:orphan:

.. include:: /_templates/include-file/page-prefix.rst

.. meta::
   :description: A step-by-step derivation of the exact first-passage law behind the RiskyMAD waiting-time forecast --- every algebraic line shown, three independent routes to the same answer, and every published figure recomputed.
   :keywords: first passage, absorbing Markov chain, hypoexponential, Laplace transform, Michaelis-Menten, RiskyMAD, waiting time, accidental nuclear winter, BEST Names, AuditTheMath
   :author: Laurence Loewe of Laodicea, Claude Opus 5, Everyone (as in https://balospe.com/en/about/authorship/)

.. _hell-mm-b16-extra-waiting-time-derivation:

.. only:: latex

   .. raw:: latex

      % --- Per-paper header / footer values (see conf.py preamble) -----------
      \renewcommand{\paperheaderleft}{\scriptsize Balospe.com/study}
      \renewcommand{\paperheadercenter}{\scriptsize Matheo-b16-extra}
      \renewcommand{\paperheaderright}{\scriptsize Variant MMv3r0p0\ $|$\ 2026m09d23}
      \renewcommand{\paperfooterleft}{\scriptsize The waiting-time law, derived line by line}
      \renewcommand{\paperfooterright}{\scriptsize Jonah License, CC-BY 4.0}
      % -----------------------------------------------------------------------
      \begin{titlepage}
      \thispagestyle{normal}
      \begin{center}
      {\LARGE\bfseries
      The Waiting-Time Law,\\
      Derived Line by Line\par}
      \vspace{4mm}
      {\large\itshape The exact first-passage mathematics behind\\
      the RiskyMAD forecast of accidental nuclear winter\par}
      \vspace{8mm}
      {\large Laurence Loewe of Laodicea\textsuperscript{1,2,3,4,5,6,7}\par}
      \vspace{2mm}
      {\small Companion to Study b16 in the Matheo Study Series\ $|$\ appendices SIA-4, SIA-5, SIA-8\par}
      \end{center}

      \vspace{3mm}
      \begin{flushleft}\scriptsize
      \textsuperscript{1}\,Balospe and Evolvix Research (Balospe.com)\\[1pt]
      \textsuperscript{2}\,Formerly Laboratory of Genetics and Wisconsin Institute for Discovery, UW-Madison\\[1pt]
      \textsuperscript{3}\,Email: \href{mailto:LLoL@balospe.org}{LLoL@balospe.org}\ $|$\ ORCID: \href{https://orcid.org/0000-0002-6253-9269}{0000-0002-6253-9269}\ $|$\ \href{https://scholar.google.com/citations?user=lBchRzQAAAAJ}{Google\,Scholar\,(lBchRzQAAAAJ)}\\[1pt]
      \textsuperscript{4-7}\,See \textit{Declarations} below for more essential background.\\
      \end{flushleft}

      \vspace{3mm}
      \begin{center}\bfseries What this document is\end{center}
      \begin{quote}\footnotesize
           Matheo-b16 forecasts how long the world should expect to wait before an
           accidental nuclear winter begins, using a three-state model small enough to
           print on a poster. Its appendix SIA-4 states the exact mathematics in about
           one page, and one page is the right length there: a reader who wants the
           forecast does not want a derivation, and a professional applied mathematician
           does not need one.

           This document is for everyone in between. It shows {\bfseries every step},
           including the ones a textbook omits as obvious, and it names its symbols so
           that a reader can tell what each one {\itshape does} without holding a lookup
           table in their head.

           Three commitments shape it. {\bfseries Nothing is asserted that is not
           derived} \textemdash{} where b16 says "solving the equations gives", this document solves
           them, line by line. {\bfseries Every result is obtained more than once, by
           routes that share no machinery} \textemdash{} the central law is derived three ways (a
           transform, elementary counting with no transform at all, and the eigenvalues
           of a matrix), and a fourth instrument, simulation that consults no formula,
           checks the answer numerically. They agree to the last digit printed.
           {\bfseries Where an assumption enters, it is named at the moment it enters,
           and its cost is stated} \textemdash{} and section 10 then shows that the headline
           result, the mean waiting time, survives the removal of the strongest
           assumption in the model. That was not known when b16 was written.

           All 52 published figures of b16 Table 3 and Table 9 were recomputed from
           these derivations and {\bfseries HELD} (0 BREACH), then cross-checked by
           three further instruments. Three points of prose imprecision were found in
           b16 and are listed in section 13. The companion code is three short Python
           files beside this one, standard library only, short enough to retype.

           What this document deliberately does {\itshape not} do is argue that the
           model is right, that the rates are correctly calibrated, or that the forecast
           should be believed. Those arguments are in b16 \textemdash{} and the calibration, not
           the algebra, is where a critical reader should push. This document
           establishes something narrower and more durable: {\bfseries given the model,
           the numbers follow, and here is why.}

           Offered not to be believed, but to be checked. \#AuditTheMath
      \end{quote}

      \vspace{3mm}
      \begin{center}\bfseries Declarations\end{center}
      \begin{flushleft}\scriptsize
      \textsuperscript{4}\,"of Laodicea" indicates taking responsibility to undo personal complicity with disastrous Laodicean legacies like banning mathematicians from clergy (Canon 36, Council of Laodicea; two magisteria separations), enabling institutional lukewarmness, weapons of math-destruction, and slow-motion explosions of misinformation from pandemics to self-compounding interests.\\[3pt]
      \textsuperscript{5}\,LLoL stands for ridiculous luck in serendipitous discovery and a commitment to find ever more fun ways to help others uncover street-wise math that matters.\\[3pt]
      \textsuperscript{6}\,Loewe's traditional standards for co-authorship demand naming AI Claude Opus 5 Max (by Anthropic) as a co-author for many substantial contributions, as if a PhD-student. Yet, AI co-authorship is withheld here until Loewe's framework for AI co-authorship after the practical singularity (PraS) passes external human peer review (see Matheo-b21 study). Anthropic is not responsible for AI mistakes here. Loewe as senior corresponding author remains forward accountable for every number in this study, including those a machine computed.\\[3pt]
      \textsuperscript{7}\,\textit{Licensed under the Jonah License and CC-BY 4.0 for maximal flexibility (see \href{https://balospe.com/en/license/joli/}{https://balospe.com/en/license/joli/}).}\\
      \end{flushleft}

      \end{titlepage}
      \newpage
      \tableofcontents
      \newpage


###########################################################################
b16-extra --- The Waiting-Time Law, Derived Line by Line
###########################################################################

.. only:: html

   | **Laurence Loewe of Laodicea**
   | Companion to Study b16 in the Matheo Study Series --- appendices SIA-4, SIA-5, SIA-8

.. admonition:: MockupModel draft
   :class: warning

   MMv1 --- not yet reviewed by anyone. Full provenance, version and audit status are
   in :ref:`b16x-provenance` at the end.

.. epigraph::

   Offered not to be believed, but to be checked.


***************************************************************************
1. What this document is
***************************************************************************

Matheo-b16 forecasts how long the world should expect to wait before an accidental
nuclear winter begins, using a three-state model small enough to print on a poster. Its
appendix SIA-4 states the exact mathematics behind that forecast in about one page, and
one page is the right length there: a reader who wants the forecast does not want a
derivation, and a professional applied mathematician does not need one --- the result is
a standard first-passage calculation that such a reader can reconstruct in ten minutes.

This document is for everyone in between. It shows **every step**, including the ones a
textbook would omit as obvious, and it names its symbols so that a reader can tell what
each one *does* without holding a table in their head.

.. admonition:: What moved into b16 on 2026m09d17, and why the numbering has gaps
   :class: important

   About half of this companion was promoted into b16 itself, on one test: **does a
   reader who accepts the result need it?** Everything that serves the beginner or the
   expert moved; what serves only a reader following the algebra line by line stayed.

   **Moved** --- the model and its assumption ledger, and the BEST Names tables and
   glossary (b16 **SIA-1**); what "a rate" means and the two racing arrows, reading the
   formula, the four checks, why the mean is not enough, the law drawn, what the two
   rates are and are not, and onset being certain (b16 **SIA-4**); the
   distribution-free result and what none of this establishes (b16 **SIA-11**).

   **Stayed** --- the line-by-line algebra, the three independent routes, the transforms,
   the Michaelis--Menten substitution, the figure audit, the imprecisions found in b16,
   the parked work, and the record of how the names were chosen.

   **The section numbers are deliberately not renumbered.** The gaps --- 2, 4, 7, 11, 14
   --- mark exactly what was promoted, and every cross-reference in either document keeps
   resolving. Renumbering would have rewritten twenty-seven references for a cosmetic
   gain and destroyed the map.

Three commitments shape it.

**Nothing is asserted that is not derived.** Where b16 says "solving the equations
gives", this document solves them, line by line.

**Every result is obtained more than once, by routes that share no machinery.** The
central law is derived three ways: by a transform, by elementary counting with no
transform at all, and by the eigenvalues of a matrix. A fourth instrument --- simulation
that consults no formula --- checks the answer numerically. If a mistake were hiding in
one route, the others would not agree with it. They do agree, to the last digit printed.

**Where an assumption enters, it is named at the moment it enters, and its cost is
stated.** Section :ref:`b16x-distribution-free` then shows that the headline result ---
the mean waiting time --- survives the removal of the strongest assumption in the model.
That was not known when b16 was written, and it is the most useful thing in this
document for anyone facing a sceptical reader.

.. admonition:: What this document deliberately does not do
   :class: tip

   It does not argue that the model is right, that the rates are correctly calibrated,
   or that the forecast should be believed. Those arguments are in b16 --- and the
   calibration, not the algebra, is where a critical reader should push (see
   :ref:`b16x-not-established`). This document establishes something narrower and
   more durable: **given the model, the numbers follow, and here is why.**



***************************************************************************
5. The mean waiting time, line by line
***************************************************************************

5.2 From a MAD crisis
===========================================================================

Starting in a crisis, three things happen in order.

**Step 1 --- the crisis runs for a while.** By Fact 2, its mean duration is
:math:`\dfrac{1}{r_{M \rightarrow \text{out}}}`.

**Step 2 --- it ends one of two ways.** By Fact 3, with probability
:math:`\dfrac{r_{M \rightarrow D}}{r_{M \rightarrow \text{out}}}` it escalates, and we have arrived: nothing further
to wait for. With probability :math:`\dfrac{r_{M \rightarrow R}}{r_{M \rightarrow \text{out}}}` it de-escalates and the
world is back in Risky.

**Step 3 --- if it de-escalated, we face the original problem again.** By
memorylessness (Fact 1), the world in Risky after a crisis is in *exactly* the same
position as the world in Risky before it: no memory, no accumulated safety, no
accumulated danger. So the remaining mean wait is :math:`T_{R \rightarrow D}` --- the same :math:`T_{R \rightarrow D}` we
are trying to find.

Adding the mean duration to the mean of what follows:

.. math::

   T_{M \rightarrow D} \;=\; \underbrace{\frac{1}{r_{M \rightarrow \text{out}}}}_{\text{how long the crisis lasts}}
   \;+\; \underbrace{\frac{r_{M \rightarrow D}}{r_{M \rightarrow \text{out}}}\cdot 0}_{\text{it escalated: done}}
   \;+\; \underbrace{\frac{r_{M \rightarrow R}}{r_{M \rightarrow \text{out}}}\cdot T_{R \rightarrow D}}_{\text{it de-escalated: start over}}

Dropping the middle term, which is zero:

.. math::

   T_{M \rightarrow D} \;=\; \frac{1}{r_{M \rightarrow \text{out}}} \;+\; \frac{r_{M \rightarrow R}}{r_{M \rightarrow \text{out}}}\, T_{R \rightarrow D}

5.3 From Risky
===========================================================================

Starting in Risky, only one arrow can fire (the escape arrow has rate zero), so the
world waits a mean :math:`1/r_{R \rightarrow M}` years and then, with certainty, is in a MAD crisis:

.. math::

   T_{R \rightarrow D} \;=\; \frac{1}{r_{R \rightarrow M}} \;+\; T_{M \rightarrow D}

5.4 Solving the pair
===========================================================================

Two equations, two unknowns. Substitute the MAD-crisis equation into the Risky equation:

.. math::

   T_{R \rightarrow D} \;=\; \frac{1}{r_{R \rightarrow M}} \;+\; \frac{1}{r_{M \rightarrow \text{out}}} \;+\; \frac{r_{M \rightarrow R}}{r_{M \rightarrow \text{out}}}\, T_{R \rightarrow D}

Collect the :math:`T_{R \rightarrow D}` terms on the left:

.. math::

   T_{R \rightarrow D} \;-\; \frac{r_{M \rightarrow R}}{r_{M \rightarrow \text{out}}}\, T_{R \rightarrow D} \;=\; \frac{1}{r_{R \rightarrow M}} \;+\; \frac{1}{r_{M \rightarrow \text{out}}}

Factor the left side:

.. math::

   T_{R \rightarrow D}\left(1 - \frac{r_{M \rightarrow R}}{r_{M \rightarrow \text{out}}}\right) \;=\; \frac{1}{r_{R \rightarrow M}} \;+\; \frac{1}{r_{M \rightarrow \text{out}}}

The bracket is :math:`\dfrac{r_{M \rightarrow \text{out}}-r_{M \rightarrow R}}{r_{M \rightarrow \text{out}}} = \dfrac{r_{M \rightarrow D}}{r_{M \rightarrow \text{out}}}`,
which is exactly :math:`p_{M \rightarrow D}`. So

.. math::

   T_{R \rightarrow D} \cdot \frac{r_{M \rightarrow D}}{r_{M \rightarrow \text{out}}} \;=\; \frac{1}{r_{R \rightarrow M}} \;+\; \frac{1}{r_{M \rightarrow \text{out}}}

Multiply both sides by :math:`\dfrac{r_{M \rightarrow \text{out}}}{r_{M \rightarrow D}}`:

.. math::

   T_{R \rightarrow D} \;=\; \frac{r_{M \rightarrow \text{out}}}{r_{M \rightarrow D}}\left(\frac{1}{r_{R \rightarrow M}} + \frac{1}{r_{M \rightarrow \text{out}}}\right)
   \;=\; \frac{r_{M \rightarrow \text{out}}}{r_{R \rightarrow M}\,r_{M \rightarrow D}} \;+\; \frac{1}{r_{M \rightarrow D}}

.. math::

   \boxed{\;\; T_{R \rightarrow D} \;=\; \frac{r_{M \rightarrow \text{out}}}{r_{R \rightarrow M}\,r_{M \rightarrow D}} \;+\; \frac{1}{r_{M \rightarrow D}} \;\;}

**At base case:** :math:`T_{R \rightarrow D} = \dfrac{9}{0.1 \times 3} + \dfrac{1}{3} = 30 + 0.333 =
\mathbf{30.33}` **years.**

***************************************************************************
6. The whole distribution, not just its average
***************************************************************************

6.2 What a Laplace transform is, and what it assumes
===========================================================================

The standard tool for exactly this difficulty is the **Laplace transform**. For a
positive random quantity :math:`T`, it is defined as

.. math::

   f_T(s) \;=\; \mathbb{E}\!\left[e^{-sT}\right]
   \;=\; \int_0^{\infty} e^{-st}\, p_T(t)\, dt

where :math:`p_T` is the density. Read it as a *weighted average of* :math:`e^{-sT}`,
computed for every value of the bookkeeping variable :math:`s > 0`.

Three properties, and only these three, are used below.

**(i) It loses nothing.** Two distributions with the same transform for all
:math:`s > 0` are the same distribution (the uniqueness theorem). So moving into
transform space discards no information and can be undone.

**(ii) It turns addition into multiplication.** If :math:`A` and :math:`B` are
independent, then :math:`\mathbb{E}[e^{-s(A+B)}] = \mathbb{E}[e^{-sA}]\,
\mathbb{E}[e^{-sB}]`, because the exponential of a sum is the product of exponentials
and independence lets the average split. This is the whole reason to use it: the
hopeless operation (combining waiting times) becomes ordinary multiplication.

**(iii) Exponentials have a simple transform.** If :math:`X` is exponential with rate
:math:`\alpha`,

.. math::

   \mathbb{E}\!\left[e^{-sX}\right] \;=\; \int_0^{\infty} e^{-st}\, \alpha e^{-\alpha t}\, dt
   \;=\; \alpha \int_0^{\infty} e^{-(\alpha+s)t}\, dt
   \;=\; \frac{\alpha}{\alpha+s}

.. admonition:: "Laplace" here is a name, not an assumption
   :class: important

   A reasonable reader meets the phrase "taking the Laplace transform" and wonders
   whether something has just been assumed --- in particular whether this is connected
   to **Laplace's principle of insufficient reason** (or his rule of succession), the
   idea that in the absence of information one should treat outcomes as equally likely.
   That principle is a genuine and contestable assumption about prior uncertainty, and
   a reader who has seen it smuggled into a forecast is right to be watchful.

   **These are two different things by the same man.** Pierre-Simon Laplace
   (1749--1827) worked on both. The *principle* is a claim about what to believe before
   seeing evidence. The *transform* is a change of coordinates --- a piece of
   bookkeeping, like taking logarithms to turn multiplication into addition. It assumes
   nothing, adds nothing, and by property (i) can be reversed exactly. **No prior, no
   uncertainty assumption, and no new information enters at this step.**

   That claim is not left to be trusted. Section 8.2 derives the same distribution
   again by solving the model's own differential equations directly, with **no
   transform of any kind**, and reaches the identical answer. If the transform had
   smuggled something in, the two routes would disagree.

6.3 The same two equations, in transform space
===========================================================================

Write :math:`f(s)` for the transform of the waiting time from **Risky** and
:math:`g(s)` for the transform of the waiting time from a **MAD crisis**. The reasoning is
exactly that of section 5, but now it tracks whole distributions instead of averages.

**From Risky.** The wait is an exponential time at rate :math:`r_{R \rightarrow M}`, *followed by*
the wait from a MAD crisis, and the two are independent. By property (ii), transforms multiply;
by property (iii), the first factor is :math:`r_{R \rightarrow M}/(r_{R \rightarrow M}+s)`:

.. math::

   f(s) \;=\; \frac{r_{R \rightarrow M}}{r_{R \rightarrow M}+s}\; g(s)

**From a MAD crisis.** The wait is an exponential time at rate :math:`r_{M \rightarrow \text{out}}` --- the crisis
itself (Fact 2) --- followed by *either* nothing, with probability
:math:`r_{M \rightarrow D}/r_{M \rightarrow \text{out}}`, *or* the whole wait from Risky again, with probability
:math:`r_{M \rightarrow R}/r_{M \rightarrow \text{out}}`. Crucially, Fact 3 says the duration and the outcome are
independent, so the duration's transform multiplies a plain average over the two
outcomes:

.. math::

   g(s) \;=\; \frac{r_{M \rightarrow \text{out}}}{r_{M \rightarrow \text{out}}+s}
   \left[\frac{r_{M \rightarrow D}}{r_{M \rightarrow \text{out}}}\cdot 1 \;+\; \frac{r_{M \rightarrow R}}{r_{M \rightarrow \text{out}}}\, f(s)\right]
   \;=\; \frac{r_{M \rightarrow D} \;+\; r_{M \rightarrow R}\, f(s)}{r_{M \rightarrow \text{out}}+s}

(The "1" is the transform of a wait of length zero: :math:`\mathbb{E}[e^{-s\cdot 0}] =
1`.)

6.4 Solving for the transform
===========================================================================

Substituting the second equation into the first:

.. math::

   f(s) \;=\; \frac{r_{R \rightarrow M}}{r_{R \rightarrow M}+s}\cdot\frac{r_{M \rightarrow D} + r_{M \rightarrow R} f(s)}{r_{M \rightarrow \text{out}}+s}

Multiply both sides by :math:`(r_{R \rightarrow M}+s)(r_{M \rightarrow \text{out}}+s)`:

.. math::

   f(s)\,(r_{R \rightarrow M}+s)(r_{M \rightarrow \text{out}}+s) \;=\; r_{R \rightarrow M}\,r_{M \rightarrow D} \;+\; r_{R \rightarrow M}\,r_{M \rightarrow R}\, f(s)

Collect the :math:`f(s)` terms:

.. math::

   f(s)\,\Big[(r_{R \rightarrow M}+s)(r_{M \rightarrow \text{out}}+s) - r_{R \rightarrow M}\,r_{M \rightarrow R}\Big] \;=\; r_{R \rightarrow M}\,r_{M \rightarrow D}

Expand the bracket:

.. math::

   (r_{R \rightarrow M}+s)(r_{M \rightarrow \text{out}}+s)
   &= r_{R \rightarrow M}\,r_{M \rightarrow R} + r_{R \rightarrow M}\,r_{M \rightarrow D} + r_{R \rightarrow M} s + s\,r_{M \rightarrow R} + s\,r_{M \rightarrow D} + s^{2} \\
   &= s^{2} + s(r_{R \rightarrow M}+r_{M \rightarrow R}+r_{M \rightarrow D}) + r_{R \rightarrow M}\,r_{M \rightarrow R} + r_{R \rightarrow M}\,r_{M \rightarrow D}

Subtracting :math:`r_{R \rightarrow M}\,r_{M \rightarrow R}` removes the only term that mentions :math:`r_{M \rightarrow R}` on its
own:

.. math::

   (r_{R \rightarrow M}+s)(r_{M \rightarrow \text{out}}+s) - r_{R \rightarrow M}\,r_{M \rightarrow R} \;=\; s^{2} + s(r_{R \rightarrow M}+r_{M \rightarrow R}+r_{M \rightarrow D}) + r_{R \rightarrow M}\,r_{M \rightarrow D}

and therefore

.. math::

   \boxed{\;\; f(s) \;=\; \frac{r_{R \rightarrow M}\,r_{M \rightarrow D}}{s^{2} + s(r_{R \rightarrow M}+r_{M \rightarrow R}+r_{M \rightarrow D}) + r_{R \rightarrow M}\,r_{M \rightarrow D}} \;\;}

That quadratic denominator is the entire content of the model. Everything that follows
is reading it.

6.5 The two rates
===========================================================================

A quadratic factors through its roots. Write the denominator as
:math:`(s + r_{\text{onset}})(s + r_{\text{relax}})`. Expanding and matching coefficients
gives the two relations (Vieta's formulas) that will be used repeatedly as checks:

.. math::

   r_{\text{onset}} + r_{\text{relax}} = r_{R \rightarrow M} + r_{M \rightarrow R} + r_{M \rightarrow D}
   \qquad\text{and}\qquad
   r_{\text{onset}}\, r_{\text{relax}} = r_{R \rightarrow M}\,r_{M \rightarrow D}

This three-term sum appears throughout, and it is worth reading rather than
skipping: it collapses to
:math:`r_{R \rightarrow M} + r_{M \rightarrow \text{out}}` --- *the rate of leaving
Risky, plus the rate of leaving a MAD crisis.* Every quantity below is built from those two
numbers and their product with the escalation rate. Written out in Greek, as drafts
before 2026m09d08 had it, the same sum looked like an unmotivated lump; that is the
whole reason the arrow names were adopted.

Solving by the quadratic formula, with :math:`r_{\text{all}} = r_{R \rightarrow M}+r_{M \rightarrow R}+r_{M \rightarrow D}`:

.. math::

   r_{\text{onset}} = \frac{r_{\text{all}} - \sqrt{\Delta}}{2},
   \qquad
   r_{\text{relax}} = \frac{r_{\text{all}} + \sqrt{\Delta}}{2},
   \qquad
   \Delta = r_{\text{all}}^{2} - 4r_{R \rightarrow M}\,r_{M \rightarrow D}

**The roots are always real, for any positive rates.** Expanding the discriminant:

.. math::

   \Delta &= (r_{R \rightarrow M}+r_{M \rightarrow R}+r_{M \rightarrow D})^{2}
              - 4r_{R \rightarrow M}\,r_{M \rightarrow D} \\
     &= r_{R \rightarrow M}^{2} + r_{M \rightarrow R}^{2} + r_{M \rightarrow D}^{2}
        + 2r_{R \rightarrow M}\,r_{M \rightarrow R} \\
     &\qquad + 2r_{R \rightarrow M}\,r_{M \rightarrow D}
        + 2r_{M \rightarrow R}\,r_{M \rightarrow D}
        - 4r_{R \rightarrow M}\,r_{M \rightarrow D} \\
     &= (r_{R \rightarrow M}-r_{M \rightarrow D})^{2} + r_{M \rightarrow R}^{2}
        + 2r_{M \rightarrow R}(r_{R \rightarrow M}+r_{M \rightarrow D})

Every term on the last line is non-negative, so :math:`\Delta \geq 0` always. It vanishes
only if :math:`r_{M \rightarrow R} = 0` **and** :math:`r_{R \rightarrow M} = r_{M \rightarrow D}` simultaneously --- a knife-edge
case in which the two roots coincide and the waiting time is a Gamma distribution with
shape 2 rather than the two-rate law below, with
:math:`P(T_{R \rightarrow D} \leq t) = 1 - (1 + rt)e^{-rt}`. It does not arise for any parameters this
paper uses, and it is recorded only so that the general statement is true as stated.

**At base case** (:math:`r_{R \rightarrow M} = 0.1`, :math:`r_{M \rightarrow R} = 6`, :math:`r_{M \rightarrow D} = 3`):
:math:`r_{\text{all}} = 9.1`, :math:`\Delta = 81.61`, :math:`\sqrt{\Delta} = 9.03383`, giving

.. math::

   r_{\text{onset}} = 0.033087\ \text{/yr},
   \qquad
   r_{\text{relax}} = 9.06691\ \text{/yr}

Both Vieta checks hold: the sum is 9.1 and the product is 0.3, exactly.

6.6 Recognising the answer
===========================================================================

Because :math:`r_{\text{onset}} r_{\text{relax}} = r_{R \rightarrow M}\,r_{M \rightarrow D}`, the numerator of
:math:`f(s)` is itself the product of the two roots, so the transform can be written

.. math::

   f(s) \;=\; \frac{r_{\text{onset}}\, r_{\text{relax}}}
   {(s + r_{\text{onset}})(s + r_{\text{relax}})}
   \;=\; \underbrace{\frac{r_{\text{onset}}}{s + r_{\text{onset}}}}_{\text{an exponential}}
   \;\times\;
   \underbrace{\frac{r_{\text{relax}}}{s + r_{\text{relax}}}}_{\text{another exponential}}

By property (iii), each factor is the transform of an exponential waiting time. By
property (ii), a product of transforms is the transform of a **sum of independent
pieces**. By property (i), no other distribution has this transform. Therefore:

.. math::

   \boxed{\;\; T_{R \rightarrow D} \;\sim\; \mathrm{Exp}(r_{\text{onset}}) \;+\; \mathrm{Exp}(r_{\text{relax}}),
   \quad \text{independent} \;\;}

This shape has a name --- **hypoexponential** --- and one consequence worth stating
plainly: the waiting time is **not** exponential. It is the sum of two exponentials
with very different rates. That difference is invisible over decades and decisive over
a single year (:ref:`b16x-two-rates`).

6.7 Inverting, to get the answer in years
===========================================================================

Split the transform by partial fractions:

.. math::

   \frac{1}{(s+r_{\text{onset}})(s+r_{\text{relax}})}
   \;=\; \frac{1}{r_{\text{relax}} - r_{\text{onset}}}
   \left[\frac{1}{s+r_{\text{onset}}} - \frac{1}{s+r_{\text{relax}}}\right]

Since :math:`1/(s+r)` is the transform of :math:`e^{-rt}`, reading the line backwards
gives the density directly:

.. math::

   p_{\text{winter}}(t) \;=\; \frac{r_{\text{onset}}\, r_{\text{relax}}}{r_{\text{relax}} - r_{\text{onset}}}
   \left(e^{-r_{\text{onset}} t} - e^{-r_{\text{relax}} t}\right)

Integrating from :math:`t` to infinity gives the chance the world has **not** yet
arrived:

.. math::

   P(T_{R \rightarrow D} > t)
   &= \frac{r_{\text{onset}} r_{\text{relax}}}{r_{\text{relax}} - r_{\text{onset}}}
      \left[\frac{e^{-r_{\text{onset}} t}}{r_{\text{onset}}}
          - \frac{e^{-r_{\text{relax}} t}}{r_{\text{relax}}}\right] \\
   &= \frac{r_{\text{relax}}\, e^{-r_{\text{onset}} t}
          - r_{\text{onset}}\, e^{-r_{\text{relax}} t}}
         {r_{\text{relax}} - r_{\text{onset}}}

and finally the quantity the paper reports:

.. math::

   \boxed{\;\; P(T_{R \rightarrow D} \leq t) \;=\; 1 \;-\;
   \frac{r_{\text{relax}}\, e^{-r_{\text{onset}} t} - r_{\text{onset}}\, e^{-r_{\text{relax}} t}}
        {r_{\text{relax}} - r_{\text{onset}}} \;\;}

**At base case, one year:** :math:`P(T_{R \rightarrow D} \leq 1) = 0.0290`, that is **2.90 %**, or about
1 in 34. Every figure in b16 Table 3 comes from this expression, and all of them are
recomputed in section 12.

**Consistency with section 5.** For a sum of independent exponentials the means add, so
:math:`\mathbb{E}[T_{R \rightarrow D}] = 1/r_{\text{onset}} + 1/r_{\text{relax}}`. Using Vieta:

.. math::

   \frac{1}{r_{\text{onset}}} + \frac{1}{r_{\text{relax}}}
   = \frac{r_{\text{onset}} + r_{\text{relax}}}{r_{\text{onset}} r_{\text{relax}}}
   = \frac{r_{R \rightarrow M}+r_{M \rightarrow R}+r_{M \rightarrow D}}{r_{R \rightarrow M}\,r_{M \rightarrow D}}
   = \frac{r_{M \rightarrow \text{out}}}{r_{R \rightarrow M}\,r_{M \rightarrow D}} + \frac{1}{r_{M \rightarrow D}}

which is exactly the boxed mean of section 5.4 --- **derived twice, by two
different routes, without assuming the second.** ✓ (30.2229 + 0.1103 = 30.3333 years.)


.. _b16x-fig-cdf:

***************************************************************************
8. The same answer, by two routes that share no machinery
***************************************************************************

A single derivation is a single point of failure. Below, the same law is obtained twice
more. The three routes share no step: one uses transforms, one uses counting, one uses
calculus. A mistake would have to be replicated three different ways to survive.

8.1 By counting crises (no calculus)
===========================================================================

Let :math:`n_{MAD}` be the number of crises until one escalates. Each crisis
escalates independently with probability :math:`p_{M \rightarrow D}`, so
:math:`n_{MAD}` is **geometric**, with mean :math:`1/p_{M \rightarrow D} = 3`.

The waiting time is then

.. math::

   T_{R \rightarrow D} \;=\; \underbrace{\sum_{i=1}^{n_{MAD}} X_i}_{\text{waiting in Risky}}
   \;+\; \underbrace{\sum_{i=1}^{n_{MAD}} Y_i}_{\text{inside crises}},
   \qquad X_i \sim \mathrm{Exp}(r_{R \rightarrow M}),\quad Y_i \sim \mathrm{Exp}r_{M \rightarrow \text{out}}

.. admonition:: Lemma --- a geometric number of exponentials is one exponential
   :class: note

   If :math:`X_1, X_2, \ldots` are independent :math:`\mathrm{Exp}(\alpha)` and
   :math:`N` is geometric with success probability :math:`p`, then
   :math:`\sum_{i=1}^{N} X_i \sim \mathrm{Exp}(\alpha p)`.

   *Proof, without algebra.* Picture a Poisson process of rate :math:`\alpha`: its
   gaps are exactly the :math:`X_i`. Toss an independent coin at each point, keeping it
   with probability :math:`p`. **Thinning** a Poisson process of rate :math:`\alpha`
   by keeping each point with probability :math:`p` produces a Poisson process of rate
   :math:`\alpha p`. The time to the first kept point is therefore
   :math:`\mathrm{Exp}(\alpha p)` --- and it is also, by construction, the sum of a
   geometric number of gaps. ∎

Applying the lemma twice:

.. math::

   \sum_{i=1}^{n_{MAD}} X_i \sim \mathrm{Exp}(r_{R \rightarrow M}\, p_{M \rightarrow D}),
   \qquad
   \sum_{i=1}^{n_{MAD}} Y_i \sim \mathrm{Exp}\big(r_{M \rightarrow \text{out}}\, p_{M \rightarrow D}\big)
   = \mathrm{Exp}(r_{M \rightarrow D})

because :math:`r_{M \rightarrow \text{out}} \cdot \dfrac{r_{M \rightarrow D}}{r_{M \rightarrow \text{out}}} = r_{M \rightarrow D}` exactly. Adding
the two means (which is legitimate whether or not they are independent):

.. math::

   \mathbb{E}[T_{R \rightarrow D}] \;=\; \frac{1}{r_{R \rightarrow M}\, p_{M \rightarrow D}} + \frac{1}{r_{M \rightarrow D}}
   \;=\; \frac{r_{M \rightarrow \text{out}}}{r_{R \rightarrow M}\,r_{M \rightarrow D}} + \frac{1}{r_{M \rightarrow D}}

--- the boxed mean of section 5.4, recovered with **no differential equations, no
transforms, and no algebra beyond one cancellation.** ✓

Pushing the same route to the full distribution requires the generating function of
:math:`n_{MAD}`; doing so reproduces the transform of section 6.4 exactly.
Section 7.3 explains why the two exponentials found here are *not* the two in the
closed form.

8.2 By solving the model's differential equations (no transforms)
===========================================================================

This route uses nothing from sections 6 or 8.1. Let :math:`p_R(t)` and :math:`p_M(t)`
be the probabilities of being in Risky and in a MAD crisis at time :math:`t`. Probability flows
along the arrows, so

.. math::

   \frac{dp_R}{dt} &= -r_{R \rightarrow M}\, p_R \;+\; r_{M \rightarrow R}\, p_M \\
   \frac{dp_M}{dt} &= \phantom{-}r_{R \rightarrow M}\, p_R \;-\; r_{M \rightarrow \text{out}}\, p_M

with :math:`p_R(0) = 1`, :math:`p_M(0) = 0`. These are the **Kolmogorov forward
equations**, and they are also exactly what the Evolvix model compiles to.

A linear system like this has solutions built from :math:`e^{-rt}`, where the
:math:`-r` are the eigenvalues of the coefficient matrix

.. math::

   A = \begin{pmatrix} -r_{R \rightarrow M} & r_{M \rightarrow R} \\
       r_{R \rightarrow M} & -r_{M \rightarrow \text{out}} \end{pmatrix}

.. math::

   \det(A - xI)
   &= (r_{R \rightarrow M}+x)(r_{M \rightarrow \text{out}}+x)
      - r_{R \rightarrow M}\,r_{M \rightarrow R} \\
   &= x^{2} + x(r_{R \rightarrow M}+r_{M \rightarrow R}+r_{M \rightarrow D})
      + r_{R \rightarrow M}\,r_{M \rightarrow D}

**That is the same quadratic as section 6.4**, arrived at with no transform anywhere.
Its roots are :math:`-r_{\text{onset}}` and :math:`-r_{\text{relax}}`.

Writing :math:`p_R = A_1 e^{-r_{\text{onset}}t} + A_2 e^{-r_{\text{relax}}t}` and
recovering :math:`p_M = (p_R' + r_{R \rightarrow M} p_R)/r_{M \rightarrow R}` mode by mode, the initial conditions
:math:`A_1 + A_2 = 1` and :math:`p_M(0) = 0` give

.. math::

   A_1 = \frac{r_{\text{relax}}-r_{R \rightarrow M}}{r_{\text{relax}}-r_{\text{onset}}},
   \qquad
   A_2 = \frac{r_{R \rightarrow M} - r_{\text{onset}}}{r_{\text{relax}}-r_{\text{onset}}}

The chance of not yet having arrived is :math:`P(T_{R \rightarrow D} > t) = p_R(t) + p_M(t)`, and
substituting, then simplifying with the two Vieta relations, gives the coefficient
identities

.. math::

   (r_{\text{relax}}-r_{R \rightarrow M})
   (r_{M \rightarrow R}+r_{R \rightarrow M}-r_{\text{onset}})
   &= r_{M \rightarrow R}\, r_{\text{relax}}, \\
   (r_{R \rightarrow M}-r_{\text{onset}})
   (r_{M \rightarrow R}+r_{R \rightarrow M}-r_{\text{relax}})
   &= -r_{M \rightarrow R}\, r_{\text{onset}}

so that

.. math::

   P(T_{R \rightarrow D} > t) \;=\; \frac{r_{\text{relax}} e^{-r_{\text{onset}} t}
   - r_{\text{onset}} e^{-r_{\text{relax}} t}}{r_{\text{relax}} - r_{\text{onset}}}

**identical to section 6.7.** ✓ Two routes, no shared machinery, one answer.

8.3 By running the model (no algebra at all)
===========================================================================

The fourth instrument consults no formula. It generates world histories from the
model's own rules --- draw a waiting time, draw a crisis duration, toss a coin weighted
:math:`p_{M \rightarrow D}`, repeat --- and counts how many arrive by each horizon. This is
the method of b16 SIA-2 (LLoL's original SD1 analysis), run at 400,000 runs instead of
40 so that it can resolve the short-horizon figures the 40-run version explicitly could
not.

.. list-table:: Table E. Simulation against closed form, base case
   :header-rows: 1
   :widths: 34 22 22 22

   * - quantity
     - simulated
     - exact
     - agreement
   * - mean wait
     - 30.345 yr
     - 30.333 yr
     - 0.04 %
   * - crises before onset
     - 2.9986
     - 3.0000
     - 0.05 %
   * - :math:`P(T_{R \rightarrow D} \leq 1\,\text{yr})`
     - 2.900 %
     - 2.900 %
     - exact to 3 digits
   * - :math:`P(T_{R \rightarrow D} \leq 20\,\text{yr})`
     - 48.216 %
     - 48.216 %
     - exact to 3 digits
   * - :math:`P(T_{R \rightarrow D} \leq 80\,\text{yr})`
     - 92.864 %
     - 92.888 %
     - 0.02 pp

**The simulator, in full.** The world-generator is eight lines, and it consults
nothing:

.. code-block:: python

   import random

   def one_world(rMADcrisisStarts, rMADcrisisCalms, rMADcrisisToWinter):
       """One random history: the year accidental nuclear winter begins."""
       t = 0.0
       while True:
           t += random.expovariate(rMADcrisisStarts)                  # wait in Risky
           t += random.expovariate(rMADcrisisCalms + rMADcrisisToWinter)    # the crisis runs
           if random.random() < rMADcrisisToWinter / (rMADcrisisCalms + rMADcrisisToWinter):
               return t                                          # it escalated

   random.seed(20260907)                              # so this reproduces exactly
   runs = [one_world(0.10, 6, 3) for _ in range(400_000)]
   print(sum(runs) / len(runs))                       # 30.345  vs exact 30.333
   print(sum(t <= 1 for t in runs) / len(runs))       #  0.0290 vs exact 0.0290

Nothing in it knows the closed form, the Laplace transform, or the two rates. It draws
the three random quantities the model describes and counts. That it lands on the same
numbers as forty lines of algebra is the check --- and it is the check a reader who
distrusts the algebra can run without reading any of it. The version in
``code/crosschecks.py`` is the same loop with the horizon tallies and the
variance/covariance instrument added.


.. _b16x-mm:

***************************************************************************
9. Michaelis--Menten, derived rather than asserted
***************************************************************************

b16 SIA-7 claims that this model is not merely *analogous* to enzyme kinetics but
**identical** to it. That is a strong claim, and it is cheap to check.

Take the reciprocal of the mean wait --- onsets per year, the *velocity* of the process
--- and simplify using the boxed mean of section 5.4:

.. math::

   v(r_{R \rightarrow M})
   &\;=\; \frac{1}{T_{R \rightarrow D}}
   \;=\; \frac{1}
      {\dfrac{r_{M \rightarrow \text{out}}}{r_{R \rightarrow M}\,r_{M \rightarrow D}}
       + \dfrac{1}{r_{M \rightarrow D}}} \\[4pt]
   &\;=\; \frac{1}
      {\dfrac{r_{M \rightarrow \text{out}} + r_{R \rightarrow M}}
             {r_{R \rightarrow M}\,r_{M \rightarrow D}}}
   \;=\; \frac{r_{R \rightarrow M}\,r_{M \rightarrow D}}
      {r_{R \rightarrow M} + r_{M \rightarrow R} + r_{M \rightarrow D}}

Now write the Michaelis--Menten rate law of enzyme kinetics beside it:

.. math::

   v \;=\; \frac{V_{max}\, S}{K_m + S}

These are the **same function**, under one substitution, at every value of
:math:`r_{R \rightarrow M}` --- not an approximation valid in a limit.

.. _b16x-mm-dictionary:

.. note:: **Table F is superseded by** :ref:`Table B-MM <b16x-names-mm>` **and
   should be replaced by it.** Table B-MM carries the same five mappings plus
   :math:`k_{on}[S]` and :math:`k_{off}`, plus the Explicit names, and it is the
   single place where the enzyme-kinetics reading of each quantity is stated. Table
   F is left standing here only until that move is made, so that section 9 is not
   left without its dictionary in the meantime. **One edit, not a rewrite:** move
   Table B-MM here and delete this note together with the table below.

.. list-table:: Table F. The dictionary
   :header-rows: 1
   :widths: 24 32 16 28

   * - Enzyme kinetics
     - RiskyMAD
     - base value
     - meaning here
   * - substrate concentration :math:`S`
     - crisis rate :math:`r_{R \rightarrow M}`
     - 0.1 /yr
     - how often the world enters a crisis
   * - :math:`K_m`
     - :math:`r_{M \rightarrow \text{out}}`
     - 9 /yr
     - how fast a crisis resolves, either way
   * - :math:`V_{max}`
     - :math:`r_{M \rightarrow D}`
     - 3 /yr
     - the committed step --- **you cannot die faster than this**
   * - :math:`k_{cat}/K_m`
     - :math:`r_{M \rightarrow D}/r_{M \rightarrow \text{out}}`
     - 1/3
     - the specificity constant
   * - commitment to catalysis
     - :math:`p_{M \rightarrow D}`
     - 1/3
     - the share of crises that go all the way

**The defining property holds exactly.** Michaelis--Menten's :math:`K_m` is the
substrate level at which the rate is half its maximum. Here, setting
:math:`r_{R \rightarrow M} = r_{M \rightarrow \text{out}} = 9`:

.. math::

   v(9) \;=\; \frac{9 \times 3}{9 + 9} \;=\; 1.5 \;=\; \frac{r_{M \rightarrow D}}{2} \;=\; \frac{V_{max}}{2}
   \qquad ✓

**Where this system sits.** :math:`S/K_m = r_{R \rightarrow M}/r_{M \rightarrow \text{out}} = 0.1/9 = 0.011` ---
**ninety times below half-saturation.** The system is deep in the first-order regime,
which is precisely *why* the naive product :math:`r_{R \rightarrow M} \cdot p_{M \rightarrow D}` works
as a rate at all: it is the low-substrate limit of a saturating law, and it agrees with
the exact velocity to 1.1 %. Saturation would require a nuclear crisis roughly every
six weeks.

**Why this is worth a section rather than a footnote.** The value of Michaelis--Menten
kinetics has never been that enzymes are simple. An enzyme is a large machine wandering
an enormous configuration space, and how it finds its substrate is, in general,
hopeless to model. The achievement is that *you do not need to*: all of that complexity
is absorbed into one measurable encounter rate. The same structure holds here. The
"configuration space" is the space of historical causal chains --- every way the world
can arrange itself into a nuclear crisis --- which nobody can model. It need not be
modelled, because the encounter rate can be **counted instead**, from the record of
times the world did in fact enter that state. The near-miss record plays exactly the
role a measured :math:`k_{on}[S]` plays in enzymology: the empirical stand-in for a
mechanism too complex to derive.

**And one prediction falls out for free.** The share of time the system spends in the
dangerous state is about :math:`r_{R \rightarrow M}/(r_{R \rightarrow M}+r_{M \rightarrow R}+r_{M \rightarrow D}) = 1.1\,\%`. A process that
is quiescent 99 % of the time and lethal one third of the time it is not will produce
long calm stretches **as a matter of course**. At base case the median wait is 21 years
--- yet roughly one run in forty survives past 127 years. Both come from the same
model. An observer inside the long run concludes the risk is small; an observer inside
the short one does not; neither observation moves the underlying rate.

**The eighty quiet years since 1945 are therefore weak evidence about the parameters,
not reassurance about them.** That is a mathematical property of first-passage times in
a saturating two-step scheme, and it is checkable by anyone who runs the model.


.. _b16x-distribution-free:

***************************************************************************
10. What survives without the exponential assumption
***************************************************************************

Assumption **A2** --- constant rates, memoryless waiting --- is the strongest thing the
model asks for, and it is the first thing a trained sceptic attacks. *"Real crises are
not memoryless. Real crisis arrivals are not Poisson. Your whole result is an artefact
of a mathematical convenience."*

The objection is well aimed. It is also, for the headline number, **wrong** --- and
this section shows why. The result below is not in b16.

10.2 The proof
===========================================================================

Let :math:`n_{MAD}` be the number of crises until one escalates. Each crisis
escalates independently with probability :math:`p_{M \rightarrow D}`, so
:math:`n_{MAD}` is geometric with :math:`\mathbb{E}[n_{MAD}] =
1/p_{M \rightarrow D} < \infty` --- and this uses only *independence between crises*, not
any distributional form.

The waiting time is the sum of the gaps plus the sum of the durations. For a random
number of independent pieces, **Wald's identity** applies whenever the count is a
stopping rule and has finite mean:

.. math::

   \mathbb{E}\!\left[\sum_{i=1}^{n_{MAD}} X_i\right]
   = \mathbb{E}[n_{MAD}] \cdot \mathbb{E}[X]
   = \frac{1}{p_{M \rightarrow D}} \cdot \frac{1}{r_{R \rightarrow M}}

Here the count is decided by the coin tosses that end each crisis, which are
independent of how long the gaps are --- so Wald applies. The same argument on the
durations gives :math:`\tau/p_{M \rightarrow D}`, and therefore

.. math::

   \boxed{\;\; \mathbb{E}[T_{R \rightarrow D}] \;=\; \frac{1}{p_{M \rightarrow D}}
   \left(\frac{1}{r_{R \rightarrow M}} + \tau\right) \;\;}

Substituting the exponential values :math:`p_{M \rightarrow D} = r_{M \rightarrow D}/r_{M \rightarrow \text{out}}` and
:math:`\tau = 1/r_{M \rightarrow \text{out}}` recovers section 5.4 exactly. ∎

If crisis duration depends on how the crisis ends --- plausible, since escalating
crises may be shorter --- the formula becomes
:math:`\mathbb{E}[T_{R \rightarrow D}] = \frac{1}{p_{M \rightarrow D}\,r_{R \rightarrow M}} +
\left(\frac{1}{p_{M \rightarrow D}}-1\right)\tau_{\text{de-esc}} + \tau_{\text{esc}}`,
which still needs only means.

***************************************************************************
12. Every published figure, recomputed
***************************************************************************

The derivations above were implemented independently of b16 --- nothing is imported
from the paper --- and every figure the paper prints was recomputed and compared at the
paper's own printed precision.

.. list-table:: Table H. Audit of b16 Table 3 and Table 9
   :header-rows: 1
   :widths: 40 20 40

   * - What was checked
     - Result
     - Instrument
   * - b16 Table 3 --- 4 scenarios |times| 8 figures
     - **32 of 32 HELD**
     - closed form, re-derived
   * - b16 Table 9 --- 4 rows |times| 5 figures
     - **20 of 20 HELD**
     - closed form, re-derived
   * - **total**
     - **52 of 52 HELD, 0 BREACH**
     - ---
   * - the same figures again
     - agree to 3 digits
     - 400,000-run Monte Carlo
   * - the inversion step
     - agree to 5 digits
     - numerical integration of the density
   * - the two root relations
     - exact
     - Vieta identities

**Reproduction, for a reader who wants to run it rather than read it:**

.. code-block:: bash

   cd source/matheology/hell/mm/b/16/extra/code
   python3 reproduce_tables.py     # prints HELD/BREACH for all 52 figures
   python3 crosschecks.py          # four independent instruments

The three files use the Python standard library only --- no installation, nothing to
download beyond the files themselves. The whole closed form is short enough to retype:

.. code-block:: python

   from math import exp, sqrt

   def onset_probability(t, rMADcrisisStarts, rMADcrisisCalms, rMADcrisisToWinter):
       """P(accidental nuclear winter has begun by year t)."""
       total = rMADcrisisStarts + rMADcrisisCalms + rMADcrisisToWinter
       root  = sqrt(total*total - 4*rMADcrisisStarts*rMADcrisisToWinter)
       rWinterSettled, rRiskWarmsUp = (total - root)/2, (total + root)/2
       return 1 - (rRiskWarmsUp*exp(-rWinterSettled*t)
                   - rWinterSettled*exp(-rRiskWarmsUp*t))/(rRiskWarmsUp - rWinterSettled)

   onset_probability(1,  0.10, 6, 3)   # 0.029003  -> 2.90 %  (1 in 34)
   onset_probability(20, 0.10, 6, 3)   # 0.482161  -> 48.22 %

Seven lines. That is the entire quantitative claim of b16, and anyone who disagrees with
the answer must disagree with :math:`r_{R \rightarrow M}`, :math:`r_{M \rightarrow R}`, or
:math:`r_{M \rightarrow D}` --- not with the arithmetic. The parameter names are the
Explicit column of Table B, so the snippet, the companion code and the Evolvix model on
the poster all say the same word for the same rate.


***************************************************************************
13. Three points of imprecision found in b16
***************************************************************************

The audit found **no error in any published number**. It found three places where the
prose is slightly looser than the mathematics, all cheap to fix and all cheap for a
hostile reader to find. They are listed here so that b16 can adopt them or reject them
deliberately.

.. list-table::
   :header-rows: 1
   :widths: 6 30 64

   * - #
     - Where
     - The fix
   * - 1
     - SIA-4, introducing :math:`r_2`
     - b16 calls :math:`r_2 = 9.067`/yr "the fast stage, **the crisis itself**, mean
       duration 40 days". The crisis clock is :math:`r_{M \rightarrow \text{out}} = 9.000`/yr
       (40.58 days); :math:`r_{\text{relax}}` is 0.74 % larger (40.28 days), and
       :ref:`b16-why-the-smaller-root-is-the-onset-rate` proves the inequality is strict. Both round to 40 days, so no
       number moves. Suggested: "the fast mode --- the crisis clock
       :math:`b+c`, shifted by less than one percent".
   * - 2
     - SIA-4, "overstates by ~12 percent **in every scenario**"
     - Measured: 12.3 %, 11.8 %, 11.5 %, 10.6 % across Table 3's four scenarios.
       Suggested: "by 11--12 percent (10.6 % in the most pessimistic scenario)".
   * - 3
     - SIA-8, ":math:`r_{R \rightarrow M}/(r_{R \rightarrow M}+b+c) \approx 0.011`"
     - Two different quantities both equal |approx| 0.011 and the text names them
       alike: the stationary share of the non-absorbing surrogate chain (0.010989) and
       the true quasi-stationary share of the absorbing chain (0.011029). Since the
       "long quiet stretches" argument leans on this, one clause naming which is meant
       closes the gap.

None of the three changes a conclusion. Together they are a fifteen-minute edit.


.. _b16x-not-established:

***************************************************************************
15. Parked, deliberately --- and the rule for when to unpark it
***************************************************************************

The following would all be genuinely useful, and none of them is being built now.

.. list-table::
   :header-rows: 1
   :widths: 28 42 30

   * - Idea
     - What it would add
     - What it would cost
   * - A downloadable Evolvix bundle (as done for b19 on Zenodo)
     - lets a reader re-run the *simulation* rather than the arithmetic
     - packaging, a DOI, a maintenance obligation
   * - A Python toolkit with plots and arbitrary parameter sweeps
     - lets a reader explore rather than check
     - dependency management, documentation, support burden
   * - A phone app or interactive figure
     - broad reach; possibly a faster adoption curve
     - substantial build, design, and hosting effort
   * - More scenarios and parameter combinations
     - completeness
     - little --- section 12 already shows the direction is invariant
   * - Time-varying :math:`r_{R \rightarrow M}(t)` worked through numerically
     - quantifies b16 SIA-X2's rising-rate argument
     - moderate; the certainty result already covers the qualitative claim

.. admonition:: The stopping rule
   :class: warning

   Each of these makes the work more convincing *in principle*, and each competes for
   time with delivering what already exists. The rule proposed here, to be applied
   before any of them is started:

   **Build it only when a specific named reader has said, in their own words, that its
   absence is what is stopping them.**

   Not "an audience might want it". Not "it would help adoption". A named person, a
   stated blocker. Until then the seven lines of code in section 12 are the whole
   reproducibility story, and they are enough for anyone whose objection is genuinely
   about the arithmetic.

   The reason for the rule is that "one more angle of proof" is unbounded, feels
   productive, and is the most comfortable possible way to postpone delivery. The
   forecast in b16 is either worth acting on now or it is not; a toolkit does not
   change which.


***************************************************************************
16. Summary
***************************************************************************

- The waiting time to accidental nuclear winter is
  :math:`\mathrm{Exp}(r_{\text{onset}}) + \mathrm{Exp}(r_{\text{relax}})`, with
  :math:`r_{\text{onset}} + r_{\text{relax}} = r_{R \rightarrow M}+r_{M \rightarrow R}+r_{M \rightarrow D}` and
  :math:`r_{\text{onset}}\, r_{\text{relax}} = r_{R \rightarrow M}\,r_{M \rightarrow D}`. At base case that is
  0.0331/yr and 9.067/yr.
- Its mean is :math:`\dfrac{r_{M \rightarrow \text{out}}}{r_{R \rightarrow M}\,r_{M \rightarrow D}} + \dfrac{1}{r_{M \rightarrow D}}` = 30.3
  years; the factor :math:`r_{M \rightarrow \text{out}}/r_{M \rightarrow D} = 3` **is** the de-escalation loop, not
  an omission of it.
- Practically, the law is *an exponential of rate 0.0331/yr that starts forty days
  late*. The delay matters at one year (2.90 %, not 3.24 %) and nowhere else.
- The identity with Michaelis--Menten kinetics is exact, not analogical, and it is why
  long quiet stretches are the model's ordinary behaviour rather than evidence of
  safety.
- **The mean needs no exponential assumption at all** --- only a mean gap, a mean
  duration, and an escalation probability. Substituting Cuba's actual thirteen days
  moves it by 0.7 %.
- **Two questions, two answers, both exact.** Asked "we are not in a crisis today", the
  one-year figure is **2.90 %**; asked "in a typical year", the right initial condition
  is the quasi-stationary one, the law is a *pure* exponential at
  :math:`r_{\text{onset}}`, and the answer is **3.25 %**. b16 reports the smaller, more
  conservative one (:ref:`b16-which-question-is-being-asked-and-what-we-are-not-in`). This also diagnoses the retired exponential shortcut:
  it was the right answer to the wrong question.
- Onset is certain for any positive escalation rate. Rates set the schedule, not the
  outcome: a tenfold reduction in crises plus a threefold reduction in escalation buys
  900 years and changes nothing else.
- **The crisis rate and the per-crisis chance are not separately identified.** With the
  escalation rate held fixed, the mean depends on them **only through their product** ---
  exactly. Fixing the chance at 1/3 is what defines which near-misses get counted, so a
  critic who disputes the 1/3 must also move the crisis rate (:ref:`b16-the-crisis-rate-and-the-escalation-probability-are`).
- **The death fraction 0.3 is a floor, not a central estimate** --- it takes the smallest
  modelled exchange's figure for a full exchange, and stops counting at two years while
  the climate disruption runs a decade or more. The scenario-matched value, 0.63, would
  roughly double every multiplier (:ref:`b16-what-the-death-fraction-counts-and-what-happens-after`).
- All 52 published figures of b16 Table 3 and Table 9 **HELD**, checked four ways.
  Three points of prose imprecision were found and are listed in section 13.
- Marginal effort is now worth far more on the calibration than on the mathematics.

Offered not to be believed, but to be checked. **#AuditTheMath**


.. _b16x-provenance:

***************************************************************************
Appendix P. Provenance, version, and audit status
***************************************************************************

.. note:: **Dropped on 2026m09d23: the naming deliberation.** Sections 3.5--3.7 held the
   first naming proposal and why it was rejected, every review comment and its resolution,
   and the runner-up names with the reason each lost --- 670 lines, more than a third of a
   document whose title promises a derivation. They are **not lost**: they stand in this
   file's git history up to commit ``5819763``. The names themselves, and their full
   explanation, are in **b16 SIA-1**.

Everything a later reader needs in order to tell which draft this is, what was
machine-checked in it, and where the session record lives. Kept at the back on purpose:
it is bookkeeping, not argument.

.. list-table:: Table P. What this document is, formally
   :class: longtable
   :header-rows: 1
   :widths: 22 78

   * - Field
     - Value
   * - **Designation**
     - Matheo-b16-extra --- a spin-off from Matheo-b16, the RiskyMAD waiting-time study
   * - **Variant**
     - MMv1r1p0 --- a MockupModel draft. **Not yet reviewed by anyone.**
   * - **VVN**
     - | ``b16-extra-waiting-time-derivation-``
       | ``dv_ClaOp50Max_MMv3r0p0_2026m09d23_10h45``
   * - **Written**
     - 2026m09d07_18h20, in the session logged at
       :ref:`hell-ll-study-b16-extra-analytical-derivation`
   * - **Nomenclature**
     - The BEST Names of :ref:`Table B <b16x-names-table>` were adopted throughout on
       2026m09d08, replacing the Greek symbols of the first draft. Section 3 keeps the
       old symbols on purpose --- it is the record of *how* the names were chosen, and
       its review comments quote what was actually said. The two right-hand columns of
       Table B map old to new, for anyone holding a printout from before that date. The
       substitution is specified, and its exclusions listed, in
       ``extra/00-transform-spec_2026m09d08.txt``; it was run as one pass across this
       document, b16 OOv3 and the companion code, and is logged at
       :ref:`hell-ll-study-b16-nomenclature-transform`.
   * - **Companion code**
     - ``code/`` beside this file --- three short Python files, standard library only:
       ``riskymad_exact.py`` (one function per formula), ``reproduce_tables.py`` (the
       audit), ``crosschecks.py`` (four independent instruments). Parameter names are
       the Explicit column of Table B, so the poster, the paper and the code say the
       same word for the same rate.
   * - **Status of the mathematics**
     - Every published figure of b16 Table 3 and Table 9 was recomputed from these
       derivations and **HELD** --- 52 of 52, 0 BREACH --- then cross-checked by three
       further instruments (a 400,000-run Monte Carlo built from the model's rules
       alone, numerical integration of the density, and the Vieta identities). The
       *arithmetic* has been checked. The *calibration* it runs on has not, and cannot
       be, by any amount of algebra --- see :ref:`b16x-not-established`.
   * - **License**
     - Jonah License, CC-BY 4.0. Balospe.com --- #AuditTheMath


.. _b16x-references:

***************************************************************************
References
***************************************************************************

.. bibliography::
   :style: apa
   :filter: docname in docnames
