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yunli · 2021年12月29日

关于dummy的数量

  1. FRM I
  2. Quantitative

NO.PZ2020010801000025

问题如下：

A model was estimated using daily data from the S&P 500 from 1977 until 2017 which included five day-of-the-week dummies (n = 10,087). The R2R^2 from this regression was 0.000599. Is there evidence that the mean varies with the day of the week?

解释：

The model estimated is

Yi=β1D1+β2D2+β3D3+β4D4+β5D5+ϵiY_i = \beta_1D_1 + \beta_2D_2 + \beta_3D_3 + \beta_4D_4 + \beta_5D_5 + \epsilon_i,

where Di is a dummy that takes the value 1 if the index of the weekday is i (e.g., Monday = 1, Tuesday = 2, c). The restriction is that

H0:β1=β2=β3=β4=β5H_0:\beta_1 = \beta_2 = \beta_3 = \beta_4 = \beta_5

so there this is no day-of-the-week effect. This model can be equivalently written as

Yi=μ+δ2D2+δ3D3+δ4D4+δ5D5+ϵiY_i = \mu + \delta_2D_2 + \delta_3D_3 + \delta_4D_4 + \delta_5D_5 + \epsilon_i,

therefore, here the null is

H0:δ2=δ3=δ4=δ5H_0:\delta_2 = \delta_3 = \delta_4 = \delta_5.

In the two models, μ=β1\mu = \beta_1, and μ+δi=βi\mu + \delta_i = \beta_i. The second form of the model is a more standard null for an F-stat.

The F-stat of the regression is

(R2−0)/4(1−R2)/(n−5)=0.000599/4(1−0.000599)/(10087−5)=1.51\frac{(R^2-0)/4}{(1-R^2)/(n-5)}=\frac{0.000599/4}{(1-0.000599)/(10087-5)}=1.51

The distribution is an F4,10082F_{4,10082} and the critical value using a 5% size is 2.37. The test statistic is less than the critical value, therefore, the null that all effects are 0 is not rejected.

题干中直接说了5dummies为什么列回归式的时候要简化成4个呀。虽然可以理解一周5个工作日，所以只要4个dummy，但是题干中的说法有点迷惑，还以为什么特殊情况考虑一天交易6天。请问考试中遇到类似问题是都要基于常识判断而不是根据题目直接表达的含义吗？谢谢
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3 个答案

李坏_品职助教 · 2022年01月10日

嗨，努力学习的PZer你好：


嗯，你说的H0检测所有系数都等于0，那个是多元线性回归里面的系数。


这里的F检验是检验dummy variable的，目的是为了检验几个哑变量是不是都一样的（都一样就说明不存在weekday effect）。至于说是不是都等于0其实不重要，无论是不是都等于0，或者都等于1，model都是Yi = 一个常数项。

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就算太阳没有迎着我们而来，我们正在朝着它而去，加油！

李坏_品职助教 · 2021年12月30日

嗨，从没放弃的小努力你好：


FRM原版书本身的深度是超过了考试难度的，从考试的角度出发不建议直接看原版书。还是以讲义和视频课为主比较好一些。


原版书的作用是查缺补漏，比如有哪个知识点你自己想多学习一些的，可以翻一翻。

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虽然现在很辛苦，但努力过的感觉真的很好，加油！

李坏_品职助教 · 2021年12月29日

嗨，努力学习的PZer你好：


题干给出的第一个model里面是五个variable，但实际上只需要4个dummy就可以表示5个工作日了（考试中如果问你需要几个dummy，就选4个）。为了构造F检验，我们需要用到讲义上的公式：


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努力的时光都是限量版，加油！

he123456 · 2021年12月30日

老师，为什么感觉原版书题目这么难，这些应用上课老师都没讲啊感觉，这个题目还是不会

jacqie · 2022年01月10日

想问一下为什么H0不是等式最后=0？而只是4个系数相等？F测试不是算所有系数都最后等于=0的么？

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NO.PZ2020010801000025 问题如下 A mol westimateusing ily ta from the S P 500 from 1977 until 2017 whiinclufive y-of-the-week mmies (n = 10,087). The R2R^2R2 from this regression w0.000599. Is there evinththe mevaries with the y of the week? The mol estimateis Yi=β1+β2+β3+β4+β5+ϵiY_i = \beta_11 + \beta_22 + \beta_33 + \beta_44 + \beta_55 + \epsilon_iYi​=β1​​+β2​​+β3​​+β4​​+β5​​+ϵi​, where is a mmy thtakes the value 1 if the inx of the weeky is i (e.g., Mony = 1, Tuesy = 2, c). The restriction is thH0:β1=β2=β3=β4=β5H_0:\beta_1 = \beta_2 = \beta_3 = \beta_4 = \beta_5H0​:β1​=β2​=β3​=β4​=β5​ so there this is no y-of-the-week effect. This mol cequivalently written Yi=μ+δ2+δ3+δ4+δ5+ϵiY_i = \mu + \lta_22 + \lta_33 + \lta_44 + \lta_55 + \epsilon_iYi​=μ+δ2​​+δ3​​+δ4​​+δ5​​+ϵi​, therefore, here the null is H0:δ2=δ3=δ4=δ5H_0:\lta_2 = \lta_3 = \lta_4 = \lta_5H0​:δ2​=δ3​=δ4​=δ5​. In the two mols, μ=β1\mu = \beta_1μ=β1​, anμ+δi=βi\mu + \lta_i = \beta_iμ+δi​=βi​. The seconform of the mol is a more stanrnull for F-stat. The F-stof the regression is(R2−0)/4(1−R2)/(n−5)=0.000599/4(1−0.000599)/(10087−5)=1.51\frac{(R^2-0)/4}{(1-R^2)/(n-5)}=\frac{0.000599/4}{(1-0.000599)/(10087-5)}=1.51(1−R2)/(n−5)(R2−0)/4​=(1−0.000599)/(10087−5)0.000599/4​=1.51The stribution is F4,10082F_{4,10082}F4,10082​ anthe criticvalue using a 5% size is 2.37. The test statistic is less ththe criticvalue, therefore, the null thall effects are 0 is not rejecte 请问一下F检验分子不是应该等于(UnrestricteR 2 - RestricteR2)/q 吗。题目说了unrestricteR2但并没有说restricteR2是多少。为什么公式里面直接就把restricteR2 忽略掉了呢

2024-05-24 11:45 1 · 回答

NO.PZ2020010801000025 问题如下 A mol westimateusing ily ta from the S P 500 from 1977 until 2017 whiinclufive y-of-the-week mmies (n = 10,087). The R2R^2R2 from this regression w0.000599. Is there evinththe mevaries with the y of the week? The mol estimateis Yi=β1+β2+β3+β4+β5+ϵiY_i = \beta_11 + \beta_22 + \beta_33 + \beta_44 + \beta_55 + \epsilon_iYi​=β1​​+β2​​+β3​​+β4​​+β5​​+ϵi​, where is a mmy thtakes the value 1 if the inx of the weeky is i (e.g., Mony = 1, Tuesy = 2, c). The restriction is thH0:β1=β2=β3=β4=β5H_0:\beta_1 = \beta_2 = \beta_3 = \beta_4 = \beta_5H0​:β1​=β2​=β3​=β4​=β5​ so there this is no y-of-the-week effect. This mol cequivalently written Yi=μ+δ2+δ3+δ4+δ5+ϵiY_i = \mu + \lta_22 + \lta_33 + \lta_44 + \lta_55 + \epsilon_iYi​=μ+δ2​​+δ3​​+δ4​​+δ5​​+ϵi​, therefore, here the null is H0:δ2=δ3=δ4=δ5H_0:\lta_2 = \lta_3 = \lta_4 = \lta_5H0​:δ2​=δ3​=δ4​=δ5​. In the two mols, μ=β1\mu = \beta_1μ=β1​, anμ+δi=βi\mu + \lta_i = \beta_iμ+δi​=βi​. The seconform of the mol is a more stanrnull for F-stat. The F-stof the regression is(R2−0)/4(1−R2)/(n−5)=0.000599/4(1−0.000599)/(10087−5)=1.51\frac{(R^2-0)/4}{(1-R^2)/(n-5)}=\frac{0.000599/4}{(1-0.000599)/(10087-5)}=1.51(1−R2)/(n−5)(R2−0)/4​=(1−0.000599)/(10087−5)0.000599/4​=1.51The stribution is F4,10082F_{4,10082}F4,10082​ anthe criticvalue using a 5% size is 2.37. The test statistic is less ththe criticvalue, therefore, the null thall effects are 0 is not rejecte 请问第二个公式是怎么推导出来的？为什么少了一个regressor

2024-05-21 15:27 3 · 回答

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2023-09-05 17:44 1 · 回答

NO.PZ2020010801000025 问题如下 A mol westimateusing ily ta from the S P 500 from 1977 until 2017 whiinclufive y-of-the-week mmies (n = 10,087). The R2R^2R2 from this regression w0.000599. Is there evinththe mevaries with the y of the week? The mol estimateis Yi=β1+β2+β3+β4+β5+ϵiY_i = \beta_11 + \beta_22 + \beta_33 + \beta_44 + \beta_55 + \epsilon_iYi​=β1​​+β2​​+β3​​+β4​​+β5​​+ϵi​, where is a mmy thtakes the value 1 if the inx of the weeky is i (e.g., Mony = 1, Tuesy = 2, c). The restriction is thH0:β1=β2=β3=β4=β5H_0:\beta_1 = \beta_2 = \beta_3 = \beta_4 = \beta_5H0​:β1​=β2​=β3​=β4​=β5​ so there this is no y-of-the-week effect. This mol cequivalently written Yi=μ+δ2+δ3+δ4+δ5+ϵiY_i = \mu + \lta_22 + \lta_33 + \lta_44 + \lta_55 + \epsilon_iYi​=μ+δ2​​+δ3​​+δ4​​+δ5​​+ϵi​, therefore, here the null is H0:δ2=δ3=δ4=δ5H_0:\lta_2 = \lta_3 = \lta_4 = \lta_5H0​:δ2​=δ3​=δ4​=δ5​. In the two mols, μ=β1\mu = \beta_1μ=β1​, anμ+δi=βi\mu + \lta_i = \beta_iμ+δi​=βi​. The seconform of the mol is a more stanrnull for F-stat. The F-stof the regression is(R2−0)/4(1−R2)/(n−5)=0.000599/4(1−0.000599)/(10087−5)=1.51\frac{(R^2-0)/4}{(1-R^2)/(n-5)}=\frac{0.000599/4}{(1-0.000599)/(10087-5)}=1.51(1−R2)/(n−5)(R2−0)/4​=(1−0.000599)/(10087−5)0.000599/4​=1.51The stribution is F4,10082F_{4,10082}F4,10082​ anthe criticvalue using a 5% size is 2.37. The test statistic is less ththe criticvalue, therefore, the null thall effects are 0 is not rejecte Means 和R Square 是个什么关系？该题目的解题思路是什么？ 这两者如何关联上？考了哪几个知识点？非常感谢

2023-02-05 02:23 1 · 回答

NO.PZ2020010801000025问题如下A mol westimateusing ily ta from the S P 500 from 1977 until 2017 whiinclufive y-of-the-week mmies (n = 10,087). The R2R^2R2 from this regression w0.000599. Is there evinththe mevaries with the y of the week? The mol estimateis Yi=β1+β2+β3+β4+β5+ϵiY_i = \beta_11 + \beta_22 + \beta_33 + \beta_44 + \beta_55 + \epsilon_iYi​=β1​​+β2​​+β3​​+β4​​+β5​​+ϵi​, where is a mmy thtakes the value 1 if the inx of the weeky is i (e.g., Mony = 1, Tuesy = 2, c). The restriction is thH0:β1=β2=β3=β4=β5H_0:\beta_1 = \beta_2 = \beta_3 = \beta_4 = \beta_5H0​:β1​=β2​=β3​=β4​=β5​ so there this is no y-of-the-week effect. This mol cequivalently written Yi=μ+δ2+δ3+δ4+δ5+ϵiY_i = \mu + \lta_22 + \lta_33 + \lta_44 + \lta_55 + \epsilon_iYi​=μ+δ2​​+δ3​​+δ4​​+δ5​​+ϵi​, therefore, here the null is H0:δ2=δ3=δ4=δ5H_0:\lta_2 = \lta_3 = \lta_4 = \lta_5H0​:δ2​=δ3​=δ4​=δ5​. In the two mols, μ=β1\mu = \beta_1μ=β1​, anμ+δi=βi\mu + \lta_i = \beta_iμ+δi​=βi​. The seconform of the mol is a more stanrnull for F-stat. The F-stof the regression is(R2−0)/4(1−R2)/(n−5)=0.000599/4(1−0.000599)/(10087−5)=1.51\frac{(R^2-0)/4}{(1-R^2)/(n-5)}=\frac{0.000599/4}{(1-0.000599)/(10087-5)}=1.51(1−R2)/(n−5)(R2−0)/4​=(1−0.000599)/(10087−5)0.000599/4​=1.51The stribution is F4,10082F_{4,10082}F4,10082​ anthe criticvalue using a 5% size is 2.37. The test statistic is less ththe criticvalue, therefore, the null thall effects are 0 is not rejecte这种难度和类型的题考试会考吗

2022-03-21 15:55 1 · 回答