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比如世界 · 2020年04月08日

问一道题: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?

选项:

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

(R20)/4(1R2)/(n5)=0.000599/4(10.000599)/(100875)=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.

1、这道题题目的意思不是很懂;2、答案解析里面为什么又要先构造5个var 然后再用dummy变成四个?

2 个答案

袁园_品职助教 · 2022年01月04日

嗨,努力学习的PZer你好:


这五个自变量的模型不是错误的,但是不符合F检验的标准形式,所以我们用4个哑变量进行转换的。

  1. 题目中说的检验方程一开始是这种:

但是我们在F检验中使用的标准形式应该是有截距项的式子。所以在这里我们对哑变量进行变形(变形后这两个方程的R2是一样的,因为方程的整体解释力度是不变的)。第二步我们做F检验就可以用第二个方程式来做了。这里就可以用F统计量的公式来计算了

2.u=β1,是五个自变量方程和四个自变量方程之间的对比。四个自变量的截距项u就是等于五个自变量模型中D1=1的时候的β1。

3.另外,补充一个点:伍德里奇的《计量经济学导论》中提到了“虚拟变量陷阱”,称在包含截距项时,两个特征要引入1个虚拟变量,如果引入2个(就是k-1个哑变量),就是完全多重共线。但在不包含截距项时,引入两个虚拟变量(像我们题目这种的5个特征都引入了虚拟变量)则不会存在虚拟变量陷阱。所以我们这道题最初建的模型不是错误的。

4.这道题那个5个变量时个迷惑选项,很多同学,直接就会把K=5带进去,但是这个模型本身就不符合标准的F检验的模型,所以要做一个变换。这个变换又结合了哑变量的性质,所以这道题比较难(直接可以说都超纲了),难在我们要识别出k=4。但是也不用特别纠结,考试中这个难度的题目很少的,大部分都是常规的题目。



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虽然现在很辛苦,但努力过的感觉真的很好,加油!

袁园_品职助教 · 2020年04月09日

同学你好!

题目意思:有一个回归模型,计算 1977年到 2017年 标普500每天的数值。已知 R2 = 0.000599,问标普500 的数值和这一天是星期几有没有关系

因为一周五天工作日,所以周一到周五有五个自变量,但是根据 dummy variable 的性质,只需要假设 n-1 即 4个变量就够了,所以有了题目中从五个到四个的变形。

后面就是直接用F检验带公式计算。

 

he123456 · 2021年12月30日

题目中原本的五个自变量相当于是构建的模型是错的么?然后才要按照4个哑变量构建?然后比较?u=β1是为什么 ​

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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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2023-02-05 02:23 1 · 回答

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