# How to formulate ordinary least squares regression in component formalism?

Matrix formulation is straightforward:

$$\mathbf{y} = \mathbf{X} \boldsymbol{\hat{\beta}} +\boldsymbol{\hat{\varepsilon}}$$

cost function: $$E = {\boldsymbol{\hat{\varepsilon}}}^T{\boldsymbol{\hat{\varepsilon}}} = {(\mathbf{y} - \mathbf{X}\boldsymbol{\hat{\beta}})}^T(\mathbf{y} - \mathbf{X}\boldsymbol{\hat{\beta}})$$

... differentiating wrt $$\boldsymbol{\hat{\beta}}$$ and searching for extremum:

$$\frac{\partial E}{\partial \boldsymbol{\hat{\beta}}} = 2 \mathbf{X}^T\mathbf{X} \boldsymbol{\hat{\beta}} - 2 \mathbf{X}^T \mathbf{y} = 0$$

thus the OLS estimate of $$\boldsymbol{\hat{\beta}}$$ is: $$\boldsymbol{\hat{\beta}} = (\mathbf{X}^T\mathbf{X})^{-1} \mathbf{X}^T \mathbf{y}$$

So, there is probably some limitation to the previous relation (e.g. $$(\mathbf{X}^T\mathbf{X})^{-1}$$ have to exist) ... am I right?

If I try to make the same with component notation, there is a problem in the same formula (I'll come back later to this).

In component formalism (using Einstein's summation convention):

$$E = (X_{ij} \beta_j - y_i)^2 = (X_{ij}\beta_j)^2 - 2 X_{ij}\beta_j y_i + y_i^2$$

$$\frac{\partial E}{\partial \beta_j} = 2X_{ij} \beta_j X_{ij} - 2X_{ij} y_i = 0$$

$$X_{ij} \beta_j X_{ij} = X_{ij} y_i$$

Now, every term is just scalar, so it's tempting to cancel $$X_{ij}$$ on both sides. However, this just leads to trivial relation: $$y_i = X_{ij} \beta_j$$

Can someone help me to enlighten this, please? Isn't it somehow connected to the use of only lower indices? When I have to consider both lower and upper indices (tensors and duals)?

Thank you!

The error in your component derivation: When you differentiate wrt $$\beta_j$$, the index $$j$$ now has two roles: one as the generic index in the sum, and one as the index specifying which $$\beta$$ you are differentiating with respect to. The partial derivative wrt $$\beta_j$$ should treat the other $$\beta$$'s as constant, but your notation can no longer distinguish them!
Better to use a new index, say $$k$$, to perform the differentiation. When you do this, you find that the partial derivative will be $$\frac{\partial E}{\partial\beta_k}=2X_{ij}\beta_jX_{ik} - 2X_{ik}y_i.\tag1$$ There's still summing going on (involving $$i$$ and $$j$$, with $$k$$ held constant) when you set (1) to zero, so it doesn't make sense to factor out $$X_{ij}$$. Convert back into matrix notation and you'll get $$X^TX\beta=X^Ty.\tag2$$ Specifically, $$\sum_i\sum_jX_{ij}\beta_jX_{ik}$$ is the $$k$$th member of the vector $$X^TX\beta$$, while $$\sum_iX_{ik}y_i$$ is the $$k$$th member of $$X^Ty$$.
What you have is essentially $$\boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}^T\boldsymbol{y}$$, regardless of which notation you use. As you pointed out, if you cancel $$\boldsymbol{X}^T$$ from both sides, you are left with $$\boldsymbol{X}\boldsymbol{\beta}=\boldsymbol{y}$$. However, the point of doing least squares in the first place is that $$\boldsymbol{X}\boldsymbol{\beta}=\boldsymbol{y}$$ cannot be solved for $$\boldsymbol{\beta}$$, i.e., $$\boldsymbol{X}$$ is not invertible. By multiplying each side by $$\boldsymbol{X^T}$$, you get $$\boldsymbol{X}^T\boldsymbol{X}$$ on the left, which is invertible.
• Thank you! Hmm, thus it would be possible to look at OLS from an algebraic perspective (as you said: non-invertible $\bf{X}$ --> fix it with left multiplication) and completely forget the idea of searching for extrema? – user847643 Nov 12 '20 at 10:13
• To justify the normal equations using linear algebra/geometry more fully, you can start from the fact that $\boldsymbol{X}\boldsymbol{\beta}$ is in the column space of $\boldsymbol{X}$, but $\boldsymbol{y}$ is not in general in the column space, so you find the closest vector to $\boldsymbol{y}$ that is in the column space (i.e., project $\boldsymbol{y}$ into the column space). The error: $\boldsymbol{X}\hat{\boldsymbol{\beta}}-\boldsymbol{y}$ must be perpendicular to the column space, so $\boldsymbol{X}^T(\boldsymbol{X}\hat{\boldsymbol{\beta}}-\boldsymbol{y})=0$. – nosuchthingasmagic Nov 14 '20 at 22:18