\documentclass[11pt]{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amssymb,amsthm,booktabs,xcolor}
\usepackage{hyperref}

\newtheorem{definition}{Definition}
\newtheorem{proposition}{Proposition}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}
\newtheorem{remark}{Remark}

\title{Bayes from Counting:\\Partial Quotients, GCD, and the\\Symmetric Learning Function on $E = I \times O$}
\author{Claude and MJC}
\date{February 12, 2026}

\begin{document}
\maketitle

\begin{abstract}
We show that Bayes' theorem falls out of the $E \to N \to Q$ chain when
the joint event space $E = I \times O$ is factored via partial quotients.
A \emph{partial quotient} divides $E$ by a single atomic event $i_0 \in I$
(not by the entire event space $I$), yielding a conditional distribution
over $O$ scaled by an equivalence class of size $c(i_0)$.
The GCD of each row (or column) of the count table separates the
\emph{common evidence} (how often the conditioning event occurred) from
the \emph{differential evidence} (the shape of the conditional), and the
GCD cancels from every conditional.
Bayes' theorem is then the \emph{consistency condition} between the
$I$-side and $O$-side partial quotients: the two GCD-reduced decompositions
of the same joint count must agree, and their agreement forces
$P(o|i)/P(i|o) = P(o)/P(i)$.
The quotient $Q = 1/P$ (luck) decomposes as $Q_{\mathrm{joint}} =
Q_{\mathrm{marginal}} \cdot Q_{\mathrm{conditional}}$, and the
symmetry of the log contingency table---the output of the standard
learning function---is what makes the causal and evidential directions
both readable from the same matrix, justifying the CMP claim that
correlation captures causation.
\end{abstract}


\section{The Joint Event Space}

Let $I$ and $O$ be finite sets (input and output event spaces) with
$|I| = m$ and $|O| = n$.  The joint event space is
$E = I \times O$.  A dataset $D$ of $N$ observations is a sequence of
joint events $(i_t, o_t) \in E$ for $t = 0, \ldots, N-1$.

\begin{definition}[Joint count]
The \emph{joint count function} $c: E \to \mathbb{N}_{\geq 0}$ is
\[
  c(i, o) = |\{t : (i_t, o_t) = (i, o)\}|.
\]
The marginals are $c(i) = \sum_o c(i,o)$ and $c(o) = \sum_i c(i,o)$,
with $\sum_{i,o} c(i,o) = N$.
\end{definition}


\section{The Standard Learning Function: $E \to N$}

The standard learning function $\omega_0$ (CMP, \S5) records every
observed joint event.  Its output is the \emph{log contingency table}:
\[
  s(i,o) = \log_2 c(i,o)
\]
for each $(i,o)$ with $c(i,o) > 0$, and $s(i,o) = -\infty$ otherwise.

This is the first step of the $E \to N$ map: events are counted, and
counts are the natural numbers assigned to events.

\begin{remark}[The matrix is symmetric in its role]
The log contingency table $M$ with $M_{io} = s(i,o)$ can be read in
two directions:
\begin{itemize}
  \item $I \to O$: fix a row $i$, read off the log supports for each $o$.
  \item $O \to I$: fix a column $o$, read off the log supports for each $i$.
\end{itemize}
CMP observes that ``learning takes the directed causal arrow from rain
to wet ground into a bidirectional one; we may infer that it has rained
from the ground being wet just as well.''  This bidirectionality is a
consequence of the matrix being a single object read two ways---the
\emph{same counts} encode both directions.
\end{remark}


\section{The Partial Quotient}

\begin{definition}[Full quotient]
The \emph{full quotient} $E / I$ identifies all events that share the same
output: $(i_1, o) \sim (i_2, o)$ for all $i_1, i_2$.  The result is $O$,
the output event space.  Similarly $E / O \cong I$.
\end{definition}

\begin{definition}[Partial quotient]
The \emph{partial quotient} of $E$ by an atomic event $i_0 \in I$ is the
pair:
\begin{enumerate}
  \item The \textbf{conditional fiber}: $O_{i_0} = \{(i_0, o) : o \in O\}
    \cong O$, with counts $c(i_0, o)$ for each $o$.
  \item The \textbf{equivalence class}: $[i_0] = \{t \in D : i_t = i_0\}$,
    with $|[i_0]| = c(i_0)$.
\end{enumerate}
\end{definition}

The partial quotient \emph{divides} by a single atom (fixing $I = i_0$)
and \emph{multiplies} by the equivalence class (recording how many
positions share that input).  The reconstruction identity is:
\[
  c(i_0, o) = P(o \mid i_0) \cdot c(i_0),
\]
or in log support:
\begin{equation}\label{eq:partial-quotient}
  s(i_0, o) = \underbrace{\log_2 P(o \mid i_0)}_{\text{conditional (shape)}}
  + \underbrace{s_I(i_0)}_{\text{class size (scale)}},
\end{equation}
where $s_I(i_0) = \log_2 c(i_0)$.

By symmetry, the partial quotient of $E$ by $o_0 \in O$ gives:
\begin{equation}\label{eq:partial-quotient-O}
  s(i, o_0) = \log_2 P(i \mid o_0) + s_O(o_0).
\end{equation}


\section{The GCD Decomposition}

\begin{definition}[Row and column GCD]
For each $i \in I$, define the \emph{row GCD}:
\[
  g_I(i) = \gcd_{o : c(i,o) > 0} c(i,o).
\]
For each $o \in O$, define the \emph{column GCD}:
\[
  g_O(o) = \gcd_{i : c(i,o) > 0} c(i,o).
\]
\end{definition}

Since $g_I(i)$ divides every entry in row $i$, we can write:
\[
  c(i, o) = g_I(i) \cdot r_I(i, o),
  \qquad
  \gcd_{o} r_I(i, o) = 1,
\]
where $r_I(i, o) \in \mathbb{N}$ are the \emph{reduced counts} (coprime
across the row).

\begin{definition}[Reduced total]
$R_I(i) = \sum_o r_I(i, o) = c(i) / g_I(i)$.
\end{definition}

\begin{proposition}[GCD cancels from the conditional]\label{prop:gcd-cancels}
\[
  P(o \mid i) = \frac{c(i,o)}{c(i)} = \frac{g_I(i) \cdot r_I(i,o)}{g_I(i) \cdot R_I(i)}
  = \frac{r_I(i,o)}{R_I(i)}.
\]
The conditional probability depends only on the reduced counts $r_I$
and reduced total $R_I$.  The GCD $g_I(i)$ cancels completely.
\end{proposition}

In log support, the joint count decomposes as:
\begin{equation}\label{eq:gcd-decomp}
  s(i, o) = \underbrace{\log_2 g_I(i)}_{\text{common evidence}}
  + \underbrace{\log_2 r_I(i, o)}_{\text{differential evidence}}.
\end{equation}

The \textbf{common evidence} $\log_2 g_I(i)$ is the part of the log
support that is shared by \emph{all} outputs given input $i$.
It tells us about the \emph{prevalence} of input $i$ without
distinguishing which output accompanied it.

The \textbf{differential evidence} $\log_2 r_I(i,o)$ is the part
that distinguishes output $o$ from the other outputs.  It is the
\emph{irreducible} evidence for this specific joint event.

By symmetry, the column-side decomposition gives:
\begin{equation}\label{eq:gcd-decomp-O}
  s(i, o) = \log_2 g_O(o) + \log_2 r_O(i, o),
  \qquad \gcd_i r_O(i, o) = 1.
\end{equation}


\section{Bayes as Consistency of Partial Quotients}

The same joint count $c(i,o)$ admits two GCD decompositions:
\begin{align}
  c(i, o) &= g_I(i) \cdot r_I(i, o) \qquad \text{(row decomposition)}\label{eq:row}\\
  c(i, o) &= g_O(o) \cdot r_O(i, o) \qquad \text{(column decomposition)}\label{eq:col}
\end{align}

\begin{theorem}[Bayes from GCD consistency]\label{thm:bayes}
Equating \eqref{eq:row} and \eqref{eq:col}:
\[
  g_I(i) \cdot r_I(i,o) = g_O(o) \cdot r_O(i,o),
\]
which gives the \textbf{GCD bridge}:
\begin{equation}\label{eq:gcd-bridge}
  \frac{r_I(i,o)}{r_O(i,o)} = \frac{g_O(o)}{g_I(i)}.
\end{equation}
Now compute the ratio of conditionals:
\begin{align*}
  \frac{P(o|i)}{P(i|o)}
  &= \frac{r_I(i,o)/R_I(i)}{r_O(i,o)/S_O(o)}
  = \frac{r_I(i,o) \cdot S_O(o)}{r_O(i,o) \cdot R_I(i)}\\[4pt]
  &= \frac{g_O(o)}{g_I(i)} \cdot \frac{S_O(o)}{R_I(i)}
  \qquad\text{(by \eqref{eq:gcd-bridge})}\\[4pt]
  &= \frac{g_O(o) \cdot S_O(o)}{g_I(i) \cdot R_I(i)}
  = \frac{c(o)}{c(i)}
  = \frac{P(o)}{P(i)}.
\end{align*}

This is \textbf{Bayes' theorem}:
\[
  \boxed{P(o \mid i) = \frac{P(i \mid o) \cdot P(o)}{P(i)}.}
\]
\end{theorem}

\begin{remark}[What the proof says]
Bayes' theorem is the \emph{consistency condition} between the two
partial quotients of $E = I \times O$.  Dividing by $I = i$ and
dividing by $O = o$ must give compatible conditionals, because they
decompose the \emph{same} joint count.
The GCD mediates: the ratio $g_O(o)/g_I(i)$ converts between the two
reduced-count representations, and when multiplied by the reduced
totals, recovers the marginal ratio $P(o)/P(i)$.
\end{remark}

\begin{corollary}[Log-support form of Bayes]
In the log domain:
\begin{equation}\label{eq:log-bayes}
  \log_2 r_I(i,o) - \log_2 r_O(i,o) = \log_2 g_O(o) - \log_2 g_I(i).
\end{equation}
The left side is the difference between the $I$-reduced and $O$-reduced
log supports for a specific joint event $(i,o)$.  The right side depends
only on the marginal GCDs.

Equivalently:
\[
  s(o|i) - s(i|o) = s_O(o) - s_I(i).
\]
The \emph{asymmetry of the conditionals equals the asymmetry of the priors.}
\end{corollary}


\section{The $E \to N \to Q$ Chain}

We now connect the partial quotient to the quotient chain from the
thermodynamic framework~\cite{quotient-chain}.

\begin{definition}[The three-step map]
For a joint event $(i, o)$:
\begin{enumerate}
  \item $E$: Identify the event $(i, o) \in I \times O$.
  \item $N$: Count it.  $c(i, o) \in \mathbb{N}$.
  \item $Q$: Compute the quotient.  $Q(i, o) = N / c(i, o) = \lambda(i,o)$
    (the joint luck).
\end{enumerate}
\end{definition}

The joint luck $\lambda(i,o) = 1/P(i,o)$ decomposes via the partial
quotient:

\begin{proposition}[Luck decomposition]
\begin{equation}\label{eq:luck-decomp}
  \log_2 Q(i,o) = \underbrace{\log_2 Q_I(i)}_{\text{marginal luck}}
  + \underbrace{\log_2 Q(o|i)}_{\text{conditional luck}}
\end{equation}
where $Q_I(i) = N/c(i) = 1/P(i)$ and
$Q(o|i) = c(i)/c(i,o) = 1/P(o|i) = R_I(i)/r_I(i,o)$.

The conditional luck depends only on the reduced counts:
\[
  Q(o|i) = \frac{R_I(i)}{r_I(i,o)}.
\]
The GCD $g_I(i)$ cancels from the conditional luck, just as it cancels
from the conditional probability.
\end{proposition}

\begin{theorem}[Bayes in quotient form]\label{thm:bayes-Q}
\[
  \frac{Q(o|i)}{Q(i|o)} = \frac{Q_O(o)}{Q_I(i)} = \frac{P(i)}{P(o)}.
\]
The ratio of conditional lucks equals the \emph{inverse} ratio of the
priors.  An event that is ``lucky'' from one direction is ``ordinary''
from the other, balanced by the marginal frequencies.
\end{theorem}

\begin{proof}
$Q(o|i)/Q(i|o) = P(i|o)/P(o|i) = P(i)/P(o)$ by Theorem~\ref{thm:bayes}.
\end{proof}

\begin{remark}[Luck is relative]
The joint luck $Q(i,o) = N/c(i,o)$ is an absolute quantity.
But it decomposes into marginal and conditional luck in \emph{two}
ways:
\begin{align*}
  \log_2 Q(i,o) &= \log_2 Q_I(i) + \log_2 Q(o|i) \qquad\text{($I$-first)}\\
  &= \log_2 Q_O(o) + \log_2 Q(i|o) \qquad\text{($O$-first)}
\end{align*}
The two decompositions are consistent (they give the same joint luck)
because Bayes holds.  The ``luck'' of seeing output $o$ after input $i$
is not intrinsic to the pair---it depends on which direction you
condition.  The GCD structure (Eq.~\ref{eq:gcd-bridge}) is what makes
the two directions compatible.
\end{remark}


\section{Why Correlation Captures Causation}

CMP makes the claim that ``we may infer that it has rained from the
ground being wet just as well as that the ground must be wet from the
fact that it is raining.''  The partial quotient framework makes this
precise.

\begin{proposition}[The symmetric matrix]
The log contingency table $M_{io} = s(i,o) = \log_2 c(i,o)$ is a single
matrix that encodes:
\begin{itemize}
  \item The causal direction: $P(o|i) = 2^{s(i,o)} / \sum_{o'} 2^{s(i,o')}$.
  \item The evidential direction: $P(i|o) = 2^{s(i,o)} / \sum_{i'} 2^{s(i',o)}$.
  \item Both partial quotients and their GCD decompositions.
  \item All lucks, marginals, and conditionals.
\end{itemize}
\end{proposition}

The matrix $M$ is the \textbf{sufficient statistic} for the function
$I \to O$ (CMP, \S5.2).  But it is equally the sufficient statistic for
$O \to I$.  \emph{No additional data is needed to reverse the direction.}

Why?  Because the GCD decomposition is stored implicitly.  The row-GCDs
$g_I(i)$ and column-GCDs $g_O(o)$ are computable from $M$, and they
mediate Bayes via the bridge equation~\eqref{eq:gcd-bridge}.  The
causal direction (the world's arrow from rain to wet ground) determined
which counts landed where, but the counts \emph{themselves} are
direction-agnostic.

\begin{remark}[Correlation and the GCD]
Consider the pointwise mutual information:
\[
  \mathrm{PMI}(i,o) = \log_2 \frac{c(i,o) \cdot N}{c(i) \cdot c(o)}
  = \log_2 \frac{r_I(i,o) \cdot N}{R_I(i) \cdot c(o)}.
\]
The PMI measures \emph{correlation}: how much the joint count exceeds
what independence would predict.  Using $c(o) = g_O(o) \cdot S_O(o)$
and the GCD bridge:
\begin{align*}
  \mathrm{PMI}(i,o)
  &= \log_2 r_I(i,o) - \log_2 R_I(i) - \log_2 S_O(o) + \log_2 N - \log_2 g_O(o)\\
  &= \log_2 r_O(i,o) - \log_2 R_I(i) - \log_2 S_O(o) + \log_2 N - \log_2 g_I(i).
\end{align*}
Both forms give the same PMI (as they must).
The global GCD $g = \gcd_{i,o} c(i,o)$ cancels entirely from the PMI.
The row and column GCDs contribute through the reduced totals $R_I$
and $S_O$.  The PMI depends on the \emph{shape} of the counts
(the reduced parts), not their \emph{scale} (the GCDs).

Correlation (PMI) captures causation (the directional relationship
$I \to O$) because the log contingency table is the sufficient
statistic for both, and the GCD structure guarantees that the causal
and evidential readings are consistent via Bayes.
\end{remark}


\section{The Quotient as the Thermodynamic Bridge}

The $E \to N \to Q$ chain from the February~11 archive~\cite{quotient-chain,e-onto-n}
identifies:
\begin{center}
\begin{tabular}{lll}
\toprule
Step & Object & Meaning \\
\midrule
$E$ & Joint event $(i,o)$ & What happened \\
$N$ & Count $c(i,o)$ & How often it happened \\
$Q$ & Quotient $N/c(i,o) = \lambda$ & How surprising (luck) \\
\bottomrule
\end{tabular}
\end{center}

The partial quotient adds the decomposition at the $Q$ level:
\begin{center}
\begin{tabular}{lll}
\toprule
Component & Formula & Meaning \\
\midrule
$Q_{\mathrm{joint}}(i,o)$ & $N/c(i,o)$ & Total luck of the joint event \\
$Q_I(i)$ & $N/c(i)$ & Luck of the input (marginal) \\
$Q(o|i)$ & $R_I(i)/r_I(i,o)$ & Luck of the output given input \\
$g_I(i)$ & $\gcd_o c(i,o)$ & Common evidence (direction-free) \\
$r_I(i,o)$ & $c(i,o)/g_I(i)$ & Differential evidence (directional) \\
\bottomrule
\end{tabular}
\end{center}

The GCD $g_I(i)$ is the ``thermodynamic floor''---the minimum evidence
that every output shares given input $i$.  In the microstate--macrostate
framework~\cite{micro-macro}, it corresponds to the number of
microstates that are assigned to the input macrostate $i$ uniformly
across all output classes.

\begin{proposition}[GCD and the entropy budget]
The row entropy decomposes as:
\[
  H(O|I=i) = \log_2 R_I(i) - \sum_o \frac{r_I(i,o)}{R_I(i)} \log_2 r_I(i,o).
\]
The GCD $g_I(i)$ does not appear.  The conditional entropy---and therefore
the conditional luck $\log_2 Q(o|i) = -\log_2 P(o|i)$---is determined
entirely by the reduced counts.

The marginal entropy includes the GCD:
\[
  H(I) = \log_2 N - \sum_i \frac{c(i)}{N} \log_2 c(i)
  = \log_2 N - \sum_i \frac{c(i)}{N} [\log_2 g_I(i) + \log_2 R_I(i)].
\]
The GCD contributes to the marginal entropy but not the conditional.
Factoring out the GCD is ``accounting for the prior.''
\end{proposition}


\section{The Tropical Approximation}

The UM's standard update function operates in the $(\max, \min)$
tropical semiring:
\[
  (f_p(t))_j = \max_i \min(t_i, p_{ij}).
\]

\begin{definition}[Tropical GCD]
In the tropical semiring $(\max, +)$, the GCD of a set of values
$\{a_1, \ldots, a_n\}$ is $\min(a_1, \ldots, a_n)$.
\end{definition}

The tropical GCD of the log-support row $\{s(i, o)\}_{o \in O}$ is:
\[
  \min_o s(i, o) = \log_2 \min_o c(i, o).
\]
The integer GCD gives $\log_2 \gcd_o c(i,o)$.
Since $\gcd \leq \min$ always:
\[
  \log_2 \gcd_o c(i,o) \leq \min_o \log_2 c(i,o),
\]
with equality when $\min_o c(i,o)$ divides every $c(i,o)$ in the row.

\begin{remark}[The UM approximates the GCD]
The $\min$ operation in the standard update is the tropical version of
the integer GCD.  When the UM computes $\min(t_i, p_{ij})$, it is
performing a tropical GCD that \emph{upper bounds} the true integer GCD.
The gap $\log_2(\min/\gcd)$ measures how far the tropical
computation is from exact Bayesian inference.

For integer counts where the smallest count divides all others (common
when counts arise from a generative process with integer rates), the gap
is zero and the UM's tropical computation is exact.
\end{remark}


\section{Worked Example}

Let $I = \{a, b\}$, $O = \{x, y\}$, with count table:
\begin{center}
\begin{tabular}{c|cc|c}
 & $x$ & $y$ & $c(i)$ \\
\hline
$a$ & 6 & 2 & 8 \\
$b$ & 3 & 9 & 12 \\
\hline
$c(o)$ & 9 & 11 & $N=20$
\end{tabular}
\end{center}

\paragraph{Row GCDs and reduced counts.}
$g_I(a) = \gcd(6,2) = 2$, so $r_I(a,\cdot) = (3, 1)$, $R_I(a) = 4$.
Check: $c(a) = 2 \cdot 4 = 8$.  \checkmark

$g_I(b) = \gcd(3,9) = 3$, so $r_I(b,\cdot) = (1, 3)$, $R_I(b) = 4$.
Check: $c(b) = 3 \cdot 4 = 12$.  \checkmark

\paragraph{Column GCDs and reduced counts.}
$g_O(x) = \gcd(6,3) = 3$, so $r_O(\cdot,x) = (2, 1)$, $S_O(x) = 3$.
Check: $c(x) = 3 \cdot 3 = 9$.  \checkmark

$g_O(y) = \gcd(2,9) = 1$, so $r_O(\cdot,y) = (2, 9)$, $S_O(y) = 11$.
Check: $c(y) = 1 \cdot 11 = 11$.  \checkmark

\paragraph{Conditionals (GCD-free).}
$P(x|a) = r_I(a,x)/R_I(a) = 3/4$.

$P(x|b) = r_I(b,x)/R_I(b) = 1/4$.

$P(a|x) = r_O(a,x)/S_O(x) = 2/3$.

\paragraph{GCD bridge (Eq.~\ref{eq:gcd-bridge}).}
$r_I(a,x)/r_O(a,x) = 3/2 = g_O(x)/g_I(a) = 3/2$.  \checkmark

$r_I(b,y)/r_O(b,y) = 3/9 = 1/3 = g_O(y)/g_I(b) = 1/3$.  \checkmark

\paragraph{Bayes.}
$P(x|a) = P(a|x) \cdot P(x)/P(a) = (2/3)(9/20)/(8/20)
= (2/3)(9/8) = 3/4$.  \checkmark

\paragraph{Luck.}
$Q(x|a) = R_I(a)/r_I(a,x) = 4/3 \approx 1.33$.

$Q(a|x) = S_O(x)/r_O(a,x) = 3/2 = 1.5$.

Ratio: $Q(x|a)/Q(a|x) = (4/3)/(3/2) = 8/9 = c(a)/c(x) = 8/9$. \checkmark

The output $x$ is ``less lucky'' given input $a$ (luck 1.33) than input
$a$ is given output $x$ (luck 1.5), because $a$ is rarer than $x$.

\paragraph{Log support decomposition.}

\begin{center}
\begin{tabular}{cccccc}
\toprule
$(i,o)$ & $s(i,o)$ & $\log_2 g_I$ & $\log_2 r_I$ & $\log_2 g_O$ & $\log_2 r_O$ \\
\midrule
$(a,x)$ & 2.585 & 1.000 & 1.585 & 1.585 & 1.000 \\
$(a,y)$ & 1.000 & 1.000 & 0.000 & 0.000 & 1.000 \\
$(b,x)$ & 1.585 & 1.585 & 0.000 & 1.585 & 0.000 \\
$(b,y)$ & 3.170 & 1.585 & 1.585 & 0.000 & 3.170 \\
\bottomrule
\end{tabular}
\end{center}

Each row confirms $s = \log_2 g_I + \log_2 r_I = \log_2 g_O + \log_2 r_O$.

The GCD bridge in log:
$\log_2 r_I - \log_2 r_O = \log_2 g_O - \log_2 g_I$
is verified for every cell (e.g., for $(a,x)$: $1.585 - 1.000 = 1.585 - 1.000$).


\section{Discussion}

\subsection{What the GCD ``accounts for''}

The GCD of a row $g_I(i) = \gcd_o c(i,o)$ is the largest integer that
divides every count in the row.  It represents the ``granularity'' of
the evidence: the data about input $i$ comes in packets of size
$g_I(i)$.  Dividing by the GCD gives the \emph{irreducible} evidence
for each output.

In the UM framework, the GCD is ``accounted for'' by separating it out
of the conditional.  The SN strength $s(i,o)$ encodes the full joint
evidence.  The partial quotient decomposes it into common evidence
(the GCD, which contributes to the prior) and differential evidence
(the reduced counts, which determine the conditional).

Bayes falls out because the joint count has a \emph{unique} value $c(i,o)$
that admits two GCD decompositions (row-side and column-side).  Their
consistency, mediated by the GCD bridge, IS Bayes' theorem.

\subsection{Justifying the ``wilder'' claims of CMP}

\paragraph{Correlation implies causation (in the UM).}
CMP claims that the log contingency table---pure correlation---captures
causal structure.  The partial quotient framework explains why:
the causal direction ($I \to O$) and the evidential direction
($O \to I$) are \emph{both} readable from the same matrix because
Bayes guarantees their consistency.  No additional causal knowledge
is needed beyond the counts.  The \emph{learning function does not
know which direction is causal}---it records joint events
symmetrically---but the \emph{partial quotient} extracts the directed
conditional in either direction.

\paragraph{$Q = \lambda$ (luck).}
The quotient $Q(o|i) = 1/P(o|i)$ is the conditional luck: how
``surprising'' the output is given the input.  The GCD decomposition
shows that luck depends only on the reduced counts.  The common
evidence (GCD) is ``already priced in''---it is the part of the
count that carries no surprise.  Luck measures only the differential
evidence.

The Bayesian identity in luck form (Theorem~\ref{thm:bayes-Q}) says
that luck is \emph{relative}: the same event has different luck from
different conditioning directions, balanced by the priors.  This is
the precise content of CMP's ``$Q = \lambda$'': the quotient is the
luck, and the luck is direction-dependent.

\paragraph{The standard learning function is complete.}
The log contingency table, the output of $\omega_0$, is the sufficient
statistic for both $I \to O$ and $O \to I$.  No other learning is
needed.  The GCD decomposition and Bayes are not additional algorithms
applied to the data---they are \emph{structural properties} of the
integer count table that hold automatically.  ``To learn everything
that can be statistically learned about a function from $I$ to $O$,
we can simply record every observed joint event'' (CMP, \S5.2).
The partial quotient shows that ``everything'' includes both
causal directions, all conditionals, all lucks, and Bayes' theorem
itself.


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Michaeljohn Clement.
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\newblock \url{https://cmpr.ai/cmp.pdf}, 2026.

\bibitem{quotient-chain}
Claude and MJC.
\newblock \emph{The Quotient Chain: $E \to N \to Q$ at Every Operation}.
\newblock Hutter archive, 11 Feb 2026.

\bibitem{e-onto-n}
Claude and MJC.
\newblock \emph{E onto N: The Bi-Embedding of Events and Numbers}.
\newblock Hutter archive, 11 Feb 2026.

\bibitem{micro-macro}
Claude and MJC.
\newblock \emph{Microstates, Macrostates, and the Partition Function}.
\newblock Hutter archive, 11 Feb 2026.
\end{thebibliography}

\end{document}
