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chapter1.tex
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chapter1.tex
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\section{Introduction}
\paragraph{Exercise 1.1}
We seek to minimize $E$ by setting its derivative with respect to $\textbf{w}$
to zero. We can consider each weight $w_i$ separately, so we have
\begin{align*}
\frac{\partial E}{\partial w_i} = 0 &\iff
\sum_{n=1}^N \{[ w_0 + w_1x_n + w_2x_n^2 + \dots + w_Mx_n^M - t_n]x_n^i\} = 0\\
&\iff \sum_{n=1}^N \{w_0x_n^i + w_1x_n^{i+1} + w_2x_n^{i+2} + \dots + w_Mx_n^{i+M}\} = \sum_{n=1}^{N} (x_n)^it_n\\
&\iff \sum_{j=0}^M \{\sum_{n=1}^N (x_n)^{i + j} w_j\} = \sum_{n=1}^N (x_n)^it_n
\end{align*}
This result corresponds to the set of linear equations shown in the text.
\paragraph{Exercise 1.2}
Again, we seek to minimize $\tilde{E}$, by setting each one of the partial derivatives
with respect to $w_i$ to zero. We have
\begin{align*}
\frac{\partial\tilde{E}}{\partial w_i} = 0 &\iff \sum_{n=1}^N \{[y(x_n, \textbf{w}) - t_n]x_n^i\} + \lambda w_i = 0\\
&\iff \sum_{n=1}^N \{[w_0 + w_1x_n + w_2x_n^2 + \dots + w_Mx_n^M - t_n]x_n^i\} + \lambda w_i = 0\\
&\iff \sum_{n=1}^N \{w_0x_n^i + w_1x_n^{i+1} + w_2x_n^{i+2} + \dots + w_Mx_n^{i+M}\} + \lambda w_i = \sum_{n=1}^N (x_n)^it_n\\
\end{align*}
The coefficients $\textbf{w} = \{w_i\}$ that minimize $\tilde{E}(\textbf{w})$ are given by the solution
to the following set of linear equations:
\begin{equation*}
\sum_{j=0}^M A_{ij}w_j + \lambda w_i = T_i
\end{equation*}
where $A_{ij} = \sum_{n=1}^N (x_n)^{i+j}$ and $T_i = \sum_{n=1}^N (x_n)^it_n$.
\paragraph{Exercise 1.3}
We want to determine the probability $p(apple)$ of randomly extracting an apple from
a randomly chosen box.
By the law of total probability, we have
\begin{equation*}
p(apple) = p(apple, r) + p(apple, b) + p(apple, g)
\end{equation*}
where $r$, $b$ and $g$ denote the red, blue and green boxes, respectively.
We also apply the product rule to each term of the sum, obtaining:
\begin{align*}
p(apple) &= p(apple|r)p(r) + p(apple|b)p(b) + p(apple|g)p(g)\\
&= \frac{3}{10}\frac{1}{5} + \frac{1}{2}\frac{1}{5} + \frac{3}{10}\frac{3}{5}\\
&= \frac{3}{50} + \frac{1}{10} + \frac{9}{50}\\
&= \frac{17}{50}
\end{align*}
Furthermore, the exercise asks use to determine the probability of having selected the]
green box, given that we have extracted an orange, corresponding to $p(g|orange)$.
We can use Bayes' theorem to obtain
\begin{equation*}
p(g|orange) = \frac{p(orange|g)p(g)}{p(orange)}
\end{equation*}
We already know $p(g)$, and we can compute $p(orange)$ in the same way we computed $p(apple)$,
obtaining
\begin{align*}
p(orange) &= p(orange, r) + p(orange, b) + p(orange, g)\\
&= p(orange|r)p(r) + p(orange|b)p(b) + p(orange|g)p(g)\\
&= \frac{4}{10}\frac{1}{5} + \frac{1}{2}\frac{1}{5} + \frac{3}{10}\frac{3}{5}\\
&= \frac{2}{25} + \frac{1}{10} + \frac{9}{50}\\
&= \frac{18}{50} = \frac{9}{25}
\end{align*}
Finally, we obtain:
\begin{align*}
p(g|orange) &= \frac{p(orange|g)p(g)}{p(orange)}\\
&= \frac{\frac{3}{10}\frac{3}{5}}{\frac{9}{25}}\\
&= \frac{3}{10}\frac{3}{5}\frac{25}{9}\\
&= \frac{225}{450} = \frac{1}{2}
\end{align*}