# Multilinear regression relation to univariate regression

I read the following from a text book

1. Initiate $$z_0 = x_0 = 1$$

2. For $$j = 1, \dots, p$$:

Regress $$x_j$$ on $$z_0, \dots, z_{j-1}$$ to produce coefficients $$\hat{\gamma}_{kj} = \langle z_l, xj\rangle / \langle z_l, z_l \rangle$$ and residual vector $$z_j = x_j - \sum\limits_{k=0}^{j-1} \hat{\gamma}_{kj}z_{k}$$

3. Regress $$y$$ on the residual $$z_p$$ to give the estimate $$\hat{\beta}_p$$.

The result of this algorithm is

$$\hat{\beta}_p = \langle z_p, y\rangle / \langle z_p, z_p \rangle$$ (3.28)

The book then goes on to assure the following:

Re-arranging the residual in step 2, we can see that each of the xj is a linear combination of the $$z_k$$, $$k \leq j$$. Since the $$z_j$$ are all orthogonal, they form a basis for the column space of $$X$$, and hence the least squares projection onto this subspace is $$\hat{y}$$. Since $$z_p$$ alone involves $$x_p$$ (with coefficient 1), we see that the coefficient (3.28) is indeed the multiple regression coefficient of $$y$$ on $$x_p$$.

I do no believe the last sentences makes sense

I think the coefficient of $$y$$ regressed on $$x$$ should be something like

$$\sum\limits_{i=1}^p \eta\beta_i$$.

for some constants $$\eta$$

$$x_j = z_j + \sum\limits_{k=0}^{j-1} \hat{\gamma}_{kj}z_{k}$$

Hence $$z = \Gamma x$$, where $$\Gamma$$ is a upper triangular matrix with diagonal 1.

Hence $$\hat{\beta} z = \hat{\beta} \Gamma^{-1} x$$ where $$\Gamma^{-1}$$ is a lower triangular matrix.

Hence the regression of $$y$$ on $$x_i$$

$$\sum\limits_{i=p}^{n} \eta\hat{\beta}_i$$.

where $$\eta$$ is obtained by inverting $$\Gamma$$ and the fact the inverse of an upper triangular matrix is a lower triangular one. Hence I disagree with the assertion in the book.

This is "The element of statistical learning" on page 54.

However, we can permute each coefficient to make it the last one, hence it holds for every coefficient that the coefficient of multilinear regression is the same as applying Gram-Schimdt to that variable with respect to all others and then regressing the $$y$$ as against the result.