Add ML4CV CLIP + generative models intro

src/year2/machine-learning-for-computer-vision/sections/_generative_models.tex

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+\chapter{Generative models}
+
+
+\begin{description}
+    \item[Generative task] \marginnote{Generative task}
+        Given the training data $\{ x^{(i)} \}$, learn the distribution of the data so that a model can sample new examples:
+        \[ \hat{x}^{(i)} \sim p_\text{gen}(x; \matr{\theta}) \]
+
+        \begin{figure}[H]
+            \centering
+            \includegraphics[width=0.4\linewidth]{./img/generative_task.png}
+        \end{figure}
+
+        \begin{remark}
+            Generative tasks are hard as natural images lay on a low dimensional subspace (i.e., only a tiny subset of all the possible RGB images makes sense).
+
+            \begin{figure}[H]
+                \centering
+                \includegraphics[width=0.6\linewidth]{./img/image_manifold.png}
+            \end{figure}
+        \end{remark}
+
+    \item[Latent vector] \marginnote{Latent vector}
+        Low-dimensional representation to encode an image.
+
+        \begin{example}
+            Face expression depends on 42 muscles. A latent representation for different poses of a face can be represented with a 42-dimensional vector.
+        \end{example}
+
+        It is assumed that the factors of a latent vector are independent or mildly correlated, and can be sampled from a known distribution.
+
+    \item[Generative model] \marginnote{Generative model}
+        Model that takes as input a latent representation and maps it into an output image.
+        \begin{figure}[H]
+            \centering
+            \includegraphics[width=0.7\linewidth]{./img/latent_for_generation.png}
+        \end{figure}
+
+        \begin{remark}
+            An ideal generative model should have the following properties:
+            \begin{itemize}
+                \item Be computationally efficient when sampling.
+                \item Produce high-quality samples.
+                \item Represent the entire training distribution.
+                \item Produce a plausible output from any latent input. Smooth changes to the input should be reflected on the output.
+                \item Have a disentangled latent space (i.e., changing a dimension of the latent space corresponds to interpretable changes in the output image).
+                \item Have the possibility to calculate the probability of the produced images (when the model is probabilistic). 
+            \end{itemize}
+        \end{remark}
+\end{description}
+
+
+
+\section{Metrics}
+
+\begin{description}
+    \item[Expectation/Expected value] \marginnote{Expectation/Expected value}
+        Informally, it is the generalization of the weighted average:
+        \[ 
+            \mathbb{E}_{x \sim p}[ f(\cdot) ] = \sum_{x \in \mathbb{X}} \prob{x} f(x) 
+            \qquad
+            \mathbb{E}_{x \sim p}[ f(\cdot) ] = \int_{x \in \mathbb{X}} \prob{x} f(x) \,dx
+        \]
+
+        \begin{description}
+            \item[Monte Carlo approximation] \marginnote{Monte Carlo approximation}
+                Approximation for expectation using $N$ i.i.d. samples drawn from $p(x)$:
+                \[ \mathbb{E}_{x \sim p}[f(\cdot)] \approx \frac{1}{N} \sum_{x_i \sim p(x)} f(x_i) \]
+        \end{description}
+
+    \item[Self-information] \marginnote{Self-information}
+        Given a probability mass function of an event, the self-information of an event $x$ is defined as:
+        \[ I(x) = -\log_b(\prob{x}) = \log_b\left( \frac{1}{\prob{x}} \right) \]
+        Intuitively, it can be seen as a measure of surprise.
+
+        \begin{example}
+            Consider the toss of a fair coin. The self-information for the outcomes are:
+            \[ I(\texttt{heads}) = I(\texttt{tails}) = \log_2\left( \frac{1}{0.5} \right) = 1 \]
+            If the coin is loaded toward tails with probability:
+            \[ 
+                \prob{\texttt{heads}} = 0.05 
+                \qquad 
+                \prob{\texttt{heads}} = 0.95 
+            \]
+            The self-information is:
+            \[ 
+                I(\texttt{heads}) = \log_2\left( \frac{1}{0.05} \right) = 4.31
+                \qquad
+                I(\texttt{tails}) = \log_2\left( \frac{1}{0.95} \right) = 0.07
+            \]
+        \end{example}
+\end{description}