$\require{mathtools}
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The entropy of a random variable is a quantitative notion of how uncertain we are about its outcome. More precisely, it’s the “average surprise” you get from samples, where surprise is measured as $\log\p{\f1{\text{probability of getting that outcome}}}$:
\[
\H[\BX] \ce \E_{x \sim \BX}\b{\log \frac{1}{\Pr[\BX=x]}}.
\]
It’s the average “amount of information” you get from observing $\BX$: the number of bits you need to describe draws from $\BX$, amortized over many draws.
By extension, we define the entropy of a distribution $P$ as
\[
\H\p{P} \ce \E_{x \sim P}\b{\log \f{1}{P(x)}}.
\]
Properties
- Entropy ranges between $0$ (if $\BX$ is deterministic) and $\log |\CX|$ (if $\BX$ is uniform), where $\CX$ is the support of $\BX$.
- It’s subadditive: $\H[\BX,\BY] \leq \H[\BX] + \H[\BY]$, with equality iff $\BX$ and $\BY$ are independent.
- Intuitively, you can’t be more surprised by the combination of $\BX$ and $\BY$ than you were by seeing both outcomes separately.
- The “correlated part” of $\BX$ and $\BY$ is counted once in $\H[\BX,\BY]$ but twice in $\H[\BX] + \H[\BY]$.
We will prove $\H[X] \leq \log |\CX|$ by looking at entropy deficit, and subadditivity by looking at mutual information.
Binary entropy
In particular, let $H(p) \ce p\log\frac{1}{p} + (1-p)\log\frac{1}{1-p}$ denote the entropy of a Bernoulli with mean $p$. Then
- $H(p)$ is concave,
- $H(p)$ is symmetric around $\frac12$: $H(p) = H(1-p)$,
- when $p\approx \frac12$, $H(p) = 1-\Theta\p{\p{p-\frac{1}{2}}^2}$,
- when $p \le \frac12$, $H(p) = \Theta\p{p \log\frac{1}{p}}$.
Conditional entropy
Similarly, we can define conditional entropy as the average surprise you get from $\BY$ if you’ve already seen $\BX$, or the number of bits you need to describe draws from $\BY$ to someone who knows $\BX$:
\[
\al{
\H\bco{\BY}{\BX}
\ce&\ \E_{x \sim \BX}\b{\H[\at{\BY}{\BX=x}]}\\
=&\ \E_{(x,y) \sim (\BX,\BY)}\b{\log \frac{1}{\Pr\bco{\BY=y}{\BX=x}}}
}
\]
where $\at{\BY}{\BX=x}$ denotes the random variable obtained by conditioning $\BY$ on $\BX=x$. Confusingly, $\H\bco\BY\BX$ denotes a fixed value, in contrast with the conditional expectation $\E\bco{\BY}{\BX}$, which is a random variable depending on $\BX$.
Chain rule
The conditional entropy gives a chain rule:
\[
\al{
\H[\BX,\BY]
&= \E_{(x,y) \sim (\BX,\BY)}\b{\log\frac{1}{\Pr[\BX = x \and \BY=y]}}\\
&= \E_{(x,y) \sim (\BX,\BY)}\b{\log\frac{1}{\Pr[\BX = x]\Pr\bco{\BY=y}{\BX=x}}}\\
&= \H[\BX] + \H\bco\BY\BX,
}
\]
which immediately shows that adding data can only increase the entropy
\[
\H[\BX,\BY] \ge \H[\BX].
\]
Mixtures and concavity
Together with subadditivity, this shows that conditioning can only decrease entropy
\[
\al{
\H\b{\BY}
&\ge \H\b{\BX,\BY} - \H[\BX]\tag{subadditivity}\\
&= \H\bco{\BY}{\BX},\tag{chain rule}
}
\]
with equality iff $\BX$ and $\BY$ are independent.
If we see $\BY$ as a mixture $\BY_{\BX}$ of the variables $\BY_x \ce \at{\BY}{\BX=x}$ weighted by $\BX$, then this is saying that entropy is strictly concave:
\[
\ob{\E_{x \sim \BX}\b{\H\b{\BY_x}}}^{\H\bco{\BY}{\BX}}_\text{average of entropies}\le \ob{\H\b{\BY_{\BX}}}^{\H\b{\BY}}_\text{entropy of mixture} \le \ob{\H\b{\BX,\BY_\BX}}^{\H\b{\BX,\BY}}_\text{entropy of ``disjoint mixture''},
\]
with equality respectively iff the variables $\BY_x$ are all indentical, and iff they have disjoint supports.
#to-write
- the “amount of information” thing makes sense? because it’s the ==information you gain== if you started from a prior of $\BX$’s distribution then you actually observe $\BX$? i.e. $\E_{\Bx \sim [\BX]}\b{\D\bpa{\at{\BX}{\BX = \Bx}}{\BX}}$
- okay I’m now thoroughly KL-pilled, it’s the source of everything. $\D\ppa{Q}{P}$ is the information you’ve gained if you started from a prior of $P$ and somehow your prior is now $Q$
- but I want to understand the sense if which $\D\ppa{Q}{P}$ is a lower bound on the “number of questions you must have asked in order to get there”?
- i.e. the information $\BX$ gains you about $\BX$ is limited by the information that it gives about itself
- i.e. $\E_{\Bx \sim [\BX]}\b{\D\bpa{\at{\BY}{\BX=\Bx}}{\BY}} \ge \H[\BX]$?
- actually I believe you’ve just rediscovered $\I[\BX;\BY] \ge \I[\BX;\BX] = \H[\BX]$
- darn information theory is so cool
- and btw you can write this as $\E\b{\D\bpa{\cond{\BY}{\BX}}{\BY}}$
- and this directly shows that it’s equal to $\I[\BX;\BY] = \H[\BY] - \H\bco{\BY}{\BX}$ I guess
- though maybe you should swallow the pill and not talk about entropy ever again!
- note: I guess it also means $\I[\BX;\BY] = \I[\BY;\BY] - \I\bco{\BY;\BY}{\BX} = -\I[\BY;\BY;\BX]$?
- cool story bro
- wait is it the case that in general if you add a variable that’s already present it just flips the sign?
- $\I[\BX;\BY;\BZ;\BZ] = \I\bco{\BX;\BY;\BZ}{\BZ} - \I\b{\BX;\BY;\BZ}$
- and somehow I’m supposed to just know that $\I\bco{\BX;\BY;\BZ}{\BZ} = 0$?
- I guess this just suggests that if any variable is a constant then the interaction is $0$, which would make a whole lot of sense tbh
- very similar to Wick in that case!
- really, entropy should be negative…
- okay I’m this close to just replacing all occurrences of $\D$ to $\I$
- and to have mutual information as a more core concept than entropy
- and to literally just define $\H[\BX]$ as $\I[\BX;\BX]$
- this would explain why you have to somehow bother to use two copies of $\BX$ in the definition of entropy!
- (and same thing in basically all of the power entropies)
- ultimately ==information is about something providing information about another thing==
- there is no such thing as “the information of one thing”
- and the information divergence is the core of what information means
- and it’s also what makes information theory so nice: it provides all the nontrivial inequalities!
- does $\I[\BX;\BX;\BX]$ make sense? it’s $\I\bco{\BX;\BX}{\BX} - \I[\BX;\BX] = 0-\H[\BX] = -\H[\BX]$ hmm maybe this is slightly cursed?
- and $\I[\BX;\BX;\BX;\BX] = \I\bco{\BX;\BX;\BX}{\BX} - \I[\BX;\BX;\BX] = \H[\BX]$? so it keeps alternating, hmmm…
- actually this is support for defining $\I[\BX] \ce -\H[\BX]$ instead of $\D[\BX]$
- indeed, $\I[\BX;\BX] = \I\bco{\BX}{\BX} - \I\b{\BX}$
- and I’m not sure what this first quantity $\I\bco{\BX}{\BX}$ is supposed to mean but it sure seems like it should be $0$?
- if you set $\I[\BX] \ce \D[\BX]$ then this quantity is $\H[\BX] + \D[\BX] =$ entropy of the prior on $\BX$?
- actually very confusing
- and $\I[\BX] = \I\bco{}{\BX} - \I\b{}$
- i.e. “how much does $\BX$ help you predict nothing”??
- so we’d have to accept that $\I\bco{}{\BX} = \I[\BX] + C = C-\H[\BX]$
- i.e. $\BX$ is actively detrimental to the task of predicting nothing? (???)
- actually feels like we should have $\I\bco{}{\BX} = \I[]$ given that $\BX$ is not dependent with anything in the LHS (this is definitely a rule right?)
- which would mean $\I[\BX] = 0$
- but then $\I\bco{\BX}{\BX} = \I\b{\BX;\BX} = \H[\BX]$, even though $\I\bco{\BX}{\BX} = \E_{\Bx \sim [\BX]}\b{\I\b{\at{\BX}{\BX = \Bx}}}$ and $\at{\BX}{\BX=x}$ is literally a point mass no matter what $x$ is
- actually that’s a general argument for $\I\bco{\BX}{\BX}$ being a constant $C$ (not clear which constant, but $0$ would seem like a reasonable choice here since we get zero whenever there is any deterministic constant in the expression?)
- so $\I[\BX]$ should be $C-\H[\BX]$
- to summarize:
- given that $\I[\BX] = C - \H[\BX]$ is the only tenable position, we would have to have $\I\bco{}{\BX} = C' - \H[\BX]$
- but then that violates the rule of “if $\BX$ is independent from the joint distribution of the things in the LHS then the information should be $0$”
- okay I vote for $\I[\BX]$ just being undefined!! $\I[\BX;\BY]$ is clearly the core quantity here
- and it seems like this makes things better in the case of continuous random variables as well
- it’s in fact somewhat baffling that the interaction information is even well-defined (that it’s consistent and symmetric)
- actually I think I’d be on board to keep $\D[\cdot]$ only for the deficit
- KL divergence is to entropy the same as squared distance (!!) is to variance?