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I have taken a course in regression analysis. I learned that the equation $\beta =(X'X)^{-1}X'y$ can be used to find the weights in a linear model.

When learning about GLMs, I came across this formula that can be used when $(X'X)$ is not invertible while $XX'$ is.

$$\beta = X'(XX')^{-1}y$$

For example, if the generalized linear model is $y(x_1,x_2) = b_0 +b_1x_1+b_2x_2+b_3x_1x_2+b_4x_1^2+b_5x_2^2$, and there are, say, four data points, then there would be correlation between the columns and so we have to use the "right inverse method".

My question is, where did the equation $\beta = X'(XX')^{-1}y$ come from? I tried searching up the derivation, but could not find anything (which is why I could not come up with a nice title for this post; I don't know what keywords to use). I suspect it would be by performing partial derivatives, as in the usual $\beta =(X'X)^{-1}X'y$ equation, but I cannot see how it works.

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Your question is a bit unclear to me. However, as far I can understand...

if the generalized linear model is $y(x_1,x_2)=b_0+b_1x_1+b_2x_2+b_3x_1x_2+b_4x^2_1+b_5x^2_2$, and there are, say, four data points, then

...then your model matrix $X$ is a $4\times 6$ matrix: $$\begin{bmatrix} 1 & x_{11} & x_{21} & x_{11}x_{21} & x_{11}^2 & x_{21}^2 \\ 1 & x_{12} & x_{22} & x_{12}x_{22} & x_{12}^2 & x_{22}^2 \\ 1 & x_{13} & x_{23} & x_{13}x_{23} & x_{13}^2 & x_{23}^2 \\ 1 & x_{14} & x_{24} & x_{14}x_{24} & x_{14}^2 & x_{24}^2 \end{bmatrix}$$

If $\text{rank}(X)=4$, then $X^TX$ is a $6\times 6$ singular matrix, while $XX^T$ is a $4\times 4$ non singular matrix.

You should need $(X^TX)^{-1}$ to estimate $\beta$, but since $X^TX$ is singular you have to use a right inverse, i.e. a matrix $X_R$ such that $XX_R=I$: $$ X\beta=y,\qquad X\beta=XX_Ry,\qquad \beta=X_Ry$$ A right inverse that always guarantees a solution is: $X_R=X^T(XX^T)^{-1}$. See Wikipedia and Cherkassky & Mulier, Learning From Data: Concepts, Theory, and Methods, John Wiley & Sons, 2007, Appendix B.

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  • $\begingroup$ Thanks for the insight and references $\endgroup$
    – Suzee
    Commented Sep 11, 2020 at 4:30
  • $\begingroup$ You are welcome. You can also look at David A. Harville, Matrix Algebra from a Statistician's Perspective, Springer, 2008. $X^T(XX^T)^{-1}$ is a special case of the Moore-Penrose inverse, in Chapter 20 you can find definitions and proofs, in §20.2 the special case. $\endgroup$
    – Sergio
    Commented Sep 11, 2020 at 6:42

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