键上面的数据功能算出的标准差B也对哦，这怎么回事。这个不能用计算器算吗

2021-05-31 00:15 1 · 回答

NO.PZ2017092702000051 FunXYZ if the measure of spersion is the variance. FunAif the measure of spersion is the meabsolute viation. C is correct. The meabsolute viation (MA of FunABC’s returns is greater ththe Mof both of the other fun. MA∑in∣Xi−X‾∣nMA\frac{\splaystyle\sum_i^n{\vert Xi-\overline X\vert}}nMAni∑n​∣Xi−X∣​ where \(\overline X\) is the arithmetic meof the series. Mfor FunA= [−20−(−4)]+[23−(−4)]+[−14−(−4)]+[5−(−4)]+[−14−(−4)]5=14.4%\frac{{\lbrack-20-{(-4)}\rbrack}+{\lbrack23-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}+{\lbrack5-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}}5=14.4\%5[−20−(−4)]+[23−(−4)]+[−14−(−4)]+[5−(−4)]+[−14−(−4)]​=14.4% Mfor FunXYZ = [−33−(−10.8)]+[-12−(−10.8)]+[−12−(−10.8)]+[-8−(−10.8)]+[11−(−10.8)]5=9.8%\frac{{\lbrack-33-{(-10.8)}\rbrack}+{\lbrack\text{-12}-{(-10.8)}\rbrack}+{\lbrack-\text{12}-{(-10.8)}\rbrack}+{\lbrack\text{-8}-{(-10.8)}\rbrack}+{\lbrack\text{11}-{(-10.8)}\rbrack}}5=\text{9}\text{.8}\%5[−33−(−10.8)]+[-12−(−10.8)]+[−12−(−10.8)]+[-8−(−10.8)]+[11−(−10.8)]​=9.8% Mfor FunPQR = [−14−(−5)]+[-18−(−5)]+[6−(−5)]+[-2−(−5)]+[3−(−5)]5=8.8%\frac{{\lbrack-\text{14}-{(-\text{5})}\rbrack}+{\lbrack\text{-18}-{(-\text{5})}\rbrack}+{\lbrack\text{6}-{(-\text{5})}\rbrack}+{\lbrack\text{-2}-{(-\text{5})}\rbrack}+{\lbrack\text{3}-{(-\text{5})}\rbrack}}5=\text{8}\text{.8}\%5[−14−(−5)]+[-18−(−5)]+[6−(−5)]+[-2−(−5)]+[3−(−5)]​=8.8% A anB are incorrebecause the range anvarianof the three fun are follows: The numbers shown for varianare unrstooto in \"percent square" terms so thwhen taking the square root, the result is stanrviation in percentage terms. Alternatively, expressing stanrviation anvarianin cimform, one cavoithe issue of units; in cimform, the variances for FunABFunXYZ, anFunPQR are 0.0317, 0.0243, an0.0110, respectively.为什么用MA比较离散程度呢

2021-02-16 13:55 1 · 回答

NO.PZ2017092702000051 请问range是什么公式？

2021-02-10 11:50 1 · 回答

FunXYZ if the measure of spersion is the variance. FunAif the measure of spersion is the meabsolute viation. C is correct. The meabsolute viation (MA of FunABC’s returns is greater ththe Mof both of the other fun. MA∑in∣Xi−X‾∣nMA\frac{\splaystyle\sum_i^n{\vert Xi-\overline X\vert}}nMAni∑n​∣Xi−X∣​ where \(\overline X\) is the arithmetic meof the series. Mfor FunA= [−20−(−4)]+[23−(−4)]+[−14−(−4)]+[5−(−4)]+[−14−(−4)]5=14.4%\frac{{\lbrack-20-{(-4)}\rbrack}+{\lbrack23-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}+{\lbrack5-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}}5=14.4\%5[−20−(−4)]+[23−(−4)]+[−14−(−4)]+[5−(−4)]+[−14−(−4)]​=14.4% Mfor FunXYZ = [−33−(−10.8)]+[-12−(−10.8)]+[−12−(−10.8)]+[-8−(−10.8)]+[11−(−10.8)]5=9.8%\frac{{\lbrack-33-{(-10.8)}\rbrack}+{\lbrack\text{-12}-{(-10.8)}\rbrack}+{\lbrack-\text{12}-{(-10.8)}\rbrack}+{\lbrack\text{-8}-{(-10.8)}\rbrack}+{\lbrack\text{11}-{(-10.8)}\rbrack}}5=\text{9}\text{.8}\%5[−33−(−10.8)]+[-12−(−10.8)]+[−12−(−10.8)]+[-8−(−10.8)]+[11−(−10.8)]​=9.8% Mfor FunPQR = [−14−(−5)]+[-18−(−5)]+[6−(−5)]+[-2−(−5)]+[3−(−5)]5=8.8%\frac{{\lbrack-\text{14}-{(-\text{5})}\rbrack}+{\lbrack\text{-18}-{(-\text{5})}\rbrack}+{\lbrack\text{6}-{(-\text{5})}\rbrack}+{\lbrack\text{-2}-{(-\text{5})}\rbrack}+{\lbrack\text{3}-{(-\text{5})}\rbrack}}5=\text{8}\text{.8}\%5[−14−(−5)]+[-18−(−5)]+[6−(−5)]+[-2−(−5)]+[3−(−5)]​=8.8% A anB are incorrebecause the range anvarianof the three fun are follows: The numbers shown for varianare unrstooto in \"percent square" terms so thwhen taking the square root, the result is stanrviation in percentage terms. Alternatively, expressing stanrviation anvarianin cimform, one cavoithe issue of units; in cimform, the variances for FunABFunXYZ, anFunPQR are 0.0317, 0.0243, an0.0110, respectively.请教一下是否有这样的规律，可以总结出规律，一般A、B两组数，MA的，var也大？对待这类题目这样可以少算一组吗？

2021-02-06 09:38 1 · 回答

根据基础班课程，对比两组或以上数据的离散程度，已知均值和标准差，则用CV就可以判断离散程度的大小？ Fun117.8/4=4.45 Fun215.6/10.8=1.44 Fun310.5/5.0=2.1 所以Fun1离散程度最大 所以在这里也可以基于CV判断，不是非要用十进制方差？ 另外，老师可以把答案中的0.0317、0.0243、0.0110计算过程展示一下吗？

2020-10-18 12:11 1 · 回答