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Drink H · 2020年02月03日

问一道题：NO.PZ2020011101000025

  1. FRM I
  2. Quantitative

问题如下：

A log-linear trend model is estimated on annual euro-area GDP using data from 1995 until 2018. The estimated model is lnRGDPt=−18.15+.0136t+ϵ^tln RGDP_t = -18.15 + .0136 t + \widehat\epsilon_t, and the estimated standard deviation of ϵt\epsilon_t is 0.0322. Assuming the shocks are normally distributed, what are the point forecasts of GDP for the next three years? How do these compare with a linear model RGDPt=−234178.8+121.3∗t+ϵ^tRGDP_t = -234178.8 + 121.3 * t + \widehat\epsilon_t?

选项：

解释：

In this case,

ET[YT]=exp(ET[lnYT]+σ2/2)E_T[Y_T]=exp(E_T[ln Y_T]+\sigma^2/2)

And the error bounds on the ln are +/-1.96*0.0322, so the bounds are given in proportional terms rather than fixed values.

Bounds_Multiplier = exp({1.96 * 0.0322) = exp(\pm 0.0631) = 0.939,1.065

Calculating E[lnRGDPt]E [ln RGDP_t]:

E[lnRGDP2019]=−18.15+0.0136∗2019=9.308E[ln RGDP_{2019}] = -18.15 + 0.0136 * 2019 = 9.308

E[lnRGDP2020]=−18.15+0.0136∗2020=9.322E[ln RGDP_{2020}] = -18.15 + 0.0136 * 2020 = 9.322

E[lnRGDP2021]=−18.15+0.0136∗2020=9.336E[ln RGDP_{2021}] = -18.15 + 0.0136 * 2020 = 9.336

Furthermore,

σ2/2=0.03222/2=0.0005\sigma^2/2=0.0322^2/2=0.0005

(which will only make a small impact in this example)

So:

E[RGDP2019] = exp(9.308 + 0.0005) = 11,031.4

E [RGDP2020] = exp(9.322 + 0.0005) = 11,186.9

E[RGDP2021] = exp(9.336 + 0.0005) = 11,344.6

And the 95% confidence bands are given as:

95%CBRGDP2019=[0.939∗11031.4,1.065∗11031.4]=[10,358.5,11,748.4]95\%_{CB_{RGDP2019}}= [0.939 * 11031.4,1.065 * 11031.4] = [10,358.5,11,748.4]

95%CBRGDP2020=[10,504.5,11,914.0]95\%_{CB_{RGDP2020}}= [10,504.5,11,914.0]

95%CBRGDP2019=[10,652.6,12,082.0]95\%_{CB_{RGDP2019}} = [10,652.6,12,082.0]

In comparison with the linera model, the bands are growing in size and overall the results are a bit bigger.

这题的答案中标准差的平方除以2，是在哪个地方的知识点？

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品职答疑小助手雍 · 2020年02月07日

同学你好，因为残差项是个随便变量，而前面的-18.5+0.136*t，其实不是随机变量，所以lnRGDPt可以看做服从 N(-18.15+0.136*t, 0.0322^2），所以RGDP的数学期望就等于

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NO.PZ2020011101000025问题如下A log-linetrenmol is estimateon annueuro-area G using ta from 1995 until 2018. The estimatemol is lnRGt=−18.15+.0136t+ϵ^tln RG_t = -18.15 + .0136 t + \wihat\epsilon_tlnRGt​=−18.15+.0136t+ϵt​, anthe estimatestanrviation of ϵt\epsilon_tϵt​ is 0.0322. Assuming the shocks are normally stribute whare the point forecasts of G for the next three years? How these compare with a linemol RGt=−234178.8+121.3∗t+ϵ^tRG_t = -234178.8 + 121.3 * t + \wihat\epsilon_tRGt​=−234178.8+121.3∗t+ϵt​? In this case, ET[YT]=exp(ET[lnYT]+σ2/2)E_T[Y_T]=exp(E_T[ln Y_T]+\sigma^2/2)ET​[YT​]=exp(ET​[lnYT​]+σ2/2)Anthe error boun on the ln are +/-1.96*0.0322, so the boun are given in proportionterms rather thfixevalues.Boun_Multiplier = exp({1.96 * 0.0322) = exp(\pm 0.0631) = 0.939,1.065Calculating E[lnRGt]E [ln RG_t]E[lnRGt​]:E[lnRG2019]=−18.15+0.0136∗2019=9.308E[ln RG_{2019}] = -18.15 + 0.0136 * 2019 = 9.308E[lnRG2019​]=−18.15+0.0136∗2019=9.308E[lnRG2020]=−18.15+0.0136∗2020=9.322E[ln RG_{2020}] = -18.15 + 0.0136 * 2020 = 9.322E[lnRG2020​]=−18.15+0.0136∗2020=9.322E[lnRG2021]=−18.15+0.0136∗2020=9.336E[ln RG_{2021}] = -18.15 + 0.0136 * 2020 = 9.336E[lnRG2021​]=−18.15+0.0136∗2020=9.336Furthermore,σ2/2=0.03222/2=0.0005\sigma^2/2=0.0322^2/2=0.0005σ2/2=0.03222/2=0.0005(whiwill only make a small impain this example)So:E[RG2019] = exp(9.308 + 0.0005) = 11,031.4E [RG2020] = exp(9.322 + 0.0005) = 11,186.9E[RG2021] = exp(9.336 + 0.0005) = 11,344.6Anthe 95% confinban are given as:95%CBRG2019=[0.939∗11031.4,1.065∗11031.4]=[10,358.5,11,748.4]95\%_{CB_{RG2019}}= [0.939 * 11031.4,1.065 * 11031.4] = [10,358.5,11,748.4]95%CBRG2019​​=[0.939∗11031.4,1.065∗11031.4]=[10,358.5,11,748.4]95%CBRG2020=[10,504.5,11,914.0]95\%_{CB_{RG2020}}= [10,504.5,11,914.0]95%CBRG2020​​=[10,504.5,11,914.0]95%CBRG2019=[10,652.6,12,082.0]95\%_{CB_{RG2019}} = [10,652.6,12,082.0]95%CBRG2019​​=[10,652.6,12,082.0]In comparison with the linera mol, the ban are growing in size anoverall the results are a bit bigger.1 解析红色的没看懂，怎么算出来的？2 这种类型的题是重点吗？这部分看到好几道类似的3 这部分讲1⃣️中在具体哪一块儿？视频中哪一部分的？4 ip的答疑板块非常不好用，提问经常性闪退，分屏的时候更加闪退，已经更新app到最新版本

2024-03-09 10:30 1 · 回答

NO.PZ2020011101000025 问题如下 A log-linetrenmol is estimateon annueuro-area G using ta from 1995 until 2018. The estimatemol is lnRGt=−18.15+.0136t+ϵ^tln RG_t = -18.15 + .0136 t + \wihat\epsilon_tlnRGt​=−18.15+.0136t+ϵt​, anthe estimatestanrviation of ϵt\epsilon_tϵt​ is 0.0322. Assuming the shocks are normally stribute whare the point forecasts of G for the next three years? How these compare with a linemol RGt=−234178.8+121.3∗t+ϵ^tRG_t = -234178.8 + 121.3 * t + \wihat\epsilon_tRGt​=−234178.8+121.3∗t+ϵt​? In this case, ET[YT]=exp(ET[lnYT]+σ2/2)E_T[Y_T]=exp(E_T[ln Y_T]+\sigma^2/2)ET​[YT​]=exp(ET​[lnYT​]+σ2/2)Anthe error boun on the ln are +/-1.96*0.0322, so the boun are given in proportionterms rather thfixevalues.Boun_Multiplier = exp({1.96 * 0.0322) = exp(\pm 0.0631) = 0.939,1.065Calculating E[lnRGt]E [ln RG_t]E[lnRGt​]:E[lnRG2019]=−18.15+0.0136∗2019=9.308E[ln RG_{2019}] = -18.15 + 0.0136 * 2019 = 9.308E[lnRG2019​]=−18.15+0.0136∗2019=9.308E[lnRG2020]=−18.15+0.0136∗2020=9.322E[ln RG_{2020}] = -18.15 + 0.0136 * 2020 = 9.322E[lnRG2020​]=−18.15+0.0136∗2020=9.322E[lnRG2021]=−18.15+0.0136∗2020=9.336E[ln RG_{2021}] = -18.15 + 0.0136 * 2020 = 9.336E[lnRG2021​]=−18.15+0.0136∗2020=9.336Furthermore,σ2/2=0.03222/2=0.0005\sigma^2/2=0.0322^2/2=0.0005σ2/2=0.03222/2=0.0005(whiwill only make a small impain this example)So:E[RG2019] = exp(9.308 + 0.0005) = 11,031.4E [RG2020] = exp(9.322 + 0.0005) = 11,186.9E[RG2021] = exp(9.336 + 0.0005) = 11,344.6Anthe 95% confinban are given as:95%CBRG2019=[0.939∗11031.4,1.065∗11031.4]=[10,358.5,11,748.4]95\%_{CB_{RG2019}}= [0.939 * 11031.4,1.065 * 11031.4] = [10,358.5,11,748.4]95%CBRG2019​​=[0.939∗11031.4,1.065∗11031.4]=[10,358.5,11,748.4]95%CBRG2020=[10,504.5,11,914.0]95\%_{CB_{RG2020}}= [10,504.5,11,914.0]95%CBRG2020​​=[10,504.5,11,914.0]95%CBRG2019=[10,652.6,12,082.0]95\%_{CB_{RG2019}} = [10,652.6,12,082.0]95%CBRG2019​​=[10,652.6,12,082.0]In comparison with the linera mol, the ban are growing in size anoverall the results are a bit bigger. 下面这步骤没看懂，Anthe 95% confinban are given as:95%������2019=[0.939∗11031.4,1.065∗11031.4]=[10,358.5,11,748.4]为什么是乘e+_0.0631?

2024-01-29 21:03 1 · 回答

NO.PZ2020011101000025 问题如下 A log-linetrenmol is estimateon annueuro-area G using ta from 1995 until 2018. The estimatemol is lnRGt=−18.15+.0136t+ϵ^tln RG_t = -18.15 + .0136 t + \wihat\epsilon_tlnRGt​=−18.15+.0136t+ϵt​, anthe estimatestanrviation of ϵt\epsilon_tϵt​ is 0.0322. Assuming the shocks are normally stribute whare the point forecasts of G for the next three years? How these compare with a linemol RGt=−234178.8+121.3∗t+ϵ^tRG_t = -234178.8 + 121.3 * t + \wihat\epsilon_tRGt​=−234178.8+121.3∗t+ϵt​? In this case, ET[YT]=exp(ET[lnYT]+σ2/2)E_T[Y_T]=exp(E_T[ln Y_T]+\sigma^2/2)ET​[YT​]=exp(ET​[lnYT​]+σ2/2)Anthe error boun on the ln are +/-1.96*0.0322, so the boun are given in proportionterms rather thfixevalues.Boun_Multiplier = exp({1.96 * 0.0322) = exp(\pm 0.0631) = 0.939,1.065Calculating E[lnRGt]E [ln RG_t]E[lnRGt​]:E[lnRG2019]=−18.15+0.0136∗2019=9.308E[ln RG_{2019}] = -18.15 + 0.0136 * 2019 = 9.308E[lnRG2019​]=−18.15+0.0136∗2019=9.308E[lnRG2020]=−18.15+0.0136∗2020=9.322E[ln RG_{2020}] = -18.15 + 0.0136 * 2020 = 9.322E[lnRG2020​]=−18.15+0.0136∗2020=9.322E[lnRG2021]=−18.15+0.0136∗2020=9.336E[ln RG_{2021}] = -18.15 + 0.0136 * 2020 = 9.336E[lnRG2021​]=−18.15+0.0136∗2020=9.336Furthermore,σ2/2=0.03222/2=0.0005\sigma^2/2=0.0322^2/2=0.0005σ2/2=0.03222/2=0.0005(whiwill only make a small impain this example)So:E[RG2019] = exp(9.308 + 0.0005) = 11,031.4E [RG2020] = exp(9.322 + 0.0005) = 11,186.9E[RG2021] = exp(9.336 + 0.0005) = 11,344.6Anthe 95% confinban are given as:95%CBRG2019=[0.939∗11031.4,1.065∗11031.4]=[10,358.5,11,748.4]95\%_{CB_{RG2019}}= [0.939 * 11031.4,1.065 * 11031.4] = [10,358.5,11,748.4]95%CBRG2019​​=[0.939∗11031.4,1.065∗11031.4]=[10,358.5,11,748.4]95%CBRG2020=[10,504.5,11,914.0]95\%_{CB_{RG2020}}= [10,504.5,11,914.0]95%CBRG2020​​=[10,504.5,11,914.0]95%CBRG2019=[10,652.6,12,082.0]95\%_{CB_{RG2019}} = [10,652.6,12,082.0]95%CBRG2019​​=[10,652.6,12,082.0]In comparison with the linera mol, the ban are growing in size anoverall the results are a bit bigger. 请问超纲吗

2022-06-28 16:04 1 · 回答

您好，请问算95%confininterval的时候为什么用forecast*0.939而不是+/-1.96stanrviation呢？这个0.939又是怎么算出来的呢？谢谢

2020-02-05 19:44 1 · 回答