\def\argmax{\operatornamewithlimits{argmax}}
\def\argmin{\operatornamewithlimits{argmin}}
-\def\expect{\mathds{E}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\def\given{\,\middle\vert\,}
+\def\proba{\operatorname{P}}
+\newcommand{\seq}{{S}}
+\newcommand{\expect}{\mathds{E}}
+\newcommand{\variance}{\mathds{V}}
+\newcommand{\empexpect}{\hat{\mathds{E}}}
+\newcommand{\mutinf}{\mathds{I}}
+\newcommand{\empmutinf}{\hat{\mathds{I}}}
+\newcommand{\entropy}{\mathds{H}}
+\newcommand{\empentropy}{\hat{\mathds{H}}}
+\newcommand{\ganG}{\mathbf{G}}
+\newcommand{\ganD}{\mathbf{D}}
+\newcommand{\ganF}{\mathbf{F}}
+
+\newcommand{\dkl}{\mathds{D}_{\mathsf{KL}}}
+\newcommand{\djs}{\mathds{D}_{\mathsf{JS}}}
+
+\newcommand*{\vertbar}{\rule[-1ex]{0.5pt}{2.5ex}}
+\newcommand*{\horzbar}{\rule[.5ex]{2.5ex}{0.5pt}}
+
+\def\positionalencoding{\operatorname{pos-enc}}
+\def\concat{\operatorname{concat}}
+\def\crossentropy{\LL_{\operatorname{ce}}}
+
\begin{document}
+\vspace*{1ex}
+
+\begin{center}
+{\Large Some bits of Information Theory}
+
\today
+\end{center}
Information Theory is awesome so here is a TL;DR about Shannon's entropy.
line, you can design a coding that takes into account that the symbols
are not all as probable, and decode on the other side.
-For instance if $P('\!\!A')=1/2$, $P('\!\!B')=1/4$, and
-$P('\!\!C')=1/4$ you would transmit ``0'' for a ``A'' and ``10'' for a
+For instance if $\proba('\!\!A')=1/2$, $\proba('\!\!B')=1/4$, and
+$\proba('\!\!C')=1/4$ you would transmit ``0'' for a ``A'' and ``10'' for a
``B'' and ``11'' for a ``C'', 1.5 bits on average.
If the symbol is always the same, you transmit nothing, if they are
It has a simple analytical form:
%
\[
- H(p) = - \sum_k p(k) \log_2 p(k)
+ \entropy(p) = - \sum_k p(k) \log_2 p(k)
\]
%
where by convention $0 \log_2 0 = 0$.
called "Arithmetic coding" does it always.
From this perspective, many quantities have an intuitive
-value. Consider for instance sending pairs of symbols (X, Y).
+value. Consider for instance sending pairs of symbols $(X, Y)$.
If these two symbols are independent, you cannot do better than
sending one and the other separately, hence
%
\[
-H(X, H) = H(X) + H(Y).
+\entropy(X, Y) = \entropy(X) + \entropy(Y).
\]
However, imagine that the second symbol is a function of the first
-Y=f(X). You just have to send X since Y can be computed from it on the
+Y=f(X). You just have to send $X$ since $Y$ can be computed from it on the
other side.
Hence in that case
%
\[
-H(X, Y) = H(X).
+\entropy(X, Y) = \entropy(X).
\]
An associated quantity is the mutual information between two random
variables, defined with
%
\[
-I(X;Y) = H(X) + H(Y) - H(X,Y),
+\mutinf(X;Y) = \entropy(X) + \entropy(Y) - \entropy(X,Y),
\]
%
that quantifies the amount of information shared by the two variables.
Conditional entropy is the average of the entropy of the conditional distribution:
%
\begin{align*}
-&H(X \mid Y)\\
- &= \sum_y p(Y=y) H(X \mid Y=y)\\
- &= \sum_y P(Y=y) \sum_x P(X=x \mid Y=y) \log P(X=x \mid Y=y)
+ & \entropy(X \mid Y) \\
+ & = \sum_y \proba(Y=y) \entropy(X \mid Y=y) \\
+ & = \sum_y \proba(Y=y) \sum_x \proba(X=x \mid Y=y) \log \proba(X=x \mid Y=y)
\end{align*}
-Intuitively it is the [minimum average] number of bits required to describe X given that Y is known.
+Intuitively it is the [minimum average] number of bits required to describe $X$ given that $Y$ is known.
-So in particular, if X and Y are independent, getting the value of $Y$
+So in particular, if $X$ and $Y$ are independent, getting the value of $Y$
does not help at all, so you still have to send all the bits for $X$,
hence
%
\[
- H(X \mid Y)=H(X)
+ \entropy(X \mid Y)=\entropy(X),
\]
-
-if X is a deterministic function of Y then
+%
+and if $X$ is a deterministic function of $Y$ then
%
\[
- H(X \mid Y)=0.
+ \entropy(X \mid Y)=0.
\]
-And if you send the bits for Y and then the bits to describe X given
-that Y, you have sent (X, Y). Hence we have the chain rule:
+And if you send the bits for $Y$ and then the bits to describe $X$ given
+that $Y$, you have sent $(X, Y)$. Hence we have the chain rule:
%
\[
-H(X, Y) = H(Y) + H(X \mid Y).
+\entropy(X, Y) = \entropy(Y) + \entropy(X \mid Y).
\]
And then we get
%
\begin{align*}
-I(X;Y) &= H(X) + H(Y) - H(X,Y)\\
- &= H(X) + H(Y) - (H(Y) + H(X \mid Y))\\
- &= H(X) - H(X \mid Y).
+I(X;Y) &= \entropy(X) + \entropy(Y) - \entropy(X,Y)\\
+ &= \entropy(X) + \entropy(Y) - (\entropy(Y) + \entropy(X \mid Y))\\
+ &= \entropy(X) - \entropy(X \mid Y).
\end{align*}
\section{Kullback-Leibler divergence}
Imagine that you encode your stream thinking it comes from
-distribution $q$ while it comes from $p$. You would emit more bits than
-the optimal $H(p)$, and that supplement is $D_{KL}(p||q)$ the
-Kullback-Leibler divergence between $p$ and $q$.
+distribution $q$ while it comes from $p$. You would emit more bits
+than the optimal $\entropy(p)$, and that excess of bits is
+$\dkl(p||q)$ the Kullback-Leibler divergence between $p$ and $q$.
In particular if $p=q$
%
\[
- D_{KL}(p\|q)=0,
+ \dkl(p\|q)=0,
\]
%
and if there is a symbol $x$ with $q(x)=0$ and $p(x)>0$, you cannot encode it and
%
\[
- D_{KL}(p\|q)=+\infty.
+ \dkl(p\|q)=+\infty.
\]
Its formal expression is
%
\[
-D_{KL}(p\|q) = \sum_x p(x) \log\left(\frac{p(x)}{q(x)}\right)
+\dkl(p\|q) = \sum_x p(x) \log\left(\frac{p(x)}{q(x)}\right)
\]
%
that can be understood as a value called the cross-entropy between $p$ and $q$
%
\[
-H(p,q) = -\sum_x p(x) \log q(x)
+\entropy(p,q) = -\sum_x p(x) \log q(x)
\]
%
minus the entropy of p
\[
-H(p) = -\sum_x p(x) \log p(x).
+\entropy(p) = -\sum_x p(x) \log p(x).
\]
-Notation horror: if $X$ and $Y$ are random variables $H(X, Y)$ is the
+Notation horror: if $X$ and $Y$ are random variables $\entropy(X, Y)$ is the
entropy of their joint law, and if $p$ and $q$ are distributions,
-$H(p,q)$ is the cross-entropy between them.
+$\entropy(p,q)$ is the cross-entropy between them.
\end{document}
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