Paper replication

Side by side with the numbers reported in Munoz Moran (2023) and its companion notebook

What this page checks

Everything here is recomputed in the browser from xrisk-model.mjs (the same file the interactive page uses) and compared with (a) the values printed in the PDF and (b) the full-precision values printed by the companion notebook, VERM 3.0.ipynb. The same checks run as node --test in the repository. Cells are shaded green when the tool matches the notebook to better than one part in \(10^6\), and amber when only the paper’s three-significant-figure value is available.

Summary. The tool reproduces every number in Table 3, the closed forms in Table 2 and the appendix, the concrete example of section 2.2.1, and Figure 8, to floating-point precision, once the notebook’s conventions are known. Some of those conventions are not stated in the PDF; they are listed at the bottom, and on the departures page.

M = import(new URL("./xrisk-model.mjs", window.location.href).href)
fmt = (x, d = 3) => M.fmtSci(x, d)
rel = (a, b) => Math.abs(a / b - 1)
cls = (r, paperOnly) => paperOnly ? (r < 5e-3 ? "match-warn" : "match-bad") : (r < 1e-6 ? "match-ok" : r < 5e-3 ? "match-warn" : "match-bad")

Concrete example (section 2.2.1)

Risk 0.2229% per year for 100 years then 0.01%, \(T = 10{,}000\), \(v_c = 1\), mitigation halves risk (\(f = 0.5\)) for \(P = 5\) years.

{
  const ex = M.concreteExample();
  const rows = [["E(w)", ex.ew, 5116.53273619555, "5116.53273619555"], ["E(w')", ex.ewm, 5145.161060930257, "5145.161060930257"], ["E(M)", ex.em, 28.62832473470735, "28.6"]];
  return html`<table class="compact table"><thead><tr><th>Quantity</th><th>Tool</th><th>Notebook</th><th>Paper</th><th>Relative difference</th></tr></thead>
  <tbody>${rows.map(([n, t, nb, p]) => html`<tr><td>${n}</td><td class="${cls(rel(t, nb))}">${t}</td><td>${nb}</td><td>${p}</td><td>${rel(t, nb).toExponential(1)}</td></tr>`)}</tbody></table>`;
}

Closed forms under constant risk (Table 2 and appendix 12.2)

\(\mathbb{E}(w)\) for polynomial value with \(r_c = 0.0022289477\): unadjusted (\(\alpha = 1\), value \(t^n\)) and adjusted (\(\alpha = 0.01\), value \((0.01t + 0.99)^n\)). The numeric column sums the series to \(T = 200{,}000\).

{
  const T = 200000, r = M.riskPath(T, {type: "constant"});
  const nb = {constant: [4.476e2, 4.476e2], linear: [2.008e5, 2.451e3], quadratic: [1.800e8, 2.242e4], cubic: [2.420e11, 3.018e5]};
  const rows = [];
  for (const vt of ["constant", "linear", "quadratic", "cubic"]) for (const [i, alpha] of [1, 0.01].entries()) {
    const cf = M.closedFormConstantRisk(vt, M.PAPER.rHigh, alpha);
    const num = M.expectedValue(r, M.valuePath(T, {type: vt, alpha})).ev;
    rows.push(html`<tr><td>${vt}</td><td>${alpha}</td><td class="${cls(rel(cf, num))}">${cf.toExponential(6)}</td><td>${num.toExponential(6)}</td><td>${nb[vt][i].toExponential(3)}</td></tr>`);
  }
  return html`<table class="compact table"><thead><tr><th>Value</th><th>alpha</th><th>Closed form</th><th>Numeric sum</th><th>Notebook (3 s.f.)</th></tr></thead><tbody>${rows}</tbody></table>`;
}

Table 3: \(\mathbb{E}(M)\) for all scenarios when \(f = 0.0001\)

Computed with the notebook’s per-cell horizon \(T\) (5,000 for constant risk with constant or linear value; 10,000 for the other constant-risk cells and all exponential-decay cells; 40,000 for Time of Perils and Two Great Filters with constant value; 120,000 for the remaining ToP and filter cells). The last column recomputes each cell with four times that horizon, which shows how much of the paper’s number is a truncation artefact (mostly the 40,000-year cells).

{
  const tab = M.replicationTable3();
  const long = {};
  for (const vt of M.VALUE_TYPES) { long[vt] = {}; for (const rt of M.RISK_TYPES) {
    const T = 4 * M.PAPER.tableT[vt][rt];
    const v = M.valuePath(T, {type: vt}), r = M.riskPath(T, {type: rt});
    const base = M.expectedValue(r, v).ev; long[vt][rt] = {};
    for (const P of M.PAPER.persistences) long[vt][rt][P] = M.expectedValue(M.mitigateRisk(r, {f: M.PAPER.f, P}), v).ev - base;
  } }
  const rows = [];
  for (const P of M.PAPER.persistences) {
    rows.push(html`<tr><th colspan="7" style="background:#f0f4f8">P = ${P}</th></tr>`);
    for (const vt of M.VALUE_TYPES) for (const rt of M.RISK_TYPES) {
      const t = tab[vt][rt][P], nb = M.NOTEBOOK_TABLE3[vt][rt][P], pp = M.PAPER_TABLE3[vt][rt][P];
      const paperOnly = M.NOTEBOOK_TABLE3_PAPER_ONLY.some(([a, b, c]) => a === vt && b === rt && c === P);
      const l = long[vt][rt][P];
      rows.push(html`<tr><td>${M.RISK_LABELS[rt]}</td><td>${M.VALUE_LABELS[vt]}</td><td>${M.PAPER.tableT[vt][rt].toLocaleString()}</td>
        <td class="${cls(rel(t, nb), paperOnly)}">${fmt(t, 6)}</td><td>${paperOnly ? "(not printed)" : fmt(nb, 6)}</td><td>${fmt(pp)}</td>
        <td>${fmt(l, 4)} <span class="muted">(${((l / t - 1) * 100).toFixed(2)}%)</span></td></tr>`);
    }
  }
  return html`<table class="compact table"><thead><tr><th>Risk</th><th>Value</th><th>Notebook T</th><th>Tool</th><th>Notebook</th><th>Paper (3 s.f.)</th><th>Tool at 4T (change)</th></tr></thead><tbody>${rows}</tbody></table>`;
}

Observations:

The paper prints truncated rather than rounded figures (for example \(9.99 \times 10^{-5}\) where the notebook value is \(9.999857 \times 10^{-5}\)), so the third significant figure occasionally differs by one; the tool matches the notebook to better than \(10^{-8}\) in every printed cell.
The Time of Perils and Two Great Filters cells with constant value use \(T = 40{,}000\) in the notebook, which leaves roughly 1.5 to 2% of \(\mathbb{E}(M)\) beyond the horizon; the paper’s statement that \(T = 120{,}000\) “achieves the same behaviour as \(T \to \infty\)” is right for the 120,000-year cells but not for these.
Exponential-decay cells use \(T = 10{,}000\); with \(r_\infty = 10^{-9}\) the survival probability plateaus near \(2 \times 10^{-10}\) and polynomial value eventually makes the sum grow again (the notebook’s “phenomenon” section), so these cells are horizon-dependent in principle, though not at 4T.

Figure 6 (the “planets” graphic): \(\mathbb{E}(M)\) at \(P = 5\) in units of the constant-constant cell

{
  const tab = M.replicationTable3();
  const unit = tab.constant.constant[5];
  return html`<table class="compact table"><thead><tr><th>Risk</th>${M.VALUE_TYPES.map(v => html`<th>${M.VALUE_LABELS[v]}</th>`)}</tr></thead>
  <tbody>${M.RISK_TYPES.map(rt => html`<tr><td>${M.RISK_LABELS[rt]}</td>${M.VALUE_TYPES.map(vt => { const x = tab[vt][rt][5] / unit, p = M.PAPER_FIGURE6[rt][vt]; const r = rel(x, p); return html`<td class="${r < 0.06 ? "match-ok" : r < 0.15 ? "match-warn" : "match-bad"}">${fmt(x)}<br><span class="muted">paper: ${fmt(p)}</span></td>`; })}</tr>`)}</tbody></table>
  <p style="font-size:0.9em">All cells agree to the precision of the figure's labels, except Two Great Filters with linear value, where the figure says "1K" and the table implies about 1,480 (the same cell in Table 3, $7.37 \times 10^{-1} / 4.98 \times 10^{-4}$, gives 1,480; the graphic's label appears to be a rounding to one significant figure or a transcription slip).</p>`;
}

Figure 8: \(\mathbb{E}(M)\) versus persistence under constant risk and value

The paper’s Figure 8 (with \(f = 0.0001\)) reaches about 0.029 at \(P = 1000\), whereas Table 3 gives 0.030 already at \(P = 500\) and 0.044 at \(P = 2000\). The two are consistent once one notices that the notebook draws Figure 8 with a universe that ends at \(T = 1000\) years (plot_value_of_mitigation_different_Ps(T=1000, ...)). The tool reproduces the figure under that convention:

{
  const fig = M.figure8();
  const inf = M.figure8(fig.map(d => d.P), 120000);
  return html`<table class="compact table" style="width:auto"><thead><tr><th>P</th><th>E(M), T = 1000 (Figure 8 convention)</th><th>E(M), T = 120,000</th></tr></thead>
  <tbody>${fig.map((d, i) => html`<tr><td>${d.P}</td><td>${d.EM.toFixed(5)}</td><td>${inf[i].EM.toFixed(5)}</td></tr>`)}</tbody></table>
  <p style="font-size:0.9em">Read off the figure: about 0.0143 at P = 200, 0.022 at 400, 0.0293 at 1000. The concavity the paper discusses is partly the finite universe: with T = 120,000 the curve is also concave but roughly 50% higher at P = 1000.</p>`;
}

Statements in the text

{
  const tab = M.replicationTable3();
  const allCells = M.VALUE_TYPES.flatMap(vt => M.RISK_TYPES.flatMap(rt => M.PAPER.persistences.map(P => tab[vt][rt][P])));
  const rows = [
    ["\"E(M) under Time of Perils ... is 11 million times bigger under cubic value compared to constant\" (p. 30)", (tab.cubic.top[5] / tab.constant.top[5]).toExponential(2) + " at P = 5 (6.2 million); the same ratio at every P. Figure 6's own labels give 110M / 18 = 6.1 million. Not reproduced as stated; 110 million would be the ratio to the constant-risk, constant-value cell.", "differs"],
    ["\"Adding another filter ... only reduces it by about 25%\" (p. 30)", "Two filters / ToP = " + (tab.constant.filters[5] / tab.constant.top[5]).toFixed(3) + " for constant value (an 18% reduction); 19% for linear, quadratic, cubic and logistic at P = 5. The notebook's own note says 20%.", "close"],
    ["\"the minimum E(M) is roughly 8 orders of magnitude smaller than the maximum\" at fixed persistence (p. 30)", "max/min at P = 5: " + (tab.cubic.top[5] / tab.constant.constant[5]).toExponential(2), "matches"],
    ["\"between one ten-thousandth of a v_c ... and two billion v_c (under cubic and time of perils assuming f is one basis point)\" (pp. 31-32), following the suggestion that persistence is \"unlikely to be higher than 50 years\"", "At f = 0.0001, Time of Perils with cubic value gives " + fmt(tab.cubic.top[50]) + " at P = 50 and " + fmt(tab.cubic.top[2000]) + " at P = 2000; the largest Table 3 cell is " + fmt(Math.max(...allCells)) + ". At f = 0.5, the efficacy used for the report's figures, the same cell at P = 50 is " + fmt(M.replicationTable3(0.5, [50]).cubic.top[50]) + ", about 2.8 billion. The sentence most likely pairs an f = 0.5 number with the one-basis-point label.", "differs"],
    ["\"depending on persistence E(M) can increase by up to 30 times\" (p. 31)", "From P = 1 to P = 2000 the table's ratios run from 176 (ToP, constant value) to 1,490 (decay, cubic); Figure 8 (T = 1000) spans 293x from P = 1 to 1000. We could not find a reading that gives 30.", "not reproduced"],
    ["\"constant risk under linear value is in the thousands of v_c and Two Great Filters under logistic value is in billions of v_c\" (p. 28)", "E(w): constant/linear = 2,451; filters/logistic = 6.3 billion (T = 120,000).", "matches"]
  ];
  return html`<table class="compact table"><thead><tr><th>Claim</th><th>Tool</th><th>Status</th></tr></thead>
  <tbody>${rows.map(([c, t, s]) => html`<tr><td>${c}</td><td>${t}</td><td class="${s === "matches" ? "match-ok" : s === "close" ? "match-warn" : "match-bad"}">${s}</td></tr>`)}</tbody></table>`;
}

Conventions needed to reproduce the paper (not all stated in the PDF)

Period \(t = 1\) is the first year; \(S_t\) includes period \(t\)’s own risk; the mitigation applies to periods \(1..P\).
Risk constants: \(r_{high} = 0.0022289477\), \(r_{low} = 0.0001\) (the concrete example on p. 15 uses 0.002229 instead, which is why it must be reproduced separately).
Time of Perils: 100 high-risk years. Great Filters: eras of 100, 500, 100 years, then low risk for the remainder.
Exponential decay: \(r_t = r_0 e^{-\lambda (t-1)} + r_\infty\) with \(\lambda = 10^{-4}\) and \(r_\infty = 10^{-9}\). The PDF’s Figure 3 shows a decay that is roughly 50 times faster; the notebook’s plotting function uses the same \(\lambda\) over a 2,000-year axis, so Figure 3 appears to have been produced with a different setting. Figure 5’s decay panel likewise.
Value paths use the yearly adjustment \(v_t = (0.01 t + 0.99)^n\); the logistic path has \(c = 10^6\), \(\gamma = 0.03\) and \(v_1 = 1\). Figure 2 in the PDF plots the unadjusted paths (\(v_t = t^n\), logistic cap \(500^3\)), which are not the ones used for any reported number.
Table 3 uses a different \(T\) per cell (listed above); Figure 8 uses \(T = 1000\).
Printed numbers are truncated to three significant figures, not rounded.