<p>I am trying to understand the solution to part 4 of problem 3 from the midterm exam <a href="https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-867-machine-learning-fall-2006/exams/midterm_f03soln.pdf" rel="nofollow noreferrer">6.867 Machine learning: Mid-term exam (October 15, 2003)</a>.</p>
<p>For reproducibility, here is problem 3.</p>
<blockquote>
<p>We consider here linear and non-linear support vector machines (SVM) of the form:
<span class="math-container">$$
\begin{equation}
\begin{aligned}
\min w_{1}^{2} / 2 &amp; \text { subject to } y_{i}\left(w_{1} x_{i}+w_{0}\right)-1 \geq 0, \quad i=1, \ldots, n, \text { or } \\
\min \mathbf{w}^{T} \mathbf{w} / 2 &amp; \text { subject to } y_{i}\left(\mathbf{w}^{T} \Phi_{i}+w_{0}\right)-1 \geq 0, \quad i=1, \ldots, n
\end{aligned}
\end{equation}
$$</span>
where <span class="math-container">$\Phi_{i}$</span> is a feature vector constructed from the corresponding real-valued input <span class="math-container">$x_{i}$</span>. We wish to compare the simple linear SVM classifier <span class="math-container">$\left(w_{1} x+w_{0}\right)$</span> and the non-linear classifier
<span class="math-container">$\left(\mathbf{w}^{T} \Phi+w_{0}\right)$</span>, where <span class="math-container">$\Phi=\left[x, x^{2}\right]^{T}$</span>.</p>
</blockquote>
<p>Here is part 4 of the same problem.</p>
<blockquote>
<p>In general, is the margin we would attain using scaled feature vectors
<span class="math-container">$\Phi=\left[2 x, 2 x^{2}\right]^{T}$</span></p>
<ul>
<li>greater</li>
<li>equal</li>
<li>smaller</li>
<li>any of the above</li>
</ul>
</blockquote>
<p>The correct answer is the first (<strong>greater</strong>). Why is that the case?</p>

CC BY-SA 4.0
2021-01-09T07:35:16.547Z
@DuttaA would you clarify a bit more? I didn't get the point.
user871621
<machine-learning><classification><support-vector-machine><homework><kernel-trick>
Why is the margin attained with $\Phi=\left[2 x, 2 x^{2}\right]^{T}$ greater than the margin attained with $\Phi=\left[x, x^{2}\right]^{T}$?
user871621
# Why is the margin attained with $\Phi=\left[2 x, 2 x^{2}\right]^{T}$ greater than the margin attained with $\Phi=\left[x, x^{2}\right]^{T}$?

(Asked by: username_0 on 2021-01-08)

I am trying to understand the solution to part 4 of problem 3 from the midterm exam [6.867 Machine learning: Mid-term exam (October 15, 2003)](https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-867-machine-learning-fall-2006/exams/midterm_f03soln.pdf).

For reproducibility, here is problem 3.

> We consider here linear and non-linear support vector machines (SVM) of the form:
> $$
> \begin{equation}
> \begin{aligned}
> \min w\_{1}^{2} / 2 & \text { subject to } y\_{i}\left(w\_{1} x\_{i}+w\_{0}\right)-1 \geq 0, \quad i=1, \ldots, n, \text { or } \\
> \min \mathbf{w}^{T} \mathbf{w} / 2 & \text { subject to } y\_{i}\left(\mathbf{w}^{T} \Phi\_{i}+w\_{0}\right)-1 \geq 0, \quad i=1, \ldots, n
> \end{aligned}
> \end{equation}
> $$
> where $\Phi\_{i}$ is a feature vector constructed from the corresponding real-valued input $x\_{i}$. We wish to compare the simple linear SVM classifier $\left(w\_{1} x+w\_{0}\right)$ and the non-linear classifier
> $\left(\mathbf{w}^{T} \Phi+w\_{0}\right)$, where $\Phi=\left[x, x^{2}\right]^{T}$.

Here is part 4 of the same problem.

> In general, is the margin we would attain using scaled feature vectors
> $\Phi=\left[2 x, 2 x^{2}\right]^{T}$
>
> * greater
> * equal
> * smaller
> * any of the above

The correct answer is the first (**greater**). Why is that the case?

**Tags:** machine-learning, classification, support-vector-machine, homework, kernel-trick

> **Comments:**
> **username_0**: @DuttaA would you clarify a bit more? I didn't get the point.  

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## 0 answers

