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小王爱学习 · 2022年01月17日

不太理解题目意思,以及答案为什么从BetaD1变成了SD1

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. 题目意思,考什么?
  2. 答案看不懂
  3. 答案里的公式怎么和R结合着用,这里怎么看出来哪个是ESS哪个是RSS?
1 个答案

李坏_品职助教 · 2022年01月17日

嗨,从没放弃的小努力你好:


  1. 题干说的是,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。(就是问你从周一到周五,是否SP500的收益率均值也不一样?)
  2. 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。 可以参考讲义P221页。
  3. 这题用的不是RSS那个公式,用的是P213的PPT下面的那个用R方的公式,Ru(无限制模型的R^2)都等于0.00059了,那Rr基本就等于0了(Rr是有限制模型,也就是只包括一个常数项的, R^2可以认为是0)

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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

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