关于Sf的解释

问题如下：

Doug Abitbol is a portfolio manager for Polyi Investments, a hedge fund that trades in the United States. Abitbol manages the hedge fund with the help of Robert Olabudo, junior portfolio manager.

Abitbol looks at economists' inflation forecasts and would like to examine the relationship between the US Consumer Price Index (US CPl) consensus forecast and the actual US CPI using regression analysis. Olabudo estimates regression coefficients to test whether the consensus forecast is unbiased. If the consensus forecasts are unbiased, the intercept should be 0.0 and the slope will be equal to 1.0. Regression results are presented in Exhibit 1. Additionally, Olabudo calculates the 95% prediction interval of the actual CPI using a US CPI consensus forecast of 2.8.

Finally, Abitbol and Olabudo discuss the forecast and forecast interval:

Observation 1 For a given confidence level, the forecast interval is the same no matter the US CPI consensus forecast.

Observation 2 A larger standard error of the estimate will result in a wider confidence interval.

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:

${\rm{\hat Y=}}{{\rm{b}}_{\rm{0}}}{\rm{+}}{{\rm{b}}_{\rm{1}}}{\rm{X = 0}}{\rm{.0001+(0}}{\rm{.9830}}\times{\rm{2}}{\rm{.8)=2}}{\rm{.7525}}$

Step 2. Compute the variance of the prediction error:

$s_f^2=s_e^2\left\{{1+(1/n)+[{{({X_f}-\bar X)}^2}]/[(n-1)\times s_x^2]}\right\}$

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

$s_f^2=0.00000088$

${s_f}=0.0009$

Step 3. Compute the prediction interval:

$\hat Y\pm{t_c}\times{s_f}$

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.