问一道题：NO.PZ202001080100003903 第3小题 [ FRM I ]

NO.PZ202001080100003903 问题如下 Using the results from part a anpart anthe correlation between the two explanatory variables, estimate the parameters for the full mol: Y=α+β1X1+β2X2Y=\alpha+\beta_1X_1+\beta_2X_2Y=α+β1​X1​+β2​X2​ The key statement from the chapter is “the estimator β^1\wihat\beta_1β​1​ converges to β1+β2δ\beta_1 + \beta_2\ltaβ1​+β2​δ”So, part a gives that:β1+β2δ1=0\beta_1 + \beta_2\lta_1 = 0β1​+β2​δ1​=0Anpart b gives that:β2+β1δ2=−1.479\beta_2 + \beta_1\lta_2 = -1.479β2​+β1​δ2​=−1.479δ1=Cov[X1,X2]V[X1]=0.467/0.933=0.501\lta_1=\frac{Cov[X_1,X_2]}{V[X_1]}=0.467/0.933=0.501δ1​=V[X1​]Cov[X1​,X2​]​=0.467/0.933=0.501anδ2=0.467/0.934=0.5\lta_2=0.467/0.934=0.5δ2​=0.467/0.934=0.5Note: In this case these quantities are essentially equal, but thwill not usually the case.Plugging bain the 2 * 2 system of equations: β1+0.501β2=0\beta_1 + 0.501\beta_2 = 0β1​+0.501β2​=0β2+0.5β1=−1.479\beta_2 + 0.5\beta_1 = -1.479β2​+0.5β1​=−1.479Solving yiel thatβ1=0.989,β2=−1.9733\beta_1=0.989, \beta_2=-1.9733β1​=0.989,β2​=−1.9733Anplugging these bainto the equation to finalpha:Y‾=α+β1X‾1+β2X‾2=−2\overline Y=\alpha+\beta_1\overline X_1+\beta_2\overline X_2=-2Y=α+β1​X1​+β2​X2​=−2Y=−2+0.989X1−1.9733X2Y=-2+0.989X_1-1.9733X_2Y=−2+0.989X1​−1.9733X2​ 请问下面这道题的解法为什么和本题的不一样？