NO.PZ2023091601000019
问题如下：
A quantitative analyst is constructing a stock selection algorithm that
will be employed in making intraday trades and uses the annual returns of two
utility stocks, stock A and stock B, to test the model’s capacity to capture
dependence between stock returns. The 5 years of annual returns data for each
stock used in the test are shown in the following table:
The analyst
estimates that the sample means of the returns of stock A (μA) and
stock B (μB) are 0.146 and 0.138, respectively. What is the unbiased
estimate of the sample covariance of stocks A and B?
选项：
A.
0.003828
B.
0.003892
C.
0.004785
D.
0.004865
解释：
D is correct.
Using the formula for the sample covariance estimator but dividing by n-1 for
an unbiased estimate, we get
𝜎𝐴𝐵=1/(𝑛−1)Σ(𝑅𝐴,𝑖−𝜇𝐴)∗(𝑅𝐵,𝑖−𝜇𝐵)
Which is expanded
as
1/4[(0.18−0.146)(0.32−0.138)+(0.13−0.146)(0.22−0.138)+(0.04−0.146)(0.00−0.138)+(0.30−0.146)(0.10−0.138)+(0.08−0.146)(0.05−0.138)]=(14)∗(0.01946)=0.004865
A is incorrect.
This is the result when the two means are switched in the summation formula and
the multiplier used is 1/5.
B is incorrect.
This is the result when the multiplier used is 1/5 instead of 1/(5-1).
C
is incorrect. This is the result when the two means are switched in the
summation formula.
老师我用cov（A、B）=E（AB）- E（A）E（B）得出来是这个数字，就算用n/（n-1）调整也差距太远了，可以麻烦看看问题出在哪里吗
