问题如下图：
您好，请问算95%confidence interval的时候为什么用forecast*0.939而不是+/-1.96standard deviation呢？这个0.939又是怎么算出来的呢？谢谢
orange品职答疑助手 · 2020年02月06日
同学你好，0.939是下面这个式子的左边界，它是由1.96*0.0322得来的。
Bounds_Multiplier = exp(1.96 * 0.0322) = exp(± 0.0631) = 0.939,1.065
因为残差项是个随机变量，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）。之所以有置信区间，是由残差项这一个随机变量带来的。而这个残差项随机变量，又在指数上。所以左边端点是，E[RGDP2019]*e^(-1.96*标准差0.0322)
薛真 · 2020年03月01日
为啥后者不是随机变量
orange品职答疑助手 · 2020年03月02日
因为它是和t相乘的，t不是随机变量没有随机性
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
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?
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
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？