%%% Local Variables: %%% mode: latex %%% TeX-master: "main" %%% End: \section{Shallow Neural Networks} In order to examine some behavior of neural networks in this chapter we consider a simple class of networks, the shallow ones. These networks only contain one hidden layer and have a single output node. \begin{Definition}[Shallow neural network] For a input dimension $d$ and a Lipschitz continuous activation function $\sigma: \mathbb{R} \to \mathbb{R}$ we define a shallow neural network with $n$ hidden nodes as $\mathcal{NN}_\vartheta : \mathbb{R}^d \to \mathbb{R}$ as \[ \mathcal{NN}_\vartheta \coloneqq \sum_{k=1}^n w_k \sigma\left(b_k + \sum_{j=1}^d v_{k,j} x_j\right) + c ~~ \forall x \in \mathbb{R}^d \] with \begin{itemize} \item weights $w_k \in \mathbb{R},~k \in \left\{1,\dots,n\right\}$ \item biases $b_k \in \mathbb{R},~k \in \left\{1, \dots,n\right\}$ \item weights $v_k \in \mathbb{R}^d,~k\in\left\{1,\dots,n\right\}$ \item bias $c \in \mathbb{R}$ \item these weights and biases collected in \[ \vartheta \coloneqq (w, b, v, c) \in \Theta \coloneqq \mathbb{R}^{n \times n \times (n \times d) \times 1} \] \end{itemize} \end{Definition} % \begin{figure} % \begin{tikzpicture}[x=1.5cm, y=1.5cm] % \tikzset{myptr/.style={decoration={markings,mark=at position 1 with % % {\arrow[scale=1.5,>=stealth]{>}}},postaction={decorate}}} % \foreach \m/\l [count=\y] in {1} % \node [every neuron/.try, neuron \m/.try] (input-\m) at (0,0.5-\y) {}; % \foreach \m [count=\y] in {1,2,missing,3,4} % \node [every neuron/.try, neuron \m/.try ] (hidden-\m) at (1.25,3.25-\y*1.25) {}; % \foreach \m [count=\y] in {1} % \node [every neuron/.try, neuron \m/.try ] (output-\m) at (2.5,0.5-\y) {}; % \foreach \l [count=\i] in {1} % \draw [myptr] (input-\i)+(-1,0) -- (input-\i) % node [above, midway] {$x$}; % \foreach \l [count=\i] in {1,2,n-1,n} % \node [above] at (hidden-\i.north) {$\mathcal{N}_{\l}$}; % \foreach \l [count=\i] in {1,n_l} % \node [above] at (output-\i.north) {}; % \foreach \l [count=\i] in {1} % \draw [myptr, >=stealth] (output-\i) -- ++(1,0) % node [above, midway] {$y$}; % \foreach \i in {1} % \foreach \j in {1,2,...,3,4} % \draw [myptr, >=stealth] (input-\i) -- (hidden-\j); % \foreach \i in {1,2,...,3,4} % \foreach \j in {1} % \draw [myptr, >=stealth] (hidden-\i) -- (output-\j); % \node [align=center, above] at (0,1) {Input \\layer}; % \node [align=center, above] at (1.25,3) {Hidden layer}; % \node [align=center, above] at (2.5,1) {Output \\layer}; % \end{tikzpicture} % \caption{Shallow Neural Network with input- and output-dimension of \(d % = 1\)} % \label{fig:shallowNN} % \end{figure} As neural networks with a large amount of nodes have a large amount of parameters that can be tuned it can often fit the data quite well. If a ReLU \[ \sigma(x) \coloneqq \max{(0, x)} \] is chosen as activation function one can easily prove that if the amount of hidden nodes exceeds the amount of data points in the training data a shallow network trained on MSE will perfectly fit the data. \begin{Theorem}[sinnvoller titel] For training data of size t \[ \left(x_i^{\text{train}}, y_i^{\text{train}}\right) \in \mathbb{R}^d \times \mathbb{R},~i\in\left\{1,\dots,t\right\} \] a shallow neural network $\mathcal{NN}_\vartheta$ with $n \geq t$ hidden nodes will perfectly fit the data when minimizing squared error loss. \proof W.l.o.g. all values $x_{ij}^{\text{train}} \in [0,1],~\forall i \in \left\{1,\dots\right\}, j \in \left\{1,\dots,d\right\}$. Now we chose $v^*$ in order to calculate a unique value for all $x_i^{\text{train}}$: \[ v^*_{k,j} = v^*_{j} = 10^{j-1}, ~ \forall k \in \left\{1,\dots,n\right\}. \] Assuming $x_i^{\text{train}} \neq x_j^{\text{train}},~\forall i\neq j$ we get \[ \left(v_k^*\right)^{\mathrm{T}} x_i^{\text{train}} \neq \left(v_k^*\right)^{\mathrm{T}} x_j^{\text{train}}, ~ \forall i \neq j. \] W.l.o.g assume $x_i^{\text{train}}$ are ordered such that $\left(v_k^*\right)^{\mathrm{T}} x_i^{\text{train}} < \left(v_k^*\right)^{\mathrm{T}} x_j^{\text{train}}, ~\forall j -\left(v^*\right)^{\mathrm{T}} x_1^{\text{train}},\\ b^*_k &= -\left(v^*\right)^{\mathrm{T}} x_{k-1}^{\text{train}},~\forall k \in \left\{2, \dots, t\right\}, \\ b_k^* &\leq -\left(v^*\right)^{\mathrm{T}} x_{t}^{\text{train}},~\forall k > t. \end{align*} With \begin{align*} w_k^* &= \frac{y_k^{\text{train}} - \sum_{j =1}^{k-1} w^*_j\left(b^*_j + x_k^{\text{train}}\right)}{b_k + \left(v^*\right)^{\mathrm{T}} x_k^{\text{train}}},~\forall k \in \left\{1,\dots,t\right\}\\ w_k^* &\in \mathbb{R} \text{ arbitrary, } \forall k > t. \end{align*} and $\vartheta^* = (w^*, b^*, v^*, c = 0)$ we get \[ \mathcal{NN}_{\vartheta^*} (x_i^{\text{train}}) = \sum_{k = 1}^{i-1} w_k\left(\left(v^*\right)^{\mathrm{T}} x_i^{\text{train}}\right) + w_i\left(\left(v^*\right)^{\mathrm{T}} x_i^{\text{train}}\right) = y_i^{\text{train}}. \] As the squared error of $\mathcal{NN}_{\vartheta^*}$ is zero all squared error loss minimizing shallow networks with at least $t$ hidden nodes will perfectly fit the data. \qed \label{theo:overfit} \end{Theorem} However this behavior is often not desired as over fit models often have bad generalization properties especially if noise is present in the data. This effect can be seen in Figure~\ref{fig:overfit}. Here a network that perfectly fits the training data regarding the MSE is \todo{Formulierung} constructed and compared to a regression spline (Definition~\ref{def:wrs}). While the network fits the data better than the spline, the spline is much closer to the underlying mechanism that was used to generate the data. The better generalization of the spline compared to the network is further illustrated by the better validation error computed with new generated test data. In order to improve the accuracy of the model we want to reduce overfitting. A possible way to achieve this is by explicitly regularizing the network through the cost function as done with ridge penalized networks (Definition~\ref{def:rpnn}) where large weights $w$ are punished. In Theorem~\ref{theo:main1} we will prove that this will result in the network converging to regressions splines as the amount of nodes in the hidden layer is increased. \begin{figure} \pgfplotsset{ compat=1.11, legend image code/.code={ \draw[mark repeat=2,mark phase=2] plot coordinates { (0cm,0cm) (0.15cm,0cm) %% default is (0.3cm,0cm) (0.3cm,0cm) %% default is (0.6cm,0cm) };% } } \begin{tikzpicture} \begin{axis}[tick style = {draw = none}, width = \textwidth, height = 0.6\textwidth] \addplot table [x=x, y=y, col sep=comma, only marks,mark options={scale = 0.7}] {Plots/Data/overfit.csv}; \addplot [red, line width=0.8pt] table [x=x_n, y=s_n, col sep=comma, forget plot] {Plots/Data/overfit.csv}; \addplot [black, line width=0.8pt] table [x=x_n, y=y_n, col sep=comma] {Plots/Data/overfit.csv}; \addplot [black, line width=0.8pt, dashed] table [x=x, y=y, col sep=comma] {Plots/Data/overfit_spline.csv}; \addlegendentry{\footnotesize{data}}; \addlegendentry{\footnotesize{$\mathcal{NN}_{\vartheta^*}$}}; \addlegendentry{\footnotesize{spline}}; \end{axis} \end{tikzpicture} \caption{For data of the form $y=\sin(\frac{x+\pi}{2 \pi}) + \varepsilon,~ \varepsilon \sim \mathcal{N}(0,0.4)$ (\textcolor{blue}{blue dots}) the neural network constructed according to the proof of Theorem~\ref{theo:overfit} (black) and the underlying signal (\textcolor{red}{red}). While the network has no bias a regression spline (black dashed) fits the data much better. For a test set of size 20 with uniformly distributed $x$ values and responses of the same fashion as the training data the MSE of the neural network is 0.30, while the MSE of the spline is only 0.14 thus generalizing much better. } \label{fig:overfit} \end{figure} \clearpage \subsection{Convergence Behaviour of 1-dim. Randomized Shallow Neural Networks} In this section we will analyze the connection of randomized shallow Neural Networks with one dimensional input and regression splines. We will see that the punishment of the size of the weights in training the randomized shallow Neural Network will result in a function that minimizes the second derivative as the amount of hidden nodes is grown to infinity. In order to properly formulate this relation we will first need to introduce some definitions. \begin{Definition}[Randomized shallow neural network] For an input dimension $d$, let $n \in \mathbb{N}$ be the number of hidden nodes and $v(\omega) \in \mathbb{R}^{i \times n}, b(\omega) \in \mathbb{R}^n$ randomly drawn weights. Then for a weight vector $w$ the corresponding randomized shallow neural network is given by \[ \mathcal{RN}_{w, \omega} (x) = \sum_{k=1}^n w_k \sigma\left(b_k(\omega) + \sum_{j=1}^d v_{k, j}(\omega) x_j\right). \] \label{def:rsnn} \end{Definition} \begin{Definition}[Ridge penalized Neural Network] \label{def:rpnn} Let $\mathcal{RN}_{w, \omega}$ be a randomized shallow neural network, as introduced in ???. Then the optimal ridge penalized network is given by \[ \mathcal{RN}^{*, \tilde{\lambda}}_{\omega}(x) \coloneqq \mathcal{RN}_{w^{*, \tilde{\lambda}}(\omega), \omega} \] with \[ w^{*,\tilde{\lambda}}(\omega) :\in \argmin_{w \in \mathbb{R}^n} \underbrace{ \left\{\overbrace{\sum_{i = 1}^N \left(\mathcal{RN}_{w, \omega}(x_i^{\text{train}}) - y_i^{\text{train}}\right)^2}^{L(\mathcal{RN}_{w, \omega})} + \tilde{\lambda} \norm{w}_2^2\right\}}_{\eqqcolon F_n^{\tilde{\lambda}}(\mathcal{RN}_{w,\omega})}. \] \end{Definition} In the ridge penalized Neural Network large weights are penalized, the extend of which can be tuned with the parameter $\tilde{\lambda}$. If $n$ is larger than the amount of training samples $N$ then for $\tilde{\lambda} \to 0$ the network will interpolate the data while having minimal weights, resulting in the \textit{minimum norm network} $\mathcal{RN}_{w^{\text{min}}, \omega}$. \[ \mathcal{RN}_{w^{\text{min}}, \omega} \text{ randomized shallow Neural network with weights } w^{\text{min}}: \] \[ w^{\text{min}} \in \argmin_{w \in \mathbb{R}^n} \norm{w}, \text{ s.t. } \mathcal{RN}_{w,\omega}(x_i^{\text{train}}) = y_i^{\text{train}}, \, \forall i \in \left\{1,\dots,N\right\}. \] For $\tilde{\lambda} \to \infty$ the learned function will resemble the data less and less with the weights approaching $0$. .\par In order to make the notation more convinient in the follwoing the $\omega$ used to express the realised random parameters will no longer be explizitly mentioned. \begin{Definition} \label{def:kink} Let $\mathcal{RN}_w$ be a randomized shallow Neural Network according to Definition~\ref{def:rsnn}, then kinks depending on the random parameters can be observed. \[ \mathcal{RN}_w(x) = \sum_{k = 1}^n w_k \sigma(b_k + v_kx) \] Because we specified $\sigma(y) \coloneqq \max\left\{0, y\right\}$ a kink in $\sigma$ can be observed at $\sigma(0) = 0$. As $b_k + v_kx = 0$ for $x = -\frac{b_k}{v_k}$ we define the following: \begin{enumerate}[label=(\alph*)] \item Let $\xi_k \coloneqq -\frac{b_k}{v_k}$ be the k-th kink of $\mathcal{RN}_w$. \item Let $g_{\xi}(\xi_k)$ be the density of the kinks $\xi_k = - \frac{b_k}{v_k}$ in accordance to the distributions of $b_k$ and $v_k$. \item Let $h_{k,n} \coloneqq \frac{1}{n g_{\xi}(\xi_k)}$ be the average estmated distance from kink $\xi_k$ to the next nearest one. \end{enumerate} \end{Definition} In order to later prove the connection between randomised shallow Neural Networks and regression splines, we first take a look at a smooth approximation of the RSNN. \begin{Definition}[Smooth Approximation of Randomized Shallow Neural Network] \label{def:srsnn} Let $RS_{w}$ be a randomized shallow Neural Network according to Definition~\ref{def:RSNN} with weights $w$ and kinks $\xi_k$ with corresponding kink density $g_{\xi}$ as given by Definition~\ref{def:kink}. In order to smooth the RSNN consider following kernel for every $x$: \[ \kappa_x(s) \coloneqq \mathds{1}_{\left\{\abs{s} \leq \frac{1}{2 \sqrt{n} g_{\xi}(x)}\right\}}(s)\sqrt{n} g_{\xi}(x), \, \forall s \in \mathbb{R} \] Using this kernel we define a smooth approximation of $\mathcal{RN}_w$ by \[ f^w(x) \coloneqq \int_{\mathds{R}} \mathcal{RN}_w(x-s) \kappa_x(s) ds. \] \end{Definition} Note that the kernel introduced in Definition~\ref{def:srsnn} satisfies $\int_{\mathbb{R}}\kappa_x dx = 1$. While $f^w$ looks highly similar to a convolution, it differs slightly as the kernel $\kappa_x(s)$ is dependent on $x$. Therefore only $f^w = (\mathcal{RN}_w * \kappa_x)(x)$ is well defined, while $\mathcal{RN}_w * \kappa$ is not. Now we take a look at weighted regression splines. Later we will prove that the ridge penalized neural network as defined in Definition~\ref{def:rpnn} converges a weighted regression spline, as the amount of hidden nodes is grown to inifity. \begin{Definition}[Adapted Weighted regression spline] \label{def:wrs} Let $x_i^{\text{train}}, y_i^{\text{train}} \in \mathbb{R}, i \in \left\{1,\dots,N\right\}$ be trainig data. For a given $\lambda \in \mathbb{R}_{>0}$ and a function $g: \mathbb{R} \to \mathbb{R}_{>0}$ the weighted regression spline $f^{*, \lambda}_g$ is given by \[ f^{*, \lambda}_g :\in \argmin_{\substack{f \in \mathcal{C}^2(\mathbb{R}) \\ \supp(f) \subseteq \supp(g)}} \underbrace{\left\{ \overbrace{\sum_{i = 1}^N \left(f(x_i^{\text{train}}) - y_i^{\text{train}}\right)^2}^{L(f)} + \lambda g(0) \int_{\supp(g)}\frac{\left(f''(x)\right)^2}{g(x)} dx\right\}}_{\eqqcolon F^{\lambda, g}(f)}. \] \todo{Anforderung an Ableitung von f, doch nicht?} \end{Definition} Similary to ridge weight penalized neural networks the parameter $\lambda$ controls a trade-off between accuracy on the training data and smoothness or low second dreivative. For $g \equiv 1$ and $\lambda \to 0$ the resuling function $f^{*, 0+}$ will interpolate the training data while minimizing the second derivative. Such a function is known as cubic spline interpolation. \todo{cite cubic spline} \[ f^{*, 0+} \text{ smooth spline interpolation: } \] \[ f^{*, 0+} \coloneqq \lim_{\lambda \to 0+} f^{*, \lambda}_1 \in \argmin_{\substack{f \in \mathcal{C}^2\mathbb{R}, \\ f(x_i^{\text{train}}) = y_i^{\text{train}}}} = \left( \int _{\mathbb{R}} (f''(x))^2dx\right). \] For $\lambda \to \infty$ on the other hand $f_g^{*\lambda}$ converges to linear regression of the data. \begin{Definition}[Spline approximating Randomised Shallow Neural Network] \label{def:sann} Let $\mathcal{RN}$ be a randomised shallow Neural Network according to Definition~\ref{def:RSNN} and $f^{*, \lambda}_g$ be the weighted regression spline as introduced in Definition~\ref{def:wrs}. Then the randomised shallow neural network approximating $f^{*, \lambda}_g$ is given by \[ \mathcal{RN}_{\tilde{w}}(x) = \sum_{k = 1}^n \tilde{w}_k \sigma(b_k + v_k x), \] with the weights $\tilde{w}_k$ defined as \[ \tilde{w}_k \coloneqq \frac{h_{k,n} v_k}{\mathbb{E}[v^2 \vert \xi = \xi_k]} (f_g^{*, \lambda})''(\xi_k). \] \end{Definition} The approximating nature of the network in Definition~\ref{def:sann} can be seen by LOOKING \todo{besseres Wort finden} at the first derivative of $\mathcal{RN}_{\tilde{w}}(x)$ which is given by \begin{align} \frac{\partial \mathcal{RN}_{\tilde{w}}}{\partial x} \Big{|}_{x} &= \sum_k^n \tilde{w}_k \mathds{1}_{\left\{b_k + v_k x > 0\right\}}(v_k) = \sum_{\substack{k \in \mathbb{N} \\ \xi_k < x}} \tilde{w}_k v_k \nonumber \\ &= \frac{1}{n} \sum_{\substack{k \in \mathbb{N} \\ \xi_k < x}} \frac{v_k^2}{g_{\xi}(\xi_k) \mathbb{E}[v^2 \vert \xi = \xi_k]} (f_g^{*, \lambda})''(\xi_k). \label{eq:derivnn} \end{align} \todo{gescheite Ableitungs Notation} As the expression (\ref{eq:derivnn}) behaves similary to a Riemann-sum for $n \to \infty$ it will converge to the first derievative of $f^{*,\lambda}_g$. A formal proof of this behaviour is given in Lemma~\ref{lem:s0}. In order to formulate the theorem describing the convergence of $RN_w$ we need to make a couple of assumptions. \todo{Bessere Formulierung} \begin{Assumption}~ \label{ass:theo38} \begin{enumerate}[label=(\alph*)] \item The probability density fucntion of the kinks $\xi_k$, namely $g_{\xi}$ as defined in Definition~\ref{def:kink} exists and is well defined. \item The density function $g_\xi$ has compact support on $\supp(g_{\xi})$. \item The density function $g_{\xi}$ is uniformly continuous on $\supp(g_{\xi})$. \item $g_{\xi}(0) \neq 0$. \item $\frac{1}{g_{\xi}}\Big|_{\supp(g_{\xi})}$ is uniformly continous on $\supp(g_{\xi})$. \item The conditional distribution $\mathcal{L}(v_k|\xi_k = x)$ is uniformly continous on $\supp(g_{\xi})$. \item $\mathbb{E}\left[v_k^2\right] < \infty$. \end{enumerate} \end{Assumption} As we will prove the prorpsition in the Sobolev space, we hereby introduce it and its inuced\todo{richtiges wort?} norm. \begin{Definition}[Sobolev Space] For $K \subset \mathbb{R}^n$ open and $1 \leq p \leq \infty$ we define the Sobolev space $W^{k,p}(K)$ as the space containing all real valued functions $u \in L^p(K)$ such that for every multi-index $\alpha \in \mathbb{N}^n$ with $\abs{\alpha} \leq k$ the mixed parial derivatives \[ u^{(\alpha)} = \frac{\partial^{\abs{\alpha}} u}{\partial x_1^{\alpha_1} \dots \partial x_n^{\alpha_n}} \] exists in the weak sense and \[ \norm{u^{(\alpha)}}_{L^p} < \infty. \] \todo{feritg machen} \label{def:sobonorm} The natural norm of the sobolev space is given by \[ \norm{f}_{W^{k,p}(K)} = \begin{cases} \left(\sum_{\abs{\alpha} \leq k} \norm{f^{(\alpha)}}^p_{L^p}\right)^{\nicefrac{1}{p}},& \text{for } 1 \leq p < \infty \\ max_{\abs{\alpha} \leq k}\left\{f^{(\alpha)}\right\},& \text{for } p = \infty \end{cases} . \] \end{Definition} With these assumption in place we can formulate the main theorem. \todo{Bezug Raum} \begin{Theorem}[Ridge weight penaltiy corresponds to weighted regression spline] \label{theo:main1} For $N \in \mathbb{N}$ arbitrary training data \(\left(x_i^{\text{train}}, y_i^{\text{train}} \right)\) and $\mathcal{RN}^{*, \tilde{\lambda}}, f_g^{*, \lambda}$ according to Definition~\ref{def:rpnn} and Definition~\ref{def:wrs} respectively with Assumption~\ref{ass:theo38} it holds \begin{equation} \label{eq:main1} \plimn \norm{\mathcal{RN^{*, \tilde{\lambda}}} - f^{*, \lambda}_{g}}_{W^{1,\infty}(K)} = 0. \end{equation} With \begin{align*} g(x) & \coloneqq g_{\xi}(x)\mathbb{E}\left[ v_k^2 \vert \xi_k = x \right], \forall x \in \mathbb{R}, \\ \tilde{\lambda} & \coloneqq \lambda n g(0). \end{align*} \end{Theorem} We will proof Theo~\ref{theo:main1} by showing that \begin{equation} \label{eq:main2} \plimn \norm{\mathcal{RN}^{*, \tilde{\lambda}} - f^{w^*}}_{W^{1, \infty}(K)} = 0 \end{equation} and \begin{equation} \label{eq:main3} \plimn \norm{f^{w^*} - f_g^{*, \lambda}}_{W^{1,\infty}(K)} = 0 \end{equation} and then using the triangle inequality to follow (\ref{eq:main1}). In order to prove (\ref{eq:main2}) and (\ref{eq:main3}) we will need to introduce a number of auxiliary lemmmata, proves to these will be provided in the appendix, as they would SPRENGEN DEN RAHMEN. \begin{Lemma}[Poincar\'e typed inequality] \label{lem:pieq} Let \(f:\mathbb{R} \to \mathbb{R}\) differentiable with \(f' : \mathbb{R} \to \mathbb{R}\) Lesbeque integrable. Then for \(K=[a,b] \subset \mathbb{R}\) with \(f(a)=0\) it holds that \begin{equation*} \label{eq:pti1} \exists C_K^{\infty} \in \mathbb{R}_{>0} : \norm{f}_{w^{1,\infty}(K)} \leq C_K^{\infty} \norm{f'}_{L^{\infty}(K)}. \end{equation*} If additionaly \(f'\) is differentiable with \(f'': \mathbb{R} \to \mathbb{R}\) Lesbeque integrable then additionally \begin{equation*} \label{eq:pti2} \exists C_K^2 \in \mathbb{R}_{>0} : \norm{f}_{W^{1,\infty}(K)} \leq C_K^2 \norm{f''}_{L^2(K)}. \end{equation*} % \proof % With the fundamental theorem of calculus, if % \(\norm{f}_{L^{\infty}(K)}<\infty\) we get % \begin{equation} % \label{eq:f_f'} % \norm{f}_{L^{\infty}(K)} = \sup_{x \in K}\abs{\int_a^x f'(s) ds} \leq % \sup_{x \in K}\abs{\int_a^x \sup_{y \in K} \abs{f'(y)} ds} \leq \abs{b-a} % \sup_{y \in K}\abs{f'(y)}. % \end{equation} % Using this we can bound \(\norm{f}_{w^{1,\infty}(K)}\) by % \[ % \norm{f}_{w^{1,\infty}(K)} \stackrel{\text{Def~\ref{def:sobonorm}}}{=} % \max\left\{\norm{f}_{L^{\infty}(K)}, % \norm{f'}_{L^{\infty}(K)}\right\} % \stackrel{(\ref{eq:f_f'})}{\leq} max\left\{\abs{b-a}, % 1\right\}\norm{f'}_{L^{\infty}(K)}. % \] % With \(C_k^{\infty} \coloneqq max\left\{\abs{b-a}, 1\right\}\) we % get (\ref{eq:pti1}). % By using the Hölder inequality, we can proof the second claim. % \begin{align*} % \norm{f'}_{L^{\infty}(K)} &= \sup_{x \in K} \abs{\int_a^bf''(y) % \mathds{1}_{[a,x]}(y)dy} \leq \sup_{x \in % K}\norm{f''\mathds{1}_{[a,x]}}_{L^1(K)}\\ % &\hspace{-6pt} \stackrel{\text{Hölder}}{\leq} sup_{x % \in % K}\norm{f''}_{L^2(K)}\norm{\mathds{1}_{[a,x]}}_{L^2(K)} % = \abs{b-a}\norm{f''}_{L^2(K)}. % \end{align*} % Thus (\ref{eq:pti2}) follows with \(C_K^2 \coloneqq % \abs{b-a}C_K^{\infty}\). % \qed \end{Lemma} \begin{Lemma} \label{lem:cnvh} Let $\mathcal{RN}$ be a shallow Neural network. For \(\varphi : \mathbb{R}^2 \to \mathbb{R}\) uniformly continous such that \[ \forall x \in \supp(g_{\xi}) : \mathbb{E}\left[\varphi(\xi, v) \frac{1}{n g_{\xi}(\xi)} \vert \xi = x \right] < \infty, \] it holds, that \[ \plimn \sum_{k \in \kappa : \xi_k < T} \varphi(\xi_k, v_k) h_{k,n} =\int_{\min\left\{C_{g_{\xi}}^l, T\right\}}^{min\left\{C_{g_{\xi}}^u,T\right\}} \mathbb{E}\left[\varphi(\xi, v) \vert \xi = x \right] dx \] uniformly in \(T \in K\). % \proof % For \(T \leq C_{g_{\xi}}^l\) both sides equal 0, so it is sufficient to % consider \(T > C_{g_{\xi}}^l\). With \(\varphi\) and % \(\nicefrac{1}{g_{\xi}}\) uniformly continous in \(\xi\), % \begin{equation} % \label{eq:psi_stet} % \forall \varepsilon > 0 : \exists \delta(\varepsilon) : \forall % \abs{\xi - \xi'} < \delta(\varepsilon) : \abs{\varphi(\xi, v) % \frac{1}{g_{\xi}(\xi)} - \varphi(\xi', v) % \frac{1}{g_{\xi}(\xi')}} < \varepsilon % \end{equation} % uniformly in \(v\). In order to % save space we use the notation \((a \wedge b) \coloneqq \min\{a,b\}\) for $a$ and $b % \in \mathbb{R}$. W.l.o.g. assume \(\sup(g_{\xi})\) in an % intervall. By splitting the interval in disjoint strips of length \(\delta % \leq \delta(\varepsilon)\) we get: % \[ % \underbrace{\sum_{k \in \kappa : \xi_k < T} \varphi(\xi_k, v_k) % \frac{\bar{h}_k}{2}}_{\circled{1}} = % \underbrace{\sum_{l \in \mathbb{Z}: % \left[\delta l, \delta (l + 1)\right] \subseteq % \left[C_{g_{\xi}}^l, C_{g_{\xi}}^u \wedge T % \right]}}_{\coloneqq \, l \in I_{\delta}} \left( \, \sum_{\substack{k \in \kappa\\ % \xi_k \in \left[\delta l, \delta (l + 1)\right]}} % \varphi\left(\xi_k, v_k\right)\frac{\bar{h}_k}{2} \right) % \] % Using (\ref{eq:psi_stet}) we can approximate $\circled{1}$ by % \begin{align*} % \circled{1} & \approx \sum_{l \in I_{\delta}} \left( \, \sum_{\substack{k \in \kappa\\ % \xi_k \in \left[\delta l, \delta (l + 1)\right]}} % \left(\varphi\left(l\delta, v_k\right)\frac{1}{g_{\xi}(l\delta)} % \pm \varepsilon\right)\frac{1}{n} \underbrace{\frac{\abs{\left\{m \in % \kappa : \xi_m \in [\delta l, \delta(l + 1)]\right\}}}{\abs{\left\{m \in % \kappa : \xi_m \in [\delta l, \delta(l + 1)]\right\}}}}_{= % 1}\right) \\ % % \intertext{} % &= \sum_{l \in I_{\delta}} \left( \frac{ \sum_{ \substack{k \in \kappa\\ % \xi_k \in \left[\delta l, \delta (l + 1)\right]}} % \varphi\left(l\delta, v_k\right)} % {\abs{\left\{m \in % \kappa : \xi_m \in [\delta l, \delta(l + 1)]\right\}}}\frac{\abs{\left\{m \in % \kappa : \xi_m \in [\delta l, \delta(l + % 1)]\right\}}}{ng_{\xi}(l\delta)}\right) \pm \varepsilon .\\ % \intertext{We use the mean to approximate the number of kinks in % each $\delta$-strip, as it follows a bonomial distribution this % amounts to % \[ % \mathbb{E}\left[\abs{\left\{m \in \kappa : \xi_m \in [\delta l, % \delta(l + 1)]\right\}\right]} = n \int_{[\delta l, \delta (l + % 1)]} g_{\xi}(x)dx \approx n (\delta g_{\xi}(l\delta) \pm % \tilde{\varepsilon}). % \] % Bla Bla Bla $v_k$} % \circled{1} & \approx % \end{align*} \end{Lemma} \begin{Lemma}[Step 0] For any $\lambda > 0$, training data $(x_i^{\text{train}} y_i^{\text{train}}) \in \mathbb{R}^2$, with $ i \in \left\{1,\dots,N\right\}$ and subset $K \subset \mathbb{R}$ the spline approximating randomized shallow neural network $\mathcal{RN}_{\tilde{w}}$ converges to the regression spline $f^{*, \lambda}_g$ in $\norm{.}_{W^{1,\infty}(K)}$ as the node count $n$ increases, \begin{equation} \label{eq:s0} \plimn \norm{\mathcal{RN}_{\tilde{w}} - f^{*, \lambda}_g}_{W^{1, \infty}(K)} = 0 \end{equation} \proof Using Lemma~\ref{lem:pieq} it is sufficient to show \[ \plimn \norm{\mathcal{RN}_{\tilde{w}}' - (f^{*, \lambda}_g)'}_{L^{\infty}} = 0. \] This can be achieved by using Lemma~\ref{lem:cnvh} with $\varphi(\xi_k, v_k) = \frac{v_k^2}{\mathbb{E}[v^2|\xi = z]} (f^{*, \lambda}_w)''(\xi_k) $ thus obtaining \begin{align*} \plimn \frac{\partial \mathcal{RN}_{\tilde{w}}}{\partial x} \stackrel{(\ref{eq:derivnn})}{=} & \plimn \sum_{\substack{k \in \mathbb{N} \\ \xi_k < x}} \frac{v_k^2}{\mathbb{E}[v^2 \vert \xi = \xi_k]} (f_g^{*, \lambda})''(\xi_k) h_{k,n} \stackrel{\text{Lemma}~\ref{lem:cnvh}}{=} \\ \stackrel{\phantom{(\ref{eq:derivnn})}}{=} & \int_{\min\left\{C_{g_{\xi}}^l,T\right\}}^{min\left\{C_{g_{\xi}}^u,T\right\}} \mathbb{E}\left[\frac{v^2}{\mathbb{E}[v^2|\xi = z]} (f^{*, \lambda}_w)''(\xi) \vert \xi = x \right] dx \equals^{\text{Tower-}}_{\text{property}} \\ \stackrel{\phantom{(\ref{eq:derivnn})}}{=} & \int_{\min\left\{C_{g_{\xi}}^l, T\right\}}^{min\left\{C_{g_{\xi}}^u,T\right\}}(f^{*,\lambda}_w)''(x) dx. \end{align*} By the fundamental theorem of calculus and $\supp(f') \subset \supp(f)$, (\ref{eq:s0}) follows with Lemma~\ref{lem:pieq}. \qed \end{Lemma} \begin{Lemma}[Step 2] For any $\lambda > 0$ and training data $(x_i^{\text{train}}, y_i^{\text{train}}) \in \mathbb{R}^2, \, i \in \left\{1,\dots,N\right\}$, we have \[ \plimn F^{\tilde{\lambda}}_n(\mathcal{RN}_{\tilde{w}}) = F^{\lambda, g}(f^{*, \lambda}_g) = 0. \] \proof This can be prooven by showing \end{Lemma} \begin{Lemma}[Step 3] For any $\lambda > 0$ and training data $(x_i^{\text{train}}, y_i^{\text{train}}) \in \mathbb{R}^2, \, i \in \left\{1,\dots,N\right\}$, with $w^*$ and $\tilde{\lambda}$ as defined in Definition~\ref{def:rpnn} and Theroem~\ref{theo:main1} respectively, it holds \[ \plimn \norm{\mathcal{RN}^{*,\tilde{\lambda}} - f^{w*, \tilde{\lambda}}}_{W^{1,\infty}(K)} = 0. \] \end{Lemma} \begin{Lemma}[Step 4] For any $\lambda > 0$ and training data $(x_i^{\text{train}}, y_i^{\text{train}}) \in \mathbb{R}^2, \, i \in \left\{1,\dots,N\right\}$, with $w^*$ and $\tilde{\lambda}$ as defined in Definition~\ref{def:rpnn} and Theroem~\ref{theo:main1} respectively, it holds \[ \plimn \abs{F_n^{\lambda}(\mathcal{RN}^{*,\tilde{\lambda}}) - F^{\lambda, g}(f^{w*, \tilde{\lambda}})} = 0. \] \end{Lemma} \begin{Lemma}[Step 7] For any $\lambda > 0$ and training data $(x_i^{\text{train}}, y_i^{\text{train}}) \in \mathbb{R}^2, \, i \in \left\{1,\dots,N\right\}$, for any sequence of functions $f^n \in W^{2,2}$ with \[ \plimn F^{\lambda, g} (f^n) = F^{\lambda, g}(f^{*, \lambda}), \] it follows \[ \plimn \norm{f^n - f^{*, \lambda}} = 0. \] \end{Lemma} \textcite{heiss2019} further show a link between ridge penalized networks and randomized shallow neural networks which are trained with gradient descent which is stopped after a certain amount of iterations. \newpage \subsection{Simulations} In the following the behaviour described in Theorem~\ref{theo:main1} is visualized in a simulated example. For this two sets of training data have been generated. \begin{itemize} \item $\text{data}_A = (x_{i, A}^{\text{train}}, y_{i,A}^{\text{train}})$ with \begin{align*} x_{i, A}^{\text{train}} &\coloneqq -\pi + \frac{2 \pi}{5} (i - 1), i \in \left\{1, \dots, 6\right\}, \\ y_{i, A}^{\text{train}} &\coloneqq \sin( x_{i, A}^{\text{train}}). \phantom{(i - 1), i \in \left\{1, \dots, 6\right\}} \end{align*} \item $\text{data}_b = (x_{i, B}^{\text{train}}, y_{i, B}^{\text{train}})$ with \begin{align*} x_{i, B}^{\text{train}} &\coloneqq \pi\frac{i - 8}{7}, i \in \left\{1, \dots, 15\right\}, \\ y_{i, B}^{\text{train}} &\coloneqq \sin( x_{i, B}^{\text{train}}). \phantom{(i - 1), i \in \left\{1, \dots, 6\right\}} \end{align*} \end{itemize} For the $\mathcal{RN}$ the random weights are distributed as follows \begin{align*} \xi_i &\stackrel{i.i.d.}{\sim} \text{Unif}(-5,5), \\ v_i &\stackrel{i.i.d.}{\sim} \mathcal{N}(0, 5), \\ b_i &\stackrel{\phantom{i.i.d.}}{\sim} -\xi_i v_i. \end{align*} Note that by the choices for the distributions $g$ as defined in Theorem~\ref{theo:main1} would equate to $g(x) = \frac{\mathbb{E}[v_k^2|\xi_k = x]}{10}$. In order to utilize the smoothing spline implemented in Mathlab, $g$ has been simplified to $g \equiv \frac{1}{10}$ instead. For all figures $f_1^{*, \lambda}$ has been calculated with Matlab's ..... As ... minimizes \[ \bar{\lambda} \sum_{i=1}^N(y_i^{train} - f(x_i^{train}))^2 + (1 - \bar{\lambda}) \int (f''(x))^2 dx \] the smoothing parameter used for fittment is $\bar{\lambda} = \frac{1}{1 + \lambda}$. The parameter $\tilde{\lambda}$ for training the networks is chosen as defined in Theorem~\ref{theo:main1} and each one is trained on the full training data for 5000 iterations using gradient descent. The results are given in Figure~\ref{blblb}, here it can be seen that in the intervall of the traing data $[-\pi, \pi]$ the neural network and smoothing spline are nearly identical, coinciding with the proposition. \input{Plots/RN_vs_RS} %%% Local Variables: %%% mode: latex %%% TeX-master: "main" %%% End: