Pair statistics can't see the triangle: preserving higher-order risk in a quantum pipeline
Tone: best-paper-ready — exact statements, every number recomputable. This post condenses my paper draft (Sep 2026); the two figures are the paper’s own.
Original. “We study this distinction using a three-asset distribution family whose individual downside probabilities and pairwise joint downside probabilities remain fixed while the probability of all three assets falling ranges from zero to one percent.” — the paper’s abstract.
Essence. A quantum pipeline that compresses market data into pairwise statistics before computing risk or topology has already decided what any later algorithm can recover — no amount of classical or quantum accuracy restores a triple-crash probability that was never stored. Once that information is kept, a probability, the full event complex, and a single Betti number each need very different sample budgets. And a fair comparison must count samples, oracle calls, loading and training in the same ledger.
DIRECT — what the paper shows (locked)
1. The question
Three assets, each falling sharply in 10% of periods, each pair falling together in 1%. The joint-tail post showed these six numbers leave the triple-fall probability $t$ free in $[0,\,0.01]$. This paper asks what that gap does to a quantum pipeline that builds topology.
Define the event complex at threshold $\tau$: a set of assets $S$ is a simplex when the probability that all of $S$ fall together is at least $\tau$,
\[K_\tau = \big\{\, S \neq \varnothing : P(\text{all } i \in S \text{ fall}) \ge \tau \,\big\}.\]It is downward closed (if a trio falls together, each pair does), so it is a simplicial complex. At $\tau = 0.005$:
- Market A ($t = 0$): three vertices, three edges, no face — a hollow triangle, first Betti number $\beta_1 = 1$.
- Market B ($t = 0.01$): the face is present — a filled triangle, $\beta_1 = 0$.
Now build the complex the usual way, as the clique complex of the pairwise graph. Both markets have all three edges, so both triangles get filled and both give $\beta_1 = 0$. Any topology algorithm, classical or quantum, that receives only the pair graph cannot tell A from B. And note the direction: B is the riskier market, yet it has the smaller Betti number — more holes does not mean more risk.
2. One line
What you compress before computing decides what any algorithm can recover; once the information is kept, a probability, the complex, and one Betti number need very different sample budgets.
3. How it works, in four steps
3.1 Identification: the loss is total, not approximate
For the whole family, the retained summaries (every singleton and pair probability, the portfolio means and variances, the Z and ZZ expectations) are identical for every $t \in [0, 0.01]$, and that interval is sharp. Two consequences (the paper’s Proposition 1):
- Any procedure fed only those summaries — deterministic, randomized or quantum — has a worst-case absolute error of at least 0.005 on $t$ (the midpoint 0.005 achieves it).
- Asked to decide “A or B” with equal priors, its error is at least one half.
In qubit language, writing $z_i = 1 - 2b_i$, the event projector expands as
\[\Pi_{111} = \tfrac18\big(1 - Z_1 - Z_2 - Z_3 + Z_1Z_2 + Z_1Z_3 + Z_2Z_3 - Z_1Z_2Z_3\big),\]so pair statistics omit exactly one term, $\langle Z_1 Z_2 Z_3 \rangle$. Measuring in the Z basis does not cause the loss: the full bitstring contains it. The loss happens when the stored statistics throw it away.
3.2 Sampling: three goals, three budgets
Estimate all seven subset probabilities from the same $N$ joint samples. A simplex’s membership can flip only if its estimate moves past $\tau$, so what matters is the margin $m$ between each probability and the threshold. Hoeffding’s inequality (Hoeffding, 1963) on each of the $|F| = 7$ events plus a union bound gives a sufficient condition for recovering the whole complex with probability $1 - \delta$:
\[N \;\ge\; \frac{\ln(2|F|/\delta)}{2m^2} = \frac{\ln(2 \cdot 7 / 0.05)}{2 \cdot 0.005^2} = 112{,}695.8 \;\Rightarrow\; 112{,}696 .\]That is conservative. Simulation (500 independent sampling streams per market, nested budgets) shows how differently the three goals behave:
- Market B, 128 samples: the Betti number matches in 100% of runs, yet the whole complex is right in only 71.8% (95% Wilson interval 67.7–75.6%). The 100% is built in: in B every pair event coincides with the triple event, so every empirical complex is either edgeless or fully filled, and both have $\beta_1 = 0$. Exact recovery needs at least one triple in the sample: $1 - 0.99^{128} = 0.7237$ — matching the simulation.
- Market A, 128 samples: 35.2% (31.1–39.5%) for both.
- 2,048 samples: complex recovery 97.0% (A) and 99.4% (B).
- Triple probability within ±0.2 percentage points (B): the first budget on the power-of-two grid to reach 95% is 16,384.
A matching Betti number is not a recovered complex, and neither is an accurate probability. Ask all three questions separately.
3.3 Detection: count the right resource
The decision “is this market A or B?” can be answered under very different kinds of access. Classically, with $N$ independent joint samples:
- At least one triple crash seen. Zero false positives; power $1 - 0.99^N \ge 0.95$ needs $N \ge \ln 0.05 / \ln 0.99 = 298.1$, so 299 samples.
- Full-bitstring likelihood-ratio test. Bitstrings with exactly two ones occur only in A, $111$ only in B, and among shared outcomes the singleton count is informative. Using all of it reaches the same power with 90 samples (89 if the boundary is randomized).
Coherently, with an oracle that prepares the market’s state and its inverse (without revealing which market it is), mark the event and amplify (Brassard et al., 2002). With $\theta = \arcsin\sqrt{0.01}$, after $k$ iterations the event probability is $\sin^2\big((2k+1)\theta\big)$; at $k = 7$ that is $\sin^2(15\theta) = 0.99534$ under B and exactly 0 under A.
3.4 Hodge: a structural number without reading every edge
A separate control asks for a directional quantity: the Hodge decomposition of a flow on the triangle’s edges (Jiang et al., 2011). Filling the face moves the cyclic component of a flow from harmonic to curl, with its energy unchanged. Implemented as a coherent dilation, all 18 probabilities (three flows × two complexes × three projectors) match the classical projections to numerical precision — a final scalar read without outputting every edge value, in the spirit of quantum HodgeRank (Leditto et al., 2025). A second control shows that turning a learned expectation into a signed amplitude (the negative edge survives as a negative amplitude) works by coherent conditioning that succeeds with probability 0.28, so about 1/0.28 ≈ 3.6 attempts per accepted outcome.
4. The table — detection power by access model
| Method | Resource used | Power under B | False positives under A |
|---|---|---|---|
| At least one triple crash seen | 299 independent samples | 0.950464 | 0 |
| Deterministic full-bitstring test | 90 independent samples | 0.953690 | 0.048159 |
| Randomized full-bitstring test | 89 independent samples | 0.954013 | 0.050000 |
| Coherent amplification, $k = 7$ | 15 state-preparation or inverse calls | 0.995344 | 0 |
| Explicit probability-table access | 0 samples | 1 | 0 |
The coherent row looks best, but a sample and a coherent preparation call are different resources with different costs, which is why the paper plots them on separate axes (figure below). The last row is the warning: in this toy the simulator is handed the probability table, and anyone holding the table has already solved the problem.
5. Strengths and weaknesses
Strengths.
- The failure is exact: the lost quantity, its identified interval and the minimum error are all closed-form, so no downstream method can be blamed or credited.
- The three endpoints (probability, complex, Betti number) are separated, with a sufficient bound and exact simulation probabilities that agree with each other.
- The classical baseline is strong: the full-bitstring likelihood-ratio test uses 90 samples, not the naive 299, so the coherent method is not compared against a straw man.
Weaknesses (stated in the paper).
- A three-asset toy. The quantum circuits are dense explicit unitaries — they verify identities, they are not scalable loading, block encodings or QSVT implementations.
- The oracle is assumed. Coherent detection needs a state-preparation oracle that does not reveal which market it prepares. Quantum risk analysis by amplitude estimation (Woerner & Egger, 2019) assumes the distribution can be loaded; loading that scales with the number of scenarios can dominate a particular implementation, so any speedup needs the full ledger checked (Aaronson, 2015).
- Easy vs hard oracles. Checking whether a given bitstring is $111$ is easy to do reversibly; deciding whether a subset belongs to the complex depends on an unknown probability, and implementing that coherently needs estimation, a margin promise and uncomputation. Substituting the first for the second would hide the main difficulty.
- Training costs add up fast. An illustrative quantum-neural-network run with parameter-shift gradients costs 20,889,600 shots; SPSA under the same batch convention costs 491,520. Neither count implies equal accuracy.
- Real data is descriptive only. On NoDur, Manufacturing and Energy (Ken French 12 industries (French, 2026), 2010–2025), with thresholds fixed on 2010–2017 and 96 overlapping 252-day windows over 2018–2025, the empirical triple-fall frequency differs from the fitted pairwise maximum-entropy model’s prediction by 0.33 percentage points on average. Overlapping windows and repeated examination rule out calling this a predictive or significant finding.
- No quantum advantage is claimed.
6. What to compare next
The paper’s §9 lists what a real quantum topology experiment must report: asset count, homology degree, number of valid simplices and the probability of preparing one, access to boundary operators, spectral gap, membership margin, precision, and the exact output (an integer Betti number is not the same target as a normalized one). The competitors are the quantum TDA algorithms (Lloyd et al., 2016), their resource analyses (Berry et al., 2024), and — the comparison that decides it — the simple classical Betti estimator of (Apers et al., 2023), asked for the same output at the same precision. Classical shadows (Huang et al., 2020) are the fair baseline when many different observables are read from one state.
INDIRECT — my extensions (revisable; dated edits go below)
- The filled triangle at scale is the common shock. In the joint-tail study, pairwise links under-produce many-industry crash days across 49 industries by 2–31× while a single common factor reproduces them. That is this paper’s market B, grown up: the information pairs discard is exactly the higher-order face a common shock creates.
- When could β₁ be a risk signal? Here the riskier market has fewer holes, so β₁ is not monotone in risk. A useful signal would have to be read relative to a baseline — e.g. a hole appearing where a pairwise model predicts a filled face — rather than as a raw count.
- Where the sampling bound bites. At $m = 0.005$ the bound asks for ~113k joint samples: about 450 years of daily data. For real markets the margin, not the algorithm, decides whether the complex is recoverable at all — and more quantum shots cannot create more market history.
Formal cited bibliography (auto-generated)
2026
- Kenneth R. French Data Library: 12 and 49 Industry Portfolios (daily)2026Accessed 2026-09
2025
- Quantum HodgeRank: Topology-based rank aggregation on quantum computersPhysical Review Applied, 2025
2024
2023
2020
2019
2016
2015
2011
2002
1963
- Probability Inequalities for Sums of Bounded Random VariablesJournal of the American Statistical Association, 1963