variables.

The above formulation can be shown to induce a quadratic regression of $y$ on $X$. To build intuition consider the case in which $\sigma_j^2 = 0$ as done in [@Mike03] for the special case in which $\Omega = 0$. *The many-to-one map* $X_i = \Lambda \eta_i$ has multiple generalized inverses $\eta_i = \Lambda^T X_i + b$ such that $\Lambda b = 0$. If we substitute in the regression equation, we obtain $$\begin{aligned}
 {\mathbb{E}}(y_i|X_i) & = (\Lambda^T X_i + b)^T \omega + (\Lambda^T X_i + b)^T \Omega (\Lambda^T X_i + b) = \\ 
 & = X_i^T \Lambda \omega + X_i^T \Lambda \Omega \Lambda^T X_i + g(b)\end{aligned}$$ The following proposition gives a similar result in the non deterministic case:

Under model (1), the following are true:

-   ${\mathbb{E}}(y_i | X_i) = tr (\Omega V)+(\omega^T A) X_i + X_i^T (A^T \Omega A) X_i$,

-   $Cov(y_i,X_i) = \Lambda \omega$,

where $V = (\Lambda^T \Psi^{-1} \Lambda + I)^{-1}$ and $A = V \Lambda^T \Psi^{-1}  = (\Lambda^T \Psi^{-1} \Lambda + I)^{-1} \Lambda^T \Psi^{-1}$.

This shows that the induced regression of $y$ on $X$ from model (1) is indeed a quadratic regression. Let us define the induced main effects as $\beta_X = A^T \omega$ and the 