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seven-zhu · 2021年01月31日

expected-value

  1. FRM I
  2. Quantitative

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

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

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

这道题95%的置信区间那里，expected-value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

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小刘_品职助教 · 2021年02月01日

同学你好，

这个the 95% confidence interval is + / - 1.96*262.8 = + / - 515.1 是指残差的，对于残差来说期望是0，所以直接是正负critical-value*t，这个相当于给出了最后回归的波动范围。

答案里的AR或MA模型这句话可以直接无视，只是告诉你没有用这个方法，解题用不到这个条件。

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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

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

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