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mhjensen 40023e523f updating week 46
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<a class="navbar-brand" href="week46-bs.html">Week 46: Gradient Boosting Summary and Support Vector Machines</a>
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<!-- navigation toc: --> <li><a href="._week46-bs001.html#___sec0" style="font-size: 80%;">Overview of week 46</a></li>
<!-- navigation toc: --> <li><a href="._week46-bs002.html#___sec1" style="font-size: 80%;">Thursday</a></li>
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<!-- navigation toc: --> <li><a href="._week46-bs007.html#___sec6" style="font-size: 80%;">A \( p \)-dimensional space of features</a></li>
<!-- navigation toc: --> <li><a href="._week46-bs008.html#___sec7" style="font-size: 80%;">The two-dimensional case</a></li>
<!-- navigation toc: --> <li><a href="._week46-bs009.html#___sec8" style="font-size: 80%;">Getting into the details</a></li>
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<!-- navigation toc: --> <li><a href="._week46-bs020.html#___sec19" style="font-size: 80%;">A soft classifier</a></li>
<!-- navigation toc: --> <li><a href="._week46-bs021.html#___sec20" style="font-size: 80%;">Soft optmization problem</a></li>
<!-- navigation toc: --> <li><a href="._week46-bs022.html#___sec21" style="font-size: 80%;">Kernels and non-linearity</a></li>
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<!-- navigation toc: --> <li><a href="#___sec23" style="font-size: 80%;">The problem to solve</a></li>
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<!-- navigation toc: --> <li><a href="._week46-bs030.html#___sec29" style="font-size: 80%;">Back to the more realistic cases</a></li>
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<h2 id="___sec23" class="anchor">The problem to solve </h2>
Using our definition of the kernel We can rewrite again the Lagrangian
$$
{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{z}_j,
$$
subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \) in terms of a convex optimization problem
$$
\frac{1}{2} \boldsymbol{\lambda}^T\begin{bmatrix} y_1y_1K(\boldsymbol{x}_1,\boldsymbol{x}_1) & y_1y_2K(\boldsymbol{x}_1,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_1,\boldsymbol{x}_n) \\
y_2y_1K(\boldsymbol{x}_2,\boldsymbol{x}_1) & y_2y_2(\boldsymbol{x}_2,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_2,\boldsymbol{x}_n) \\
\dots & \dots & \dots & \dots & \dots \\
\dots & \dots & \dots & \dots & \dots \\
y_ny_1K(\boldsymbol{x}_n,\boldsymbol{x}_1) & y_ny_2K(\boldsymbol{x}_n\boldsymbol{x}_2) & \dots & \dots & y_ny_nK(\boldsymbol{x}_n,\boldsymbol{x}_n) \\
\end{bmatrix}\boldsymbol{\lambda}-\mathbb{1}\boldsymbol{\lambda},
$$
subject to \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \). Here we defined the vectors \( \boldsymbol{\lambda} =[\lambda_1,\lambda_2,\dots,\lambda_n] \) and
\( \boldsymbol{y}=[y_1,y_2,\dots,y_n] \).
If we add the slack constants this leads to the additional constraint \( 0\leq \lambda_i \leq C \).
<p>
We can rewrite this (see the solutions below) in terms of a convex optimization problem of the type
$$
\begin{align*}
&\mathrm{min}_{\lambda}\hspace{0.2cm} \frac{1}{2}\boldsymbol{\lambda}^T\boldsymbol{P}\boldsymbol{\lambda}+\boldsymbol{q}^T\boldsymbol{\lambda},\\ \nonumber
&\mathrm{subject\hspace{0.1cm}to} \hspace{0.2cm} \boldsymbol{G}\boldsymbol{\lambda} \preceq \boldsymbol{h} \hspace{0.2cm} \wedge \boldsymbol{A}\boldsymbol{\lambda}=f.
\end{align*}
$$
Below we discuss how to solve these equations. Here we note that the matrix \( \boldsymbol{P} \) has matrix elements \( p_{ij}=y_iy_jK(\boldsymbol{x}_i,\boldsymbol{x}_j) \).
Given a kernel \( K \) and the targets \( y_i \) this matrix is easy to set up. The constraint \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \) leads to \( f=0 \) and \( \boldsymbol{A}=\boldsymbol{y} \). How to set up the matrix \( \boldsymbol{G} \) is discussed later. Here note that the inequalities \( 0\leq \lambda_i \leq C \) can be split up into
\( 0\leq \lambda_i \) and \( \lambda_i \leq C \). These two inequalities define then the matrix \( \boldsymbol{G} \) and the vector \( \boldsymbol{h} \).
<p>
<p>
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