# Instead of plug in the $x_i$ to find fitted value, we can regress anther form of regression to get the fitted value

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?

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.