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claireteng · 2022年06月28日

超纲吗

NO.PZ2020011101000025

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 $ln RGDP_t = -18.15 + .0136 t + \widehat\epsilon_t$ , and the estimated standard deviation of $\epsilon_t$ 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 $RGDP_t = -234178.8 + 121.3 * t + \widehat\epsilon_t$ ?

选项：

解释：

In this case,

$E_T[Y_T]=exp(E_T[ln Y_T]+\sigma^2/2)$

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 $E [ln RGDP_t]$ :

$E[ln RGDP_{2019}] = -18.15 + 0.0136 * 2019 = 9.308$

$E[ln RGDP_{2020}] = -18.15 + 0.0136 * 2020 = 9.322$

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

Furthermore,

$\sigma^2/2=0.0322^2/2=0.0005$

(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:

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

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

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

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

请问超纲吗

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1 个答案

DD仔_品职助教 · 2022年06月29日

嗨，努力学习的PZer你好：

不属于重点内容，稍微了解一下即可。

我们的题库是有关FRM的所有题型，FRM考试和CFA不同，他的特点就是天马行空，就算已经被移出考纲的题目也很有可能会再次出现，也就意味着超纲题目有可能出现在考试中。

所以题库题目会有一些觉得陌生的题型，稍微了解一下即可，不作为重点，为了能够让同学多见一些题型，这些题目我们都会保留在题库里。

重点和易考题型请参考经典题。

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就算太阳没有迎着我们而来，我们正在朝着它而去，加油！

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