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I am reading "Deep Learning", (Goodfellow et al.) and I am trying to understand the development of the normal equations, starting from the MSE minimization

It has been way too long that I try to understand this equation, and I think I am missing one element.

The predicted output is defined as a basic linear function:

$$ \hat{y} = w^Tx \tag{5.3}$$ with $w \in\mathbb{R}^n$ a vector of parameters and $x \in\mathbb{R}^n$ the input vector.

The Mean Squared Error is defined as:

$$MSE_{test}=\frac{1}{m}\sum_{i}^{}(\hat{y}^{(test)}-y^{(test)})^2_i\tag{5.4}$$

The sum is then rewritten as an Euclidian distance for i=2:

$$MSE_{train}=\frac{1}{m}\left \|(\hat{y}^{(train)}-y^{(train)})\right \|^2_2 \tag{5.5}$$

The goal here is to find the optimal weights by minimizing the Mean Squared Error. To do so, the gradient descent technique is used:

$$\bigtriangledown _{w}MSE=0 \tag{5.6}$$ $$\bigtriangledown _{w}\frac{1}{m}\left \|(\hat{y}^{(train)}-y^{(train)})\right \|^2_2=0\tag{5.7}$$ $$ \frac{1}{m}\bigtriangledown _{w}\left \|({X}^{(train)}w-y^{(train)})\right \|^2_2=0 \tag{5.8}$$ $$\bigtriangledown _{w}{({X}^{(train)}w-y^{(train)})}^{T}({X}^{(train)}w-y^{(train)})=0 \tag{5.9}$$ $$ \bigtriangledown _{w}({w^{T}}{X^{(train)T}}{X^{(train)}}w-2{w^{T}}{X^{(train)T}}{y^{(train)}}+{y^{(train)T}}{y^{(train)})}=0 \tag{5.10}$$ $$ 2{X^{(train)T}}{X^{(train)}}w-2{X^{(train)T}}{y^{(train)}}=0 \tag{5.11}$$ $$ w=({X^{(train)T}}{X^{(train)}})^{-1}{X^{(train)T}}{y^{(train)}} \tag{5.12}$$

My questions might be simple, but it has been a while I didn't do matrices operation and I still need to exercice a bit:

  1. In the equation $(5.3)$: $\hat{y}=w^Tx$ however from equation $(5.7) \rightarrow (5.8)$ the following equality apply : $\hat{y}^{(train)}=X^{(train)}w$. Why isn't it $\hat{y}^{(train)}=w^{T}X^{(train)}$ instead.

  2. $(5.9) \rightarrow (5.10)$ When I develop it, I obtain the following equation: $$ \bigtriangledown _{w}({w^{T}}{X^{(train)T}}{X^{(train)}}w-{w^{T}}{X^{(train)T}}{y^{(train)}}-{y^{(train)T}}{X^{(train)}}w+{y^{(train)T}}{y^{(train)})}=0 $$

By identification, I see that I have the first and last term, but I don't have the $-2w^{T}{X^{(train)T}}{y^{(train)}}$. The closest I can get is by doing so:

$${w^{T}}{X^{(train)T}}{y^{(train)}}-{y^{(train)T}}{X^{(train)}}w={w^{T}}{X^{(train)T}}{y^{(train)}}-({w^{T}}{X^{(train)T}}{y^{(train)}})^T$$

I know I am almost there, but I am missing the final property that would help me close the deal.

If anyone with more experience than me could teach me this piece of knowledge, I would gladly learn.

Best regards,

Valentin

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  • $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ Commented Sep 3, 2021 at 7:14
  • $\begingroup$ Please use ImGur: if you noticed, when you asked your question you had the option to upload images. Please edit your question and use the inbuilt image uploader (or type everything in LaTex yourself) because that site is a touch ... shady, and is certainly suboptimal $\endgroup$
    – FShrike
    Commented Sep 3, 2021 at 7:20
  • $\begingroup$ It has been modified to Imgur yes. I hope someone can help me on that, it has been way too long now that I have been looking into it, but I really want to understand $\endgroup$ Commented Sep 3, 2021 at 8:40
  • $\begingroup$ So I figured out how to write directly in LateX here now I think the post is way better ^^ (even though I don't know yet how to write the numero of the equation properly, but this is for another subject) $\endgroup$ Commented Sep 3, 2021 at 10:59

1 Answer 1

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  1. $\hat{y}=w^Tx=x^Tw$ when $x$ is a vector in $\mathbb{R}^n$.

$\hat{y}^{(train)}\in \mathbb{R}^m$ is a column vector where each entry corresponds to a training instance.

$X^{(train)} \in \mathbb{R}^{m \times n}$ and $w \in \mathbb{R}^{n}$, hence $X^{(train)}w \in \mathbb{R}^m$ where each row correspodns to a training instance.

$w^T \in \mathbb{R}^{1 \times n}$, we can't compute $w^TX^{(train)}$ as there is a dimension mismatch in matrix multiplication.

  1. $w^TX^{(train)T}y^{(train)}$ is a scalar. Hence it is symmetric and equalt to its transpose. That is it is equal to $y^{(train)T}X^{(train)}w$
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  • $\begingroup$ Thank you very much for the answer ! I guess I need to do dimension analysis more often when I deal with matrices. $\endgroup$ Commented Sep 6, 2021 at 6:34

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