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Drink H · 2020年02月03日

问一道题:NO.PZ2020011101000024

问题如下:

A linear time trend model is estimated on annual real euro-area GDP, measured in billions of 2010 euros, using data from 1995 until 2018. The estimated model is RGDPt=234178.8+121.3t+ϵ^tRGDP_t = -234178.8 + 121.3 * t + \widehat\epsilon_t. The estimate of the residual standard deviation is σ^=262.8\widehat\sigma = 262.8.

Construct point forecasts and 95% confidence intervals (assuming Gaussian white noise errors) for the next three years. Note that t is the year, so that in the first observation, t = 1995, and in the last, t = 2018.

选项:

解释:

Note that there is no AR or MA component, so the variance remains constant. Therefore, the 95% confidence interval is + / - 1.96*262.8 = + / - 515.1 about the expected value.

As for the expected means:

E[RGDP2019]=234178.8+121.32019=10,725.9E[RGDP_{2019}] = -234178.8 + 121.3 * 2019 = 10,725.9

E[RGDP2020]=234178.8+121.32020=10,847.2E[RGDP_{2020}] = -234178.8 + 121.3 * 2020 = 10,847.2

E[RGDP2021]=234178.8+121.32021=10,968.5E[RGDP_{2021}] = -234178.8 + 121.3 * 2021 = 10,968.5

AR的mean、variance、autocovariances不都是constant的吗?为什么答案中“Note that there is no AR or MA component, so the variance remains constant”这句话no AR,so the variance remains constant?

2 个答案

李坏_品职助教 · 2021年09月24日

嗨,努力学习的PZer你好:


AR模型使用的前提是时间序列数据必须是平稳的,白噪声是一种特殊的平稳序列。


我理解前面助教的意思是:必须是残差服从白噪声才可以用AR,但是题目里没有使用AR,所以因变量的方差不受残差的影响,所以varicance 是constant的。



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努力的时光都是限量版,加油!

品职答疑小助手雍 · 2020年02月07日

同学你好,这题解析的意思是这个回归没有用AR模型进行自回归,因为要记得课上讲的AR模型的假设,第二条就是残差项服从白噪声(这是AR成立的前提,但不意味着用了AR它就成立)。所以我们不能保证用了AR模型就一定残差项是服从白噪声的。

所以这题解析就先加了一句,这里没用AR模型。

nanaluo · 2021年09月23日

请问为什么用了AR但不一定残差项服从白噪声?不是基于残差项服从白噪声才可以使用AR吗?

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NO.PZ2020011101000024问题如下A linetime trenmol is estimateon annureeuro-area G, measurein billions of 2010 euros, using ta from 1995 until 2018. The estimatemol is RGt=−234178.8+121.3∗t+ϵ^tRG_t = -234178.8 + 121.3 * t + \wihat\epsilon_tRGt​=−234178.8+121.3∗t+ϵt​. The estimate of the resistanrviation is σ^=262.8\wihat\sigma = 262.8σ=262.8. Construpoint forecasts an95% confinintervals (assuming Gaussiwhite noise errors) for the next three years. Note tht is the year, so thin the first observation, t = 1995, anin the last, t = 2018.Note ththere is no or MA component, so the varianremains constant. Therefore, the 95% confinintervis + / - 1.96*262.8 = + / - 515.1 about the expectevalue.for the expectemeans:E[RG2019]=−234178.8+121.3∗2019=10,725.9E[RG_{2019}] = -234178.8 + 121.3 * 2019 = 10,725.9E[RG2019​]=−234178.8+121.3∗2019=10,725.9E[RG2020]=−234178.8+121.3∗2020=10,847.2E[RG_{2020}] = -234178.8 + 121.3 * 2020 = 10,847.2E[RG2020​]=−234178.8+121.3∗2020=10,847.2E[RG2021]=−234178.8+121.3∗2021=10,968.5E[RG_{2021}] = -234178.8 + 121.3 * 2021 = 10,968.5E[RG2021​]=−234178.8+121.3∗2021=10,968.51、所以方差和95%的条件根本用不上?2、看了往期的,还是不太明白为何不用这个条件

2024-05-07 21:53 4 · 回答

NO.PZ2020011101000024 这道题95%的置信区间那里,expectevalue是等于0吗,为什么呢,答案中说不是用的AR或者MA模型呢,算CI的时候好像直接就是正负critical-value*t了

2021-01-31 20:46 1 · 回答

为什么不使用残差的标准误而是标准差去做点估计

2020-05-31 07:01 4 · 回答

答案给的不对吧

2020-03-01 23:37 1 · 回答