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Ethan橙 · 2023年10月13日

残差项不用考虑么?

NO.PZ2020011101000026

问题如下:

The seasonal dummy model Yt=δ+j=13γjljt+ϵtY_t = \delta + \sum_{j=1}^3\gamma_jl_{jt}+\epsilon_t is estimated on the quarterly growth rate of housing starts, and the estimated parameters are γ^1=6.23,γ^2=56.77,γ^3=10.61\widehat\gamma_1 = 6.23, \widehat\gamma_2 = 56.77, \widehat\gamma_3 = 10.61, and δ^=15.79\widehat\delta = -15.79 using data until the end of 2018. What are the forecast growth rates for the four quarters of 2019?

选项:

解释:

Though there is variance from quarter-to-quarter, the expected value of YtY_t is the same for any two observations of the same quarter, regardless of the year.

Accordingly:

E[YQ1]=δ+j=13γjljt=15.79+6.231+56.770+10.610=9.56E[Y_{Q1}]=\delta + \sum_{j=1}^3\gamma_jl_{jt}=-15.79 + 6.23 * 1 + 56.77 * 0+ 10.61 * 0 = -9.56

E[YQ2]=δ+j=13γjljt=15.79+6.230+56.771+10.610=40.98E[Y_{Q2}]=\delta + \sum_{j=1}^3\gamma_jl_{jt}=-15.79 + 6.23 * 0 + 56.77 * 1+ 10.61 * 0 = 40.98

E[YQ3]=δ+j=13γjljt=15.79+6.230+56.770+10.611=5.18E[Y_{Q3}]=\delta + \sum_{j=1}^3\gamma_jl_{jt}=-15.79 + 6.23 * 0 + 56.77 * 0+ 10.61 * 1 = -5.18

E[YQ4]=δ+j=13γjljt=15.79+6.230+56.770+10.610=15.79E[Y_{Q4}]=\delta + \sum_{j=1}^3\gamma_jl_{jt}=-15.79 + 6.23 * 0 + 56.77 * 0+ 10.61 * 0 = -15.79

残差项不用考虑么?

1 个答案

品职答疑小助手雍 · 2023年10月15日

同学你好,残差项均值为0,通常在回归方程的预测里不考虑的~

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NO.PZ2020011101000026 问题如下 The seasonmmy mol Yt=δ+∑j=13γjljt+ϵtY_t = \lta + \sum_{j=1}^3\gamma_jl_{jt}+\epsilon_tYt​=δ+∑j=13​γj​ljt​+ϵt​ is estimateon the quarterly growth rate of housing starts, anthe estimateparameters are γ^1=6.23,γ^2=56.77,γ^3=10.61\wihat\gamma_1 = 6.23, \wihat\gamma_2 = 56.77, \wihat\gamma_3 = 10.61γ​1​=6.23,γ​2​=56.77,γ​3​=10.61, anδ^=−15.79\wihat\lta = -15.79δ=−15.79 using ta until the enof 2018. Whare the forecast growth rates for the four quarters of 2019? Though there is varianfrom quarter-to-quarter, the expectevalue of YtY_tYt​ is the same for any two observations of the same quarter, regaress of the year. Accorngly:E[YQ1]=δ+∑j=13γjljt=−15.79+6.23∗1+56.77∗0+10.61∗0=−9.56E[Y_{Q1}]=\lta + \sum_{j=1}^3\gamma_jl_{jt}=-15.79 + 6.23 * 1 + 56.77 * 0+ 10.61 * 0 = -9.56E[YQ1​]=δ+∑j=13​γj​ljt​=−15.79+6.23∗1+56.77∗0+10.61∗0=−9.56E[YQ2]=δ+∑j=13γjljt=−15.79+6.23∗0+56.77∗1+10.61∗0=40.98E[Y_{Q2}]=\lta + \sum_{j=1}^3\gamma_jl_{jt}=-15.79 + 6.23 * 0 + 56.77 * 1+ 10.61 * 0 = 40.98E[YQ2​]=δ+∑j=13​γj​ljt​=−15.79+6.23∗0+56.77∗1+10.61∗0=40.98E[YQ3]=δ+∑j=13γjljt=−15.79+6.23∗0+56.77∗0+10.61∗1=−5.18E[Y_{Q3}]=\lta + \sum_{j=1}^3\gamma_jl_{jt}=-15.79 + 6.23 * 0 + 56.77 * 0+ 10.61 * 1 = -5.18E[YQ3​]=δ+∑j=13​γj​ljt​=−15.79+6.23∗0+56.77∗0+10.61∗1=−5.18E[YQ4]=δ+∑j=13γjljt=−15.79+6.23∗0+56.77∗0+10.61∗0=−15.79E[Y_{Q4}]=\lta + \sum_{j=1}^3\gamma_jl_{jt}=-15.79 + 6.23 * 0 + 56.77 * 0+ 10.61 * 0 = -15.79E[YQ4​]=δ+∑j=13​γj​ljt​=−15.79+6.23∗0+56.77∗0+10.61∗0=−15.79

2022-07-28 21:02 1 · 回答