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

Laurence Loewe of Laodicea

Companion to Study b16 in the Matheo Study Series — appendices SIA-4, SIA-5, SIA-8

MockupModel draft

MMv1 — not yet reviewed by anyone. Full provenance, version and audit status are in Appendix P. Provenance, version, and audit status at the end.

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.

What moved into b16 on 2026m09d17, and why the numbering has gaps

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 10. What survives without the exponential assumption 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.

What this document deliberately does not do

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 15. Parked, deliberately — and the rule for when to unpark it). 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 \(\dfrac{1}{r_{M \rightarrow \text{out}}}\).

Step 2 — it ends one of two ways. By Fact 3, with probability \(\dfrac{r_{M \rightarrow D}}{r_{M \rightarrow \text{out}}}\) it escalates, and we have arrived: nothing further to wait for. With probability \(\dfrac{r_{M \rightarrow R}}{r_{M \rightarrow \text{out}}}\) 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 \(T_{R \rightarrow D}\) — the same \(T_{R \rightarrow D}\) we are trying to find.

Adding the mean duration to the mean of what follows:

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

Dropping the middle term, which is zero:

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

5.3 From Risky#

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

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

5.4 Solving the pair#

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

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

Collect the \(T_{R \rightarrow D}\) terms on the left:

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

Factor the left side:

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

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

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

Multiply both sides by \(\dfrac{r_{M \rightarrow \text{out}}}{r_{M \rightarrow D}}\):

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

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

At base case: \(T_{R \rightarrow D} = \dfrac{9}{0.1 \times 3} + \dfrac{1}{3} = 30 + 0.333 = \mathbf{30.33}\) 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 \(T\), it is defined as

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

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

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

(i) It loses nothing. Two distributions with the same transform for all \(s > 0\) 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 \(A\) and \(B\) are independent, then \(\mathbb{E}[e^{-s(A+B)}] = \mathbb{E}[e^{-sA}]\, \mathbb{E}[e^{-sB}]\), 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 \(X\) is exponential with rate \(\alpha\),

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

“Laplace” here is a name, not an assumption

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 \(f(s)\) for the transform of the waiting time from Risky and \(g(s)\) 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 \(r_{R \rightarrow M}\), 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 \(r_{R \rightarrow M}/(r_{R \rightarrow M}+s)\):

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

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

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

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

6.4 Solving for the transform#

Substituting the second equation into the first:

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

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

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

Collect the \(f(s)\) terms:

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

Expand the bracket:

\[\begin{split}(r_{R \rightarrow M}+s)(r_{M \rightarrow \text{out}}+s) &= r_{R \rightarrow M}\,r_{M \rightarrow R} + r_{R \rightarrow M}\,r_{M \rightarrow D} + r_{R \rightarrow M} s + s\,r_{M \rightarrow R} + s\,r_{M \rightarrow D} + s^{2} \\ &= s^{2} + s(r_{R \rightarrow M}+r_{M \rightarrow R}+r_{M \rightarrow D}) + r_{R \rightarrow M}\,r_{M \rightarrow R} + r_{R \rightarrow M}\,r_{M \rightarrow D}\end{split}\]

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

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

and therefore

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

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 \((s + r_{\text{onset}})(s + r_{\text{relax}})\). Expanding and matching coefficients gives the two relations (Vieta’s formulas) that will be used repeatedly as checks:

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

This three-term sum appears throughout, and it is worth reading rather than skipping: it collapses to \(r_{R \rightarrow M} + r_{M \rightarrow \text{out}}\) — 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 \(r_{\text{all}} = r_{R \rightarrow M}+r_{M \rightarrow R}+r_{M \rightarrow D}\):

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

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

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

Every term on the last line is non-negative, so \(\Delta \geq 0\) always. It vanishes only if \(r_{M \rightarrow R} = 0\) and \(r_{R \rightarrow M} = r_{M \rightarrow D}\) 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 \(P(T_{R \rightarrow D} \leq t) = 1 - (1 + rt)e^{-rt}\). 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 (\(r_{R \rightarrow M} = 0.1\), \(r_{M \rightarrow R} = 6\), \(r_{M \rightarrow D} = 3\)): \(r_{\text{all}} = 9.1\), \(\Delta = 81.61\), \(\sqrt{\Delta} = 9.03383\), giving

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

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

6.6 Recognising the answer#

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

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

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:

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

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 (What the two rates are — and what they are not).

6.7 Inverting, to get the answer in years#

Split the transform by partial fractions:

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

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

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

Integrating from \(t\) to infinity gives the chance the world has not yet arrived:

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

and finally the quantity the paper reports:

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

At base case, one year: \(P(T_{R \rightarrow D} \leq 1) = 0.0290\), 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 \(\mathbb{E}[T_{R \rightarrow D}] = 1/r_{\text{onset}} + 1/r_{\text{relax}}\). Using Vieta:

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

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.)

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

The waiting time is then

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

Lemma — a geometric number of exponentials is one exponential

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

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

Applying the lemma twice:

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

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

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

— 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 \(n_{MAD}\); 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 \(p_R(t)\) and \(p_M(t)\) be the probabilities of being in Risky and in a MAD crisis at time \(t\). Probability flows along the arrows, so

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

with \(p_R(0) = 1\), \(p_M(0) = 0\). 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 \(e^{-rt}\), where the \(-r\) are the eigenvalues of the coefficient matrix

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

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

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

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

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

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

\[\begin{split}(r_{\text{relax}}-r_{R \rightarrow M}) (r_{M \rightarrow R}+r_{R \rightarrow M}-r_{\text{onset}}) &= r_{M \rightarrow R}\, r_{\text{relax}}, \\ (r_{R \rightarrow M}-r_{\text{onset}}) (r_{M \rightarrow R}+r_{R \rightarrow M}-r_{\text{relax}}) &= -r_{M \rightarrow R}\, r_{\text{onset}}\end{split}\]

so that

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

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 \(p_{M \rightarrow D}\), 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.

Table E. Simulation against closed form, base case#
quantity	simulated	exact	agreement
mean wait	30.345 yr	30.333 yr	0.04 %
crises before onset	2.9986	3.0000	0.05 %
\(P(T_{R \rightarrow D} \leq 1\,\text{yr})\)	2.900 %	2.900 %	exact to 3 digits
\(P(T_{R \rightarrow D} \leq 20\,\text{yr})\)	48.216 %	48.216 %	exact to 3 digits
\(P(T_{R \rightarrow D} \leq 80\,\text{yr})\)	92.864 %	92.888 %	0.02 pp

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

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.

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:

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

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

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

These are the same function, under one substitution, at every value of \(r_{R \rightarrow M}\) — not an approximation valid in a limit.

Note

Table F is superseded by Table B-MM and should be replaced by it. Table B-MM carries the same five mappings plus \(k_{on}[S]\) and \(k_{off}\), 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.

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

The defining property holds exactly. Michaelis–Menten’s \(K_m\) is the substrate level at which the rate is half its maximum. Here, setting \(r_{R \rightarrow M} = r_{M \rightarrow \text{out}} = 9\):

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

Where this system sits. \(S/K_m = r_{R \rightarrow M}/r_{M \rightarrow \text{out}} = 0.1/9 = 0.011\) — ninety times below half-saturation. The system is deep in the first-order regime, which is precisely why the naive product \(r_{R \rightarrow M} \cdot p_{M \rightarrow D}\) 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 \(k_{on}[S]\) 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 \(r_{R \rightarrow M}/(r_{R \rightarrow M}+r_{M \rightarrow R}+r_{M \rightarrow D}) = 1.1\,\%\). 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.

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 \(n_{MAD}\) be the number of crises until one escalates. Each crisis escalates independently with probability \(p_{M \rightarrow D}\), so \(n_{MAD}\) is geometric with \(\mathbb{E}[n_{MAD}] = 1/p_{M \rightarrow D} < \infty\) — 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:

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

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 \(\tau/p_{M \rightarrow D}\), and therefore

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

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

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

Table H. Audit of b16 Table 3 and Table 9#
What was checked	Result	Instrument
b16 Table 3 — 4 scenarios × 8 figures	32 of 32 HELD	closed form, re-derived
b16 Table 9 — 4 rows × 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:

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:

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 \(r_{R \rightarrow M}\), \(r_{M \rightarrow R}\), or \(r_{M \rightarrow D}\) — 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.

#	Where	The fix
1	SIA-4, introducing \(r_2\)	b16 calls \(r_2 = 9.067\)/yr “the fast stage, the crisis itself, mean duration 40 days”. The crisis clock is \(r_{M \rightarrow \text{out}} = 9.000\)/yr (40.58 days); \(r_{\text{relax}}\) is 0.74 % larger (40.28 days), and 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 \(b+c\), 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, “\(r_{R \rightarrow M}/(r_{R \rightarrow M}+b+c) \approx 0.011\)”	Two different quantities both equal ≈ 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.

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.

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 \(r_{R \rightarrow M}(t)\) worked through numerically	quantifies b16 SIA-X2’s rising-rate argument	moderate; the certainty result already covers the qualitative claim

The stopping rule

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 \(\mathrm{Exp}(r_{\text{onset}}) + \mathrm{Exp}(r_{\text{relax}})\), with \(r_{\text{onset}} + r_{\text{relax}} = r_{R \rightarrow M}+r_{M \rightarrow R}+r_{M \rightarrow D}\) and \(r_{\text{onset}}\, r_{\text{relax}} = r_{R \rightarrow M}\,r_{M \rightarrow D}\). At base case that is 0.0331/yr and 9.067/yr.
Its mean is \(\dfrac{r_{M \rightarrow \text{out}}}{r_{R \rightarrow M}\,r_{M \rightarrow D}} + \dfrac{1}{r_{M \rightarrow D}}\) = 30.3 years; the factor \(r_{M \rightarrow \text{out}}/r_{M \rightarrow D} = 3\) 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 \(r_{\text{onset}}\), and the answer is 3.25 %. b16 reports the smaller, more conservative one (Which question is being asked — and what “we are not in a crisis today” is worth). 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 (The crisis rate and the escalation probability are not separately identified).
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 (What the death fraction counts, and what happens after two years).
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

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.

Table P. What this document is, formally#
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 LLog — Matheo-b16: the exact waiting-time derivation, and whether it stays
Nomenclature	The BEST Names of Table B 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 LLog — Matheo-b16: running the nomenclature transform, and what it exposed.
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 15. Parked, deliberately — and the rule for when to unpark it.
License	Jonah License, CC-BY 4.0. Balospe.com — #AuditTheMath

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

1. What this document is#

5. The mean waiting time, line by line#

5.2 From a MAD crisis#

5.3 From Risky#

5.4 Solving the pair#

6. The whole distribution, not just its average#

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

6.3 The same two equations, in transform space#

6.4 Solving for the transform#

6.5 The two rates#

6.6 Recognising the answer#

6.7 Inverting, to get the answer in years#

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

8.1 By counting crises (no calculus)#

8.2 By solving the model’s differential equations (no transforms)#

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

9. Michaelis–Menten, derived rather than asserted#

10. What survives without the exponential assumption#

10.2 The proof#

12. Every published figure, recomputed#

13. Three points of imprecision found in b16#

15. Parked, deliberately — and the rule for when to unpark it#

16. Summary#

Appendix P. Provenance, version, and audit status#

References#