开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)|3.2.5(安卓)APP下载

薛真 · 2020年03月01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

答案给的不对吧

1 个答案

orange品职答疑助手 · 2020年03月03日

同学你好,你指的答案不对是指哪一个?是置信区间吗?

置信区间我觉得是应该加上数学期望、不应该单独是正负1.96*262.8,它答案±515.1 后面的about the expected value,连起来的意思应该就是期望±515.1 的意思

  • 1

    回答
  • 0

    关注
  • 396

    浏览
相关问题

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

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

2020-02-03 23:31 2 · 回答