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FrankSun · 2022年01月29日

老师，可以帮忙讲一下第二步骤吗？，没看懂答案

  1. CFA II
  2. Quantitative

* 问题详情，请 查看题干

NO.PZ202109130200004502

问题如下：

Based on Exhibit 1, Olabudo should calculate a prediction interval for the actual US CPI closest to:

选项：

A.2.7506 to 2.7544 B.2.7521 to 2.7529 C.2.7981 to 2.8019

解释：

A is correct. The forecast interval for inflation is calculated in three steps:

Step 1. Make the prediction given the US CPI forecast of 2.8:

Y^=b0+b1X=0.0001+(0.9830×2.8)=2.7525{\rm{\hat Y=}}{{\rm{b}}_{\rm{0}}}{\rm{+}}{{\rm{b}}_{\rm{1}}}{\rm{X = 0}}{\rm{.0001+(0}}{\rm{.9830}}\times{\rm{2}}{\rm{.8)=2}}{\rm{.7525}}

Step 2. Compute the variance of the prediction error:

sf2=se2{1+(1/n)+[(Xf−Xˉ)2]/[(n−1)×sx2]}s_f^2=s_e^2\left\{{1+(1/n)+[{{({X_f}-\bar X)}^2}]/[(n-1)\times s_x^2]}\right\}

sf2=0.0009e2{1+(1/60)+[(2.8−1.3350)2]/[(60−1)×0.75392]}s_f^2=0.0009_e^2\left\{{1+(1/60)+[{{(2.8-1.3350)}^2}]/[(60-1)\times{{0.7539}^2}]}\right\}

sf2=0.00000088s_f^2=0.00000088

sf=0.0009{s_f}=0.0009

Step 3. Compute the prediction interval:

Y^±tc×sf\hat Y\pm{t_c}\times{s_f}

2.7525±(2.0 x 0.0009)

Lower bound: 2.7525 - (2.0 x 0.0009) = 2.7506.

Upper bound: 2.7525 + (2.0 x 0.0009) = 2.7544.

So, given the US CPI forecast of 2.8, the 95% prediction interval is 2.7506 to 2.7544.

老师，可以帮忙讲一下第二步骤吗？，没看懂答案

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1 个答案

星星_品职助教 · 2022年01月30日

同学你好，

第二步为完全的套公式计算。所有的数字题干中都给了，对号入座即可。

公式可以参见讲义上的这个部分，需要记忆一下。

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NO.PZ202109130200004502问题如下 ug Abitbol is a portfolio manager for Polyi Investments, a hee funthtras in the UniteStates. Abitbol manages the hee funwith the help of Robert Olabu, junior portfolio manager. Abitbol looks economists' inflation forecasts anwoullike to examine the relationship between the US Consumer PriInx (US CPl) consensus forecast anthe actuUS CPI using regression analysis. Olabu estimates regression coefficients to test whether the consensus forecast is unbiase If the consensus forecasts are unbiase the intercept shoul0.0 anthe slope will equto 1.0. Regression results are presentein Exhibit 1. Aitionally, Olabu calculates the 95% prection intervof the actuCPI using a US CPI consensus forecast of 2.8. Finally, Abitbol anOlabu scuss the forecast anforecast interval: Observation 1 For a given confinlevel, the forecast intervis the same no matter the US CPI consensus forecast. Observation 2 A larger stanrerror of the estimate will result in a wir confininterval. Baseon Exhibit 1, Olabu shoulcalculate a prection intervfor the actuUS CPI closest to: A.2.7506 to 2.7544 B.2.7521 to 2.7529 C.2.7981 to 2.8019 A is correct. The forecast intervfor inflation is calculatein three steps:Step 1. Make the prection given the US CPI forecast of 2.8:Y^=b0+b1X=0.0001+(0.9830×2.8)=2.7525{\rm{\hY=}}{{\rm{b}}_{\rm{0}}}{\rm{+}}{{\rm{b}}_{\rm{1}}}{\rm{X = 0}}{\rm{.0001+(0}}{\rm{.9830}}\times{\rm{2}}{\rm{.8)=2}}{\rm{.7525}}Y^=b0​+b1​X=0.0001+(0.9830×2.8)=2.7525 Step 2. Compute the varianof the prection error:sf2=se2{1+(1/n)+[(Xf−Xˉ)2]/[(n−1)×sx2]}s_f^2=s_e^2\left\{{1+(1/n)+[{{({X_f}-\bX)}^2}]/[(n-1)\times s_x^2]}\right\}sf2​=se2​{1+(1/n)+[(Xf​−Xˉ)2]/[(n−1)×sx2​]}sf2=0.0009e2{1+(1/60)+[(2.8−1.3350)2]/[(60−1)×0.75392]}s_f^2=0.0009_e^2\left\{{1+(1/60)+[{{(2.8-1.3350)}^2}]/[(60-1)\times{{0.7539}^2}]}\right\}sf2​=0.0009e2​{1+(1/60)+[(2.8−1.3350)2]/[(60−1)×0.75392]}sf2=0.00000088s_f^2=0.00000088sf2​=0.00000088sf=0.0009{s_f}=0.0009sf​=0.0009 Step 3. Compute the prection interval:Y^±tc×sf\hY\pm{t_c}\times{s_f}Y^±tc​×sf​ 2.7525±(2.0 x 0.0009)Lower boun 2.7525 - (2.0 x 0.0009) = 2.7506.Upper boun 2.7525 + (2.0 x 0.0009) = 2.7544.So, given the US CPI forecast of 2.8, the 95% prection intervis 2.7506 to 2.7544. 老师，关于公式里的Sf，我想再确认一下。这个Sf，是prection error，理应用这个很复杂的公式计算出来的。但是当n特别大的时候，就会是约等于stanrerror of estimate。所以在这道题目中，用到了复杂的公式进行计算，但是其实题目中也给出了stanrerror of estimate为0.0009，其实和step 2计算出来是一样的。然后题目里面的criticvalue，应该用的是2.002吧？解答里计算的时候就直接用2.0了是嘛？

2022-03-23 22:20 1 · 回答