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RobinB27
502 lines
21 KiB
TeX
502 lines
21 KiB
TeX
\textbf{Motivation}: So far, when looking for $\hat{f}(x)$ the form was $\hat{f}(x)=w^\top x$, or $\hat{f}(x) = w^\top \phi(x)$.
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Note how the features $x, \phi(x)$ are predetermined. Why not learn them?
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\textbf{New Optimization Problem}:
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The new join-optimization problem, for $w$ and $\phi$:\\
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\subtext{$\Theta$ is a set of parameters for $\phi$}
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$$
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\hat{w} = \underset{w\in\R^m,\Theta\in\R^{m\times d}}{\text{arg min}}\Biggl( \frac{1}{n}\sum_{i=1}^n l\Bigl( w^\top \phi(x_i;\Theta),y_i \Bigr) \Biggr)
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$$
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Where $\phi(x,\Theta) = \Bigl( \phi_1(x;\theta_1),\ldots,\phi_m(x;\theta_m) \Bigr)$.\\
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\subtext{$\theta_i$ is the $i$th row of $\Theta$, i.e. $\theta_i := (\Theta)_{i,:}$}
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More compact, in terms of $\Theta$, which combines $w, \phi$:
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$$
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\Theta^* := \underset{\Theta}{\text{arg min}}\Bigl( L(\Theta; \mathcal{D}) \Bigr) = \underset{\Theta}{\text{arg min}} \Biggl( \frac{1}{n} \sum_{i=1}^{n} l\Bigl( \Theta;x_i,y_i \Bigr) \Biggr)
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$$
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\subtext{$\Theta$ may also encapsulate $w,\phi$ for multiple layers, depending on definition}
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\subsection{Definitions}
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\definition \textbf{Activation Function}\\
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We set $\phi_i(x;\theta_i) = \psi(\theta_i^\top x)$, $\psi$ is the activation function.\\
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\subtext{$\theta_i \in \R^d,\quad\psi:\R\to\R$}
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{\scriptsize
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\notation More concisely, $\phi(x;\Theta) = \psi(\Theta x)$
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}
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\begin{center}
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\begin{tabular}{ll}
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\textbf{Activation Function} & \textbf{Definition} \\
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\hline
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Identity & $\psi(z) = z$ \\
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Sigmoid & $\psi(z) = \frac{1}{1+e^{-z}}$ \\
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Hyperbolic tangent & $\psi(z) = \tanh(z)$ \\
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Rectified Linear Unit (ReLU) & $\psi(z) = \max(0,z)$
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\end{tabular}
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\end{center}
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\definition \textbf{Artificial Neural Network}\\
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\subtext{The output functions of the above problem take the form:}
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$$
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f(x;w,\theta) = \sum_{j=1}^{m}w_j\psi(\theta_j^\top x)
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$$
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{\scriptsize
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\remark Also called Multi-Layer Perceptron (MLP)
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}
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\newpage
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\textbf{What is happening here?}\\
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\smalltext{Explaining the calculation steps for such an $f$ naturally leads to the common pictorial depiction of neural networks.}
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\begin{align*}
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\text{(i)} &\quad x &=\quad& (x_1,\ldots,x_n) \in \R^d & \text{(Input Vector)} \\
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\text{(ii)} &\quad z &=\quad& \Theta x & \text{(Linear transformation)} \\
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\text{(iii)} &\quad h_i &=\quad& \psi(z_i) & \text{(Activation function)} \\
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\text{(iv)} &\quad f(x) &=\quad& \sum_{j=1}^m w_j h_j & \text{(Output)}
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\end{align*}
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\definition \textbf{Hidden Layer} $h = \psi(z)$
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\definition \textbf{Bias Term} $b \in \R^m$\\
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\subtext{Needed, as $f$ might not pass through origin. Similar to using $F_\text{lin}$ in regression, these can also be added by augmenting the input \& hidden layers.}
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\textbf{Does this work at all?}\\
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\smalltext{Yes, for most functions this does work.}
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\definition \textbf{Sigmoidal Function}
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$$
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\sigma(t) \text{ s.t. } \begin{cases}
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\sigma: \R \to \R \\
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\underset{t\to\infty}{\lim} = 1 \text{ and } \underset{t\to\infty}{\lim}
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\end{cases}
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$$
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\theorem \textbf{Universal Approximation Theorem}\\
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\smalltext{$\hat{f}$, that uniformly approximates $f$, exists and takes this form:}
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$$
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\hat{f}(x) = \textbf{W}^{(2)}\psi\Bigl( \textbf{W}^{(1)}x + b \Bigr)
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$$
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\smalltext{$f: [0,1]^d\to\R$ continuous$,\quad \psi $ sigmoidal}\\
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\subtext{$\textbf{W}^{(1)} \in\R^{m\times d},\quad \textbf{W}^{(2)}\in\R^{1\times m},\quad m \in \N$}
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Note how $m$ could be very large.\\
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\subtext{$m$ can intuitively be understood as the "width" of the ANN}
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\newpage
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\definition \textbf{Fully Connected Neural Network}
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More complex ANNs might have:
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\begin{enumerate}
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\item More hidden layers
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\item Multiple outputs
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\item Differen activation functions across layers
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\end{enumerate}
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These are called \textit{fully connected}, since every node in a layer is connected to every node in the adjacent layers.\\
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\subtext{There are also more complex architectures.}
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\begin{center}
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\includegraphics[width=0.9\linewidth]{resources/FCANN.png}\\
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\subtext{\textit{Introduction to Machine Learning (2026), p. 183}}
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\end{center}
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\notation Weights: $\textbf{W}^{(i)} := \Bigl[ w_{k,l}^{(i)} \Bigr]$, Biases: $b^{(i)}_k$\\
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and $\Theta = \Bigl(\textbf{W}^{(1)},\ldots,\textbf{W}^{(L)}, b^{(1)},\ldots,b^{(L)}\Bigr)$ (All parameters)\\
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\subtext{$w_{k,l}^{(i)}$: "Weight at layer $i$ to node $k$ from node $l$"}
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\newpage
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\subsection{Forward Propagation}
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How can we make predictions, i.e. how can $\hat{f}$ be evaluated?
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\definition \textbf{Forward Propagation}\\
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\subtext{This is just the computation for $1$-layer ANN generalized for $L$ layers}
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\begin{algorithm}
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\caption{Forward Propagation}
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$h^{(0)}\gets x$\;
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\For{$l=1,\ldots,L$}{
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$z^{(l)} = \textbf{W}^{(l)}h^{(l-1)} + b^{(l)}$ \\
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$h^{(l)} = \psi(z^{(l)})$
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}
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$f \gets \textbf{W}^{(L)}h^{(L-1)}+b^{(L)}$ \\
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\Return f
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\end{algorithm}
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\subsection{Backwards Propagation}
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How can we get all gradients needed for model training?
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\definition \textbf{Backwards Propagation}
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\textbf{Intuition}: An efficient way to get the gradients is to reuse results from forward prop. and previous steps. This works best when starting at the back, at $\nabla_{\textbf{W}^{(L)}}l$.
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\textbf{Goal}: $\nabla_{\textbf{W}^{(1)}}l,\ldots,\nabla_{\textbf{W}^{(L)}} l, \nabla_{b^{(1)}}l,\ldots,\nabla_{b^{(L)}}l$
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\textbf{Step 1}: Calculate $\nabla_{\textbf{W}^{(L)}}l$, i.e. start from the back.
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\begin{align*}
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\nabla_{\textbf{W}^{(L)}}l &= \frac{\partial l}{\partial \textbf{W}^{L}} \\
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&= \frac{\partial l}{\partial f}\cdot\frac{\partial f}{\partial \textbf{W}^{(L)}} & \text{(Chain Rule)} \\
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&= \frac{\partial l}{\partial f}\cdot\begin{bmatrix}
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\bigl( h^{(L-1)} \bigr)^\top \\
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\vdots \\
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\bigl( h^{(L-1)} \bigr)^\top
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\end{bmatrix} & (f = \textbf{W}^{(L)}h^{(L-1)} + b^{(L)}) \\
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&= \nabla_f l \cdot\begin{bmatrix}
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\bigl( h^{(L-1)} \bigr)^\top \\
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\vdots \\
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\bigl( h^{(L-1)} \bigr)^\top
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\end{bmatrix} & \Biggl(\frac{\partial l}{\partial f} = \nabla_f l\Biggr)
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\end{align*}
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Notice how $h^{(L-1)}$ was computed during forward prop.
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\newpage
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\textbf{Step 2}: Calculate $\nabla_{\textbf{W}^{(L-1)}}l$.
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\begin{align*}
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\nabla_{\textbf{W}^{(L-1)}}l &= \underbrace{\frac{\partial l}{\partial f}}_{\text{(1)}}\cdot\underbrace{\frac{\partial f}{\partial h^{(L-1)}}}_{\text{(2)}}\cdot\underbrace{\frac{\partial h^{(L-1)}}{\partial z^{(L-1)}}}_\text{(3)}\cdot\underbrace{\frac{z^{(L-1)}}{\partial \textbf{W}^{(L-1)}}}_\text{(4)} & \text{(Chain Rule)} \\
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\end{align*}
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\begin{enumerate}
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\item Already done in Step 1.
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\item Already done in forward propagation, equal to $\textbf{W}^{(L)}$:
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$$
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f \overset{\text{def}}{=} \textbf{W}^{(L)}h^{(L-1)}+b^{(L)} \implies \frac{\partial f}{\partial h^{(L-1)}} = \textbf{W}^{(L)}
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$$
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\item \textbf{Not done.} Needs to be calculated:
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\begin{align*}
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\frac{\partial h^{(L-1)}}{\partial z^{(L-1)}} &= \frac{\partial \psi\bigl( z^{(L-1)} \bigr)}{\partial z^{(L-1)}} \\
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&= \text{diag}\Bigl( \psi'\bigl( z^{(L-1)} \bigr) \Bigr) \\
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&= \begin{bmatrix}
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\psi'\Bigl(z_1^{(L-1)}\Bigr) & 0 & \cdots & 0 \\
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0 & \psi'\Bigl(z_2^{(L-1)}\Bigr) & \cdots & 0 \\
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\vdots & \vdots & \ddots & \vdots \\
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0 & 0 & \cdots & \psi\Bigl(z_n^{(L-1)}\Bigr) \\
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\end{bmatrix}
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\end{align*}
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\item Already done in forward propagation, analogous to step 1.
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$$
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\frac{\partial z^{(L-1)}}{\partial \textbf{W}^{(L-1)}} =
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\begin{bmatrix}
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\bigl( h^{(L-2)} \bigr)^\top \\
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\vdots \\
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\bigl( h^{(L-2)} \bigr)^\top
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\end{bmatrix}
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$$
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\end{enumerate}
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\textbf{Step $i \leq L$}: Calculate $\nabla_{\textbf{W}^{(L-i)}}l$ Analogoues to step 2.\\
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\subtext{The biases $\nabla_{b^{(l)}}l$ are analogous.}
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\newpage
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\subsection{Optimization}
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\textbf{Problem}: How can we train the model, i.e find $\Theta^*$?
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$$
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\Theta^* := \underset{\Theta}{\text{arg min}}\Bigl( L(\Theta; \mathcal{D}) \Bigr) = \underset{\Theta}{\text{arg min}} \Biggl( \frac{1}{n} \sum_{i=1}^{n} l\Bigl( \Theta;x_i,y_i \Bigr) \Biggr)
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$$
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{\footnotesize
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\remark $L(\Theta;\mathcal{D})$ is generally not convex.\\
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{\color{gray}
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i.e. local minima, saddle points may exist
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}
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\remark $\dim(\Theta)$ is the total param. count of NN, may be very large
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}
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\textbf{Solution}: Gradient Descent (with optimizations)
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\begin{itemize}
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\item Stochastic Gradient Descent\\
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\subtext{(Why? $\dim(\Theta)$ is very large, $\nabla_\Theta l(\Theta;x_i,y_i)$ are expensive)}
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\item Minibatch Gradient Descent\\
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\subtext{(Why? $\mathcal{D}$ may be very large, so there are \textit{many} gradients)}
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\end{itemize}
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The standard GD update for $\Theta$ is:
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$$
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\Theta^{t+1} = \Theta^t - \eta_t\cdot\nabla_\Theta L\Bigl( \Theta;\mathcal{D} \Bigr)
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$$
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In Minibatch GD, this becomes:\\
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\subtext{Where $\mathcal{S} \subset \{1,\ldots,n\}$}
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$$
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\Theta^{t+1} = \Theta^t - \eta_t\cdot\nabla_{\Theta^t} \Biggl( \frac{1}{|\mathcal{S}|}\sum_{i\in \mathcal{S}} l\Bigl( \Theta^t; x_i,y_i \Bigr) \Biggr)
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$$
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{\footnotesize
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\remark An advantage: If $\Theta^t$ approaches a stat. point (which isn't the global minimum), GD will converge, but MB-GD may not converge.
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}
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The further subsections go into more details:
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\begin{enumerate}
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\item Preventing vanishing \& exploding Gradients
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\item Choice of initial $w_i$
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\item Choice of $\mu_t$
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\end{enumerate}
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\newpage
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\subsubsection{Vanishing \& Exploding Gradients}
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$$
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\nabla_{\Theta^t} \Biggl( \frac{1}{|\mathcal{S}|}\sum_{i\in \mathcal{S}} l\Bigl( \Theta^t; x_i,y_i \Bigr) \Biggr) = \frac{1}{|\mathcal{S}|}\sum_{i \in \mathcal{S}}\Bigl( \nabla_{\Theta^t} l(\Theta^t;x_i,y_i) \Bigr)
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$$
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The terms $\nabla_{\Theta^t} l\Bigl(\Theta^t;x_i,y_i\Bigr)$ each contain $\nabla_{\textbf{W}^{(l)}}l\Bigl(\textbf{W}^{(l)};x_i,y_i\Bigr)$.
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\textbf{Problem}: Optimization might fail if:
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$$
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\Bigl\Vert \nabla_{\textbf{W}^{(l)}}l \Bigr\Vert \to \infty \qquad\text{or}\qquad \Bigl\Vert \nabla_{\textbf{W}^{(l)}}l \Bigr\Vert \to 0
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$$
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\textbf{Solution}: $\Vert \nabla_{\textbf{W}^{(l)}}l \Vert$ depends linearly on $\text{diag}\Bigl( \psi'(z^{(l)}) \Bigr)$, so the choice of $\psi$ ($\psi'$) can be used to constrain $\Vert \nabla_{\textbf{W}^{(l)}}l \Vert$\\
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\subtext{Generally, the gradient follows the behaviour of $\psi'$}
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% Script contains examples for this on Sigmoid & ReLU. Generally, the properties we need are visible by inspection of the derivatives graph.
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Which features do we want $\psi$ ($\psi'$) to fullfil?
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\begin{itemize}
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\item $\psi'$ should be fast to calculate
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\item $\psi'$ should be non-zero (and not get too close)
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\end{itemize}
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\remark The $\Bigl\Vert \nabla_{\textbf{W}^{(l)}}l \Bigr\Vert$ may still vanish for any $\psi$.
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\subsubsection{Random Weight Initialization}
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\begin{align*}
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\nabla_{\textbf{W}^{(l)}}l &= \frac{\partial l}{\partial f}\cdot\frac{\partial f}{\partial h^{(l)}}\cdot\frac{\partial h^{(l)}}{\partial z^{(l)}}\cdot\frac{\partial z^{(l)}}{\partial \textbf{W}^{(l)}} \\
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&= \frac{\partial l}{\partial f}\cdot\frac{\partial f}{\partial h^{(l)}}\cdot\text{diag}\Bigl( \psi'(z^{(l)}) \Bigr)\cdot\begin{bmatrix}
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(h^{(l-1)})^\top \\
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\vdots \\
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(h^{(l-1)})^\top
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\end{bmatrix}
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\end{align*}
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So, the gradient $\Vert \nabla_{\textbf{W}^{(l)}}l \Vert$ also depends on $\Vert h^{(l-1)} \Vert$.\\
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\subtext{For $l \in \{1,\ldots,L\}$}
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\textbf{Problem}: It might be that $\Vert h^{(l-1)} \Vert \to \infty$ or $\Vert h^{(l-1)} \Vert \to 0$.
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\textbf{Solution}: Set $h^{(l-1)}$ randomly, bound mean $\mu$ and var. $\sigma^2$.\\
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\subtext{There is no generally optimal bound for $\sigma^2$, it depends on the NN.}
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\newpage
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\remark Practical distributions for common $\psi$:\\
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\subtext{$n_\text{out},n_\text{out}$ are the node counts of the layers adjacent to $w_i$}
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% Table in script
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\begin{center}
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\begin{tabular}{l|l}
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$\psi$ & \textbf{Weights} \\
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\hline
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$\tanh$ & $w_i \sim \mathcal{N}\Bigl( 0, \frac{1}{n_{\text{in}}} \Bigr)$ \\
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$\tanh$ & $w_i \sim \mathcal{N}\Bigl( 0, \frac{2}{n_\text{in}+n_\text{out}} \Bigr)$ \\
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\text{ReLu} & $w_i \sim \mathcal{N}\Bigl( 0, \frac{2}{n_\text{in}} \Bigr)$
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\end{tabular}
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\end{center}
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\subsubsection{Learning Rate \& Weight Updates}
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$$
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\Theta^{t+1} = \Theta^{t} - \mu_t \cdot \nabla_{\Theta^t}\Biggl( \frac{1}{|\mathcal{S}|}\sum_{i\in\mathcal{S}}l\Bigl( \Theta^t; x_i,y_i \Bigr) \Biggr)
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$$
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\textbf{Problem}: How to choose $\mu$? (Learning Rate)
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\textbf{Solution}: Heuristics.\\
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\subtext{There is no generally optimal $\mu$.}
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\method \textbf{Piecewise constant $\mu_t$}
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Intuitively, it makes sense to reduce $\mu_t$ as optimization progresses, as the algorithm approaches the minimum.\\
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\subtext{Linear/cosine decay could also be used.}
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$$
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\mu_t = \begin{cases}
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1 & 0 \geq t < 3 \\
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0.5 & 3 \leq t < 6 \\
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0.25 & 6 \leq t < 9 \\
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\ldots
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\end{cases}
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$$
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\method \textbf{Weight update indicator}
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In practice, SGD often oscillates in finding the minimum. Then, a monotonic $\mu_t$ doesn't make sense.
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$$
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\frac{\Big| \nabla_{\Theta^t} L( \Theta^t;\mathcal{D} ) \Big|}{\Vert \Theta^t \Vert}
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$$
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In general, if this indicator ratio is small, a higher learning rate makes sense (and vice versa).\\
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\subtext{Intuitively: How strongly is the weight change, relative to weight size.}
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\method \textbf{Momentum}
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Combine the update direction with the previous update directions, for some weight $m > 0$, to stabilize.
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\newpage
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\subsection{Regularization}
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\textbf{Problem}: How can overfitting be avoided?
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A few methods can be applied directly to SGD:
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\method \textbf{Penalty Term}
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Similar to Ridge/LASSO Regression, a penalty term can be used, with some weight $\lambda > 0$.
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$$
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\underset{\Theta \in \R^d}{\text{arg min}}\Bigl( L(\Theta;\mathcal{D}) \Bigr) \quad \to \quad \underset{\Theta \in \R^d}{\text{arg min}}\Bigl( L(\Theta;\mathcal{D}) + \lambda \Vert\Theta\Vert^2 \Bigr)
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$$
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\method \textbf{Earlier stop}
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Choosing a different stop criterion for SGD, e.g. performance on the test set $\mathcal{D}'$.
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\remark \textbf{Validation \& Training Error}
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Overfitting occurs when the training error (on $\mathcal{D}$) continues to fall, but the test error increases (on $\mathcal{D}'$).
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\begin{center}
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\includegraphics[width=0.7\linewidth]{resources/ValidationTrainingErrors.png}\\
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\color{gray}\footnotesize
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\textit{Introduction to Machine Learning (2026), p. 196}
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\end{center}
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\newpage
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\subsubsection{Dropout Regularization}
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This is a method specific to Neural Networks.
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\method \textbf{Dropout}
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Fix some $p \in (0,1)$. For each SGD iteration in training:\\
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\textit{Drop out} each hidden unit with probability $1-p$, and skip their optimization for this iteration.
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\begin{align*}
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z_j^{(l)} &= \sum_{i=0}^{n_{i-1}}\Bigl( w_{j,i}^{(l)}h_i^{(l-1)} \Bigr) + b_j^{(l)} & \text{(Regular Neuron)} \\
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z_j^{(l)} &= \sum_{i=0}^{n_{i-1}}\Bigl( w_{j,i}^{(l)}h_i^{(l-1)} \cdot \mathbb{I}_{C_i} \Bigr) + b_j^{(l)} & \text{(With Dropout)}
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\end{align*}
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\subtext{Where $C_i:=\{\text{"Unit } h_i^{(l-1)} \text{ is kept this iter."}\}$, so $\P[C_i] = p$}
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\textbf{Problem}: For $\mathcal{D}'$, we again want to use all layers.
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\textbf{Solution}: Scale all weights with $p$
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% This makes sense but I don't get why exactly the scaling is needed
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For this, we use $\E\Bigl[ z_j^{(l)} \Bigr]$ instead of $z_j^{(l)}$.
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\begin{align*}
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\E\Bigl[ z_j^{(l)} \Bigr] = \Bigl( p\cdot w_j^{(l)} \Bigr)^\top \cdot h^{(l-1)} + b_j^{(l)}
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\end{align*}
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\subtext{By using $\E\bigl[ \mathbb{I}_{C_i} \bigr] = \P[C_i] = p$}
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\subsubsection{Batch Normalization}
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During SGD, the weight's $\sigma^2$ may again explode.\\
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\subtext{After few iterations, $w_i$ may have changed completely from init.}
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\textbf{Problem}: Internal covariate shift.\\
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\subtext{The mean $\mu$ deviates from $0$ and $\sigma^2$ might increase}
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\textbf{Solution}: Standardize $\psi$ also during training.
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\newpage
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\method \textbf{Batch Normalization}
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% The script doesn't specify how \alpha is set, unfortunately
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In pratice, only batches of $\psi$ are normalized. The core idea is to set $\mu \mapsto 0$ and $\sigma^2 \mapsto 1$ within the batch.\\
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\subtext{This isn't optimal for all problems, and can be tweaked using $\beta,\gamma$}
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The algorithm uses the parameters $\beta, \gamma$ and buffers $\mu_\text{EMA}, \sigma^2_\text{EMA}$.\\
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\subtext{$\beta,\gamma$ are learnable and can also be optimized}
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\textbf{Normalization Step} (Training set)\\
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\smalltext{For a minibatch $\mathcal{S} = \{i_1,\ldots,i_k\}$, the batch is $\{x_{i_1},\ldots,x_{i_k}\}$.}
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\textbf{Step 1}: Find current values of $\mu, \sigma^2$.
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\begin{align*}
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\mu_\mathcal{S} &:= \frac{1}{|\mathcal{S}|}\sum_{j\in\mathcal{S}}x_j & \text{\color{gray}\footnotesize(minibatch mean)} \\
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\sigma^2_\mathcal{S} &:= \frac{1}{|\mathcal{S}|}\sum_{j\in\mathcal{S}}\Bigl( x_i-\mu_\mathcal{S} \Bigr)^2 & \text{\color{gray}\footnotesize(minibatch variance)}
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\end{align*}
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\textbf{Step 2}: Update the moving average: $\mu_\text{EMA},\sigma^2_\text{EMA}$.
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\begin{align*}
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\mu_\text{EMA} &= (1-\alpha)\mu_\text{EMA} + \alpha \cdot \mu_\mathcal{S} & \text{\color{gray}\footnotesize(avg. mean update)} \\
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\sigma^2_\text{EMA} &= (1-\alpha)\sigma^2_\text{EMA}+\alpha\cdot \sigma_\mathcal{S}^2 & \text{\color{gray}\footnotesize(avg. variance update)}
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\end{align*}
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\textbf{Step 3}: Update the $x_j$.
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\begin{align*}
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\hat{x}_j &= \frac{x_j-\mu_\mathcal{S}}{\sqrt{\sigma^2_\mathcal{S} + \epsilon}} & \text{\color{gray}\footnotesize(point normalization)} \\
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\bar{x}_j &= \gamma\cdot\hat{x}_j + \beta & \text{\color{gray}\footnotesize(scale \& shift)}
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\end{align*}
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\textbf{Normalization Step} (Test set)\\
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\smalltext{Only apply step 3, now using the moving average values.}
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\begin{align*}
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\hat{x}_j &= \frac{x_j-\mu_\text{EMA}}{\sqrt{\sigma^2_\text{EMA}+\epsilon}} & \text{\color{gray}\footnotesize(point normalization)} \\
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\bar{x}_j &= \gamma\cdot\hat{x}_j + \beta & \text{\color{gray}\footnotesize(scale \& shift)}
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\end{align*}
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\newpage
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\subsection{Convolutional Neural Networks}
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In fully connected NNs:
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$$
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h^{(l)} = \psi\Bigl( \textbf{W}^{(l)} h^{(l-1)} \Bigr)
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$$
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Each unit of layer $l-1$ affects each unit of layer $l$.\\
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In CNNs, this is relaxed: not all nodes (must) interact.
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\definition \textbf{Convolutional Neural Network} (CNN)\\
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Layers are connected via convolutions.
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$$
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h^{(l)} = \psi\Bigl( w^{(l)} * h^{(l-1)} \Bigr)
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$$
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{\footnotesize
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\remark In CNNs, the weights are also called \textit{filters}.
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}
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\definition \textbf{Convolution} {\footnotesize (Discrete, 2D) } \\
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\subtext{$w \in \R^k,\quad x \in \R^d$}
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$$
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w * x := \sum_{j=\max\{1,i-d+1\}}^{\min\{i,k\}}\Biggl( w_j\cdot x_{i-j+1} \Biggr)
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$$
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Understanding this is easier by example:
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\smalltext{\textbf{Example}: $w = (w_1,w_2)^\top,\quad x=(x_1,x_2,x_3)^\top$}
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{\footnotesize
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$$
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w*x = \begin{bmatrix}
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w_1 \cdot x_1 + \color{gray}w_2 \cdot 0 \\
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w_1 \cdot x_2 + w_2 \cdot x_1 \\
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w_1 \cdot x_3 + w_2 \cdot x_2 \\
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\color{gray}w_1 \cdot 0 + \color{black}w_2 \cdot x_3
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\end{bmatrix}
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$$
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}
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\smalltext{\textbf{Example}: a CNN with $3$ inputs and $1$ hidden layer:}
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{\footnotesize
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$$
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\begin{bmatrix}
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z_1 \\
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z_2 \\
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z_3 \\
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z_4
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\end{bmatrix} = \underbrace{\begin{bmatrix}
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w_1 & 0 & 0 \\
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w_2 & w_1 & 0 \\
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0 & w_2 & w_1 \\
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0 & 0 & w_2 \\
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\end{bmatrix}}_{\textbf{W}^{(1)} \text{ in CNN}} \cdot \begin{bmatrix}
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x_1 \\
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x_2 \\
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x_3
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\end{bmatrix} = w * x
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$$
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}
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\subsubsection{Multidimensional Convolution}
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\textbf{TODO} add explanation
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% The script has a very good intutive walkthrough of how this works. |