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Suppose we have a data set and run a regression to obtain the form $$\hat{y_i}=\hat{\alpha}+\hat{\beta}x_i+\hat{\gamma}z_i$$

Suppose $x_i=3,z_i=5$ we can calculate the fitted value $\hat{y_i}$ assume to be $10$ given $\hat{\alpha},\hat{\beta},\hat{\gamma}$ is known.

However, in Eview, the programme calculate the fitted value by regress again in the form of $$\hat{y_i}^*=\hat{\alpha}^*+\hat{\beta}^*(x_i-3)+\hat{\gamma}^*(z_i-5)$$ then $\hat{\alpha}^*$ will be the fitted value $10$ as the fitted value of first equation.

I was wabbling about the intuition behind this, I feel it's right but I am surprised I cannot tell why this basic algebra, what guarantee such happens?

Appreciate any comments

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  • $\begingroup$ Could you please attach any relevant screenshots or code parts? $\endgroup$
    – Spätzle
    Oct 21, 2021 at 8:18

1 Answer 1

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Simply plug in $x_i = 3$ and $z_i = 5$ in the new regression. You will see that $\hat y_i^\ast = \hat\alpha^\ast$. The new regression simply shifts the data. The idea of this approach is that any statistical software will report standard errors for your estimate of $\alpha$, so you can obtain standard errors of the prediction.

Note that in the original regression we have that $\hat y_i = \hat\alpha + 3\hat\beta + 5\hat\gamma$. As the data is shifted, we have in the new regression (for $x,z=0$) that $\hat y_i^\ast = \hat\alpha^\ast - 3 \hat\beta^\ast - 5\hat\gamma^\ast$.

Let $\mathbf A$ be given by $$\mathbf A = \begin{bmatrix} 1 & -3 & -5 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$ When multiplied from the left with another matrix, $\mathbf A$ multiplies the first column of the other matrix with 3 and 5 and subtracts it from the second and third column, respectively. Note that $\mathbf A^{-1}\mathbf A = \mathbf I_3$ as $$\mathbf A^{-1} = \begin{bmatrix} 1 & 3 & 5 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$ Let $\mathbf X$ be the matrix of explanatory variables and $\mathbf y$ the vector of dependent variables, i.e. $$ \mathbf X = \begin{bmatrix} 1 & x_1 & z_1 \\ 1 & x_2 & z_2 \\ \vdots & \vdots & \vdots \\ 1 & x_n & z_n \end{bmatrix}\qquad\text{and}\qquad \mathbf y = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n\end{bmatrix}.$$ The OLSE in the first regression is given by $$\boldsymbol{\hat\theta} =(\mathbf X'\mathbf X)^{-1}(\mathbf X'\mathbf y).$$ Let $\hat y = ((\mathbf A^{-1})'\mathbf e_1)\cdot\boldsymbol{\hat\theta}$, where $\mathbf e_1$ is the first canonical basis vector in $\mathbb R^3$, i.e. $\mathbf e_1 = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}'$. Thus, $(\mathbf A^{-1})'\mathbf e_1$ is first row of $\mathbf A^{-1}$ (as column vector). Recall that we constructed $\mathbf A$ according to the values for which we want the prediction. The OLSE in the second model is given by $$\begin{align*}\boldsymbol{\hat\vartheta} &= ((\mathbf X\mathbf A)'(\mathbf X\mathbf A))^{-1}(\mathbf X\mathbf A)'\mathbf y \\&= \mathbf A^{-1}(\mathbf X'\mathbf X)^{-1}(\mathbf A')^{-1}\mathbf A'\mathbf X'\mathbf y \\&= \mathbf A^{-1}(\mathbf X'\mathbf X)^{-1}(\mathbf X'\mathbf y) \\&= \mathbf A^{-1}\boldsymbol{\hat\theta}.\end{align*}$$ Now observe that $$\hat y = (\mathbf A^{-1})'\mathbf e_1\cdot\boldsymbol{\hat\theta} = \mathbf e_1'\mathbf A^{-1}\boldsymbol{\hat\theta} = \mathbf e_1'\boldsymbol{\hat\vartheta}.$$ Recall that, by definition, the first element of $\boldsymbol{\hat\vartheta}$ is $\hat\alpha^\ast$ and that $\mathbf e_1$ extracts the first element of this vector.

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