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Dinny · 2020年08月15日

问一道题:NO.PZ201709270100000502

* 问题详情,请 查看题干

问题如下:

2. Based on the regression output in Exhibit 1, the first-differenced series used to run Regression 2 is consistent with:

选项:

A.

a random walk.

B.

covariance stationarity.

C.

a random walk with drift.

解释:

B is correct. The critical t-statistic at a 5% confidence level is 1.98. As a result, neither the intercept nor the coefficient on the first lag of the first-differenced exchange rate in Regression 2 differs significantly from zero. Also, the residual autocorrelations do not differ significantly from zero. As a result, Regression 2 can be reduced to yt = εt with a mean-reverting level of b0/(1 b1) = 0/1 = 0.

Therefore, the variance of yt in each period is Var(εt) = σ2. The fact that the residuals are not autocorrelated is consistent with the covariance of the times series, with itself being constant and finite at different lags. Because the variance and the mean of yt are constant and finite in each period, we can also conclude that yt is covariance stationary.

老师好,题目的逻辑,在First-Differenced AR(1) Model之后,两个系数neither the intercept nor the coefficient on the first lag of the first-differenced exchange rate in Regression 2 differs significantly from zero,然后就说他是协方差平稳的。可是,应该是 coefficient on the first lag of the first-differenced exchange rate in Regression 2 不等于1 才能断定协方差平稳吧? 感觉那么说不太严谨啊!

1 个答案

星星_品职助教 · 2020年08月15日

同学你好,

题干描述“neither the intercept nor the coefficient on the first lag of the first-differenced exchange rate in Regression 2 differs significantly from zero”,这句话的意思是方程 yt=b0+b1yt-1+εt的系数假设检验结果为:b0=0,b1=0。

所以这个时候的b1实际上是等于0的,所以从这个角度说明了b1≠1。

但仅仅是b1≠1还不够说明协方差平稳。

这道题能得出协方差平稳的结论是因为根据假设检验的结果,Regression 2 实际就简化为了yt=εt。而这个简化后的方程是满足协方差平稳的三个条件的。

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