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徐威廉 · 2024年04月07日

自由度

  1. FRM I
  2. Quantitative

NO.PZ2022062760000017

问题如下：

Using the returns of the prior 12 months, an analyst estimates the mean monthly return of stock XYZ to be -0.75% with a standard error of 2.70%.


Using the t-table above, which of the following is the 95% confidence interval for the mean return?

选项：

A.

-6.69% and 5.19%

B.

-6.63% and 5.15%

C.

-5.60% and 4.10%

D.

-5.56% and 4.06%

解释：

中文解析：

区间=mean+z值*σ

mean=-0.75%

σ=2.7%

z值的查找：自由度=12-1=11，双尾5%，单尾2.5%

z值=2.201

区间=-0.75% - 2.201*2.70% and -0.75% + 2.201*2.70%

The confidence interval is equal to the mean monthly return plus or minus the t-statistic times the standard error. To get the proper t-statistic, the 0.025 column must be used since this is a two-tailed interval. Since the mean return is being estimated using the sample observations, the appropriate degrees of freedom to use is equal to the number of sample observations minus 1, which is 11. Therefore, the proper statistic to use from the t-distribution is 2.201. The 95% confidence interval is between -0.75% - 2.201*2.70% and -0.75% + 2.201*2.70%.

关于这个自由度能不能汇总讲一下？

1.这个12月就是总体啊，为什么自由度要减1

2.还有后面章节的，自由度 n-1的，n-k-1的，k的这都是对应什么的？全混了

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李坏_品职助教 · 2024年04月07日

嗨，努力学习的PZer你好：


题目说的是Using the returns of the prior 12 months, an analyst estimates the mean monthly return，也就是利用过去12个月的数据作为样本去推测总体，而这个样本的均值是 -0.75%.


统计学都是从样本去推算总体。正是因为用的是样本数据，不是精确的总体数据，所以才需要置信区间，真正的总体平均收益率应该处在一个置信区间范围内。


自由度里面的k指的都是自变量的个数，就是回归方程中的X的个数。

1.如果是单样本的t检验，自由度=n-1. 这个可以参考Student’s T-distribution部分的内容。

2.如果是多元线性回归中的回归系数（βj），那么t检验使用n-k-1。可以参考讲义Hypothesis Test in Multiple Regression部分的内容。

3.如果是F检验：分子和分母涉及到不同的自由度：




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