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上小学 · 2023年02月05日

这个题目完全看不懂，请详细解释下步骤和原理、考试的知识点。谢谢

  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?

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

(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.

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

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1 个答案

DD仔_品职助教 · 2023年02月05日

嗨，努力学习的PZer你好：


同学你好，

1，means在题目里指的是SP500的均值收益率，也就是我们模型的因变量。R^2是我们在构建模型时，来判断模型自变量的确定是否能够很好的解释因变量，在本题中也就是这个means。R^2越高，自变量解释因变量的程度就越高，模型就越好。


2，思路就是先读题，然后根据题目可以判断出考察的是关于哑变量，以及构建模型的知识点。

题目问：利用SP500指数从1977到2017年的收益率数据，用一个含有虚拟变量的模型对收益率进行线性回归分析，R^2是0.00059，问你是否存在weekday效应（也就是问你从周一到周五，是否SP500的收益率均值也不一样？）

答案先给出一个模型，是不带常数项的，直接用5个虚拟变量（dummy）去建模。如果当前样本数据是周一，那么D1=1，其他的D等于0，如果当前是周二，那么D2=1，其他的D等于0…… 原假设H0是：不存在weekday效应，那么所有的dummy前面的系数都等于0，意思是SP500的收益率和周几没关系。 之后给出第二个模型，是带有常数项μ的。这样只需要4个虚拟变量就可以表示周一到周五了。因为可以把μ看作周一的收益率。周二就是δ2=1，其他的δ等于0。


这个题难度比较高，关于哑变量这里的确定不在我们考纲范围内了。本题主要掌握检验几个变量应该用F检验，并记住F检验的公式就够了。

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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！

Brian邵彬 · 2024年05月21日

请问采用第二个模型的原因是？逻辑是什么？

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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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2023-09-05 17:44 1 · 回答

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2022-03-21 15:55 1 · 回答