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# 计算机代写|机器学习代写Machine Learning代考|ENGG3300 Margin and Support Vector

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## 计算机代写|机器学习代写Machine Learning代考|Margin and Support Vector

Given a training set $D=\left{\left(\boldsymbol{x}_1, y_1\right),\left(\boldsymbol{x}_2, y_2\right), \ldots,\left(\boldsymbol{x}_m, y_m\right)\right}$, where $y_i \in{-1,+1}$. The basic idea of classification is to utilize the training set $D$ to find a separating hyperplane in the sample space that can separate samples of different classes. However, there could be multiple qualified separating hyperplanes, as shown in $-$ Figure 6.1, which one should be chosen?

Intuitively, we should choose the one right in the middle of two classes, that is, the red one in $\bullet$ Figure 6.1, since this separating hyperplane has the best “tolerance” to local data perturbation. For example, the samples not in the training set could be closer to the decision boundary due to the noises or limitations of the training set. As a result, many separating hyperplanes that perform well on the training set will make mistakes, whereas the red hyperplane is less likely to be affected. In other words, this separating hyperplane has the strongest generalization ability and the most robust classification results.
A separating hyperplane in the sample space can be expressed as the following linear function:
$$\boldsymbol{w}^{\top} \boldsymbol{x}+b=0$$
where $\boldsymbol{w}=\left(w_1 ; w_2 ; \ldots ; w_d\right)$ is the normal vector which controls the direction of the hyperplane, and $b$ is the bias which controls the distance between the hyperplane and the origin. The normal vector $\boldsymbol{w}$ and the bias $b$ determine the separating hyperplane, denoted by $(\boldsymbol{w}, b)$. The distance from any point $\boldsymbol{x}$ in the sample space to the hyperplane $(\boldsymbol{w}, b)$ can be written as
$$r=\frac{\left|\boldsymbol{w}^{\top} \boldsymbol{x}+b\right|}{|\boldsymbol{w}|}$$

## 计算机代写|机器学习代写Machine Learning代考|Dual Problem

We wish to solve (6.6) to obtain the maximum margin separating hyperplane model
$$f(\boldsymbol{x})=\boldsymbol{w}^{\top} \boldsymbol{x}+b,$$
where $\boldsymbol{w}$ and $b$ are the model parameters. (6.6) is a convex quadratic programming problem, which can be solved with existing optimization packages. However, there are more efficient methods.

Applying Lagrange multipliers to (6.6) leads to its dual problem. To be specific, introducing a Lagrange multiplier $\alpha_i \geqslant 0$ to each constraint in (6.6) gives the Lagrange function
$$L(\boldsymbol{w}, b, \boldsymbol{\alpha})=\frac{1}{2}|\boldsymbol{w}|^2+\sum_{i=1}^m \alpha_i\left(1-y_i\left(\boldsymbol{w}^{\top} \boldsymbol{x}i+b\right)\right),$$ where $\boldsymbol{\alpha}=\left(\alpha_1 ; \alpha_2 ; \ldots ; \alpha_m\right)$. Setting the partial derivatives of $L(\boldsymbol{w}, b, \boldsymbol{\alpha})$ with respect to $\boldsymbol{w}$ and $b$ to 0 gives \begin{aligned} &\boldsymbol{w}=\sum{i=1}^m \alpha_i y_i \boldsymbol{x}i, \ &0=\sum{i=1}^m \alpha_i y_i, \end{aligned}
Substituting (6.9) into (6.8) eliminates $\boldsymbol{w}$ from $L(\boldsymbol{w}, b, \boldsymbol{\alpha})$. Then, with the constraint (6.10), we have the dual problem of (6.6) as
$$\max \alpha \sum{i=1}^m \alpha_i-\frac{1}{2} \sum_{i=1}^m \sum_{j=1}^m \alpha_i \alpha_j y_i y_j \boldsymbol{x}i^{\top} \boldsymbol{x}_j$$ s.t. $\sum{i=1}^m \alpha_i y_i=0$,
$$\alpha_i \geqslant 0, \quad i=1,2, \ldots, m .$$

## 计算机代写|机器学习代写Machine Learning代考|Margin and Support Vector

$$\boldsymbol{w}^{\top} \boldsymbol{x}+b=0$$

$$r=\frac{\left|\boldsymbol{w}^{\top} \boldsymbol{x}+b\right|}{|\boldsymbol{w}|}$$

## 计算机代写机器学习代写Machine Learning代考|Dual Problem

$$f(\boldsymbol{x})=\boldsymbol{w}^{\top} \boldsymbol{x}+b,$$

$$L(\boldsymbol{w}, b, \boldsymbol{\alpha})=\frac{1}{2}|\boldsymbol{w}|^2+\sum_{i=1}^m \alpha_i\left(1-y_i\left(\boldsymbol{w}^{\top} \boldsymbol{x} i+b\right)\right),$$

$$\boldsymbol{w}=\sum i=1^m \alpha_i y_i \boldsymbol{x} i, \quad 0=\sum i=1^m \alpha_i y_i,$$

$$\max \alpha \sum i=1^m \alpha_i-\frac{1}{2} \sum_{i=1}^m \sum_{j=1}^m \alpha_i \alpha_j y_i y_j \boldsymbol{x} i^{\top} \boldsymbol{x}_j$$

$$\alpha_i \geqslant 0, \quad i=1,2, \ldots, m .$$

## MATLAB代写

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