Stress envelopes and reverse loss thresholds
Prove a joint-shock loss envelope and construct the smallest stress that reaches a declared limit.
38 minutes · advanced · Original Poker4Quant lessonWhat you will learn
Derive sharp loss envelopes over box and shared-budget shock sets, invert them with feasible minimal-severity witnesses, and diagnose direction switches and strict-trigger boundaries under sensitivity changes.
Prerequisites: Factor exposure and residual performance · Decision robustness
Declare a joint shock set before calculating a worst loss
A forward stress test asks what a declared adverse scenario would do to a portfolio. A reverse stress test starts with a specified damaging outcome and asks which scenarios could produce it. Neither question supplies an event probability by itself. A scenario set describes what the analyst permits, whereas a probability law additionally assigns weights. In this lesson we hold a portfolio fixed and investigate the geometry of its permitted shocks. We do not rank investments by average return or estimate a covariance matrix. The central task is to prove that no permitted scenario can exceed a claimed loss, and then exhibit a scenario that attains it.
Assume one period, initial wealth 100 fictional chips, no interim trades, no fees, no borrowing and no cash flows. Two dimensionless adverse shock coordinates x and y are nonnegative. Their units are declared normalized stress units, not percentage returns. The stipulated loss model is L(x,y)=6x+by chips, with baseline sensitivity b=4 chips per stress unit. Terminal wealth is 100−L. The model is exactly linear for this exercise; applying a locally estimated linear approximation to an actual extreme move would need separate validation. We call loss at least 15 chips a trigger, so equality counts. This is a teaching limit, not a regulatory capital ratio or an insolvency definition.
Prove two envelopes with matching witnesses
For severity t≥0, the box set permits 0≤x≤t and 0≤y≤t simultaneously. Because both coefficients are positive, 6x+by≤6t+bt=(6+b)t. The permitted corner (t,t) attains this upper bound, proving the worst loss Wbox(t)=(6+b)t. An inequality alone would establish only an upper bound; the attaining corner makes it sharp. At baseline b=4 and t=1 the worst loss is ten chips. Testing only the two single-factor scenarios (1,0) and (0,1) would report at most six, missing the simultaneous move allowed by this set.
Now replace the box with a shared-budget triangle: x≥0, y≥0 and x+y≤t. With m=max(6,b), L≤m(x+y)≤mt. If b<6, the point (t,0) attains mt; if b>6, (0,t) attains it. When b=6 every point on x+y=t attains the same loss. Thus Wbudget(t)=max(6,b)t. At b=4,t=1 its worst loss is six, not ten. Adding the separate losses six and four silently combines two scenarios into the forbidden point (1,1), whose shock budget is two. The triangle is a stipulated joint restriction, not an independence assumption, a confidence region or a universal economic law.
Invert the envelope and certify minimal severity
Let K>0 be an inclusive loss trigger. In either family the envelope has form W(t)=ct with c>0. A necessary condition for any permitted point to trigger is ct≥K, so t≥K/c. At t*=K/c the attaining point from the envelope proof has loss exactly K. These two directions prove that t* is the minimum severity, rather than just one stress that happens to work. The proof combines an impossibility certificate for every smaller severity with an explicit feasible witness at the boundary. A large arbitrary shock that breaches the limit is not a minimality proof.
For K=15,b=4 the box coefficient is ten and t*=15/10=3/2. Choose x=y=3/2. The component losses are nine and six, so total loss is fifteen and terminal wealth is 85. Independently, at any smaller box severity t, loss is at most 10t<15. In the shared-budget family the coefficient is six and t*=15/6=5/2. Choose x=5/2,y=0, which has loss fifteen and uses exactly the available budget. The box witness (3/2,3/2) is infeasible at budget severity 3/2 because its coordinates sum to three. Even at budget severity 5/2 that particular point remains infeasible. A valid loss calculation cannot rescue a violated scenario constraint.
Separate parameter sensitivity from scenario severity
Changing t expands the permitted shock set; changing b alters the valuation response to the same shocks. Keep those operations separate in a model review. Under the box, increasing b from four to eight changes the unit-severity worst loss from ten to fourteen and the reverse severity from 3/2 to 15/14. The maximizing corner remains (t,t). Under the shared budget the same coefficient change moves the worst direction from the x axis to the y axis. Its worst loss becomes 8t and its reverse threshold becomes 15/8=1.875. The attaining point is now (0,15/8). Substitution checks eight times fifteen eighths equals fifteen.
The shared-budget envelope is continuous in b but changes slope at b=6. Below six, increasing b leaves the worst loss unchanged because x remains the more damaging use of the budget. Above six, every additional unit of b raises the unit-severity worst loss by one. At exactly six the optimizer is not unique: a balanced point and either axis point are equally damaging. This exposes why perturbing a coefficient can change the worst scenario abruptly even when the reported worst loss changes smoothly. A derivative evaluated on one side does not establish global insensitivity, and a fixed previously worst point may cease to be worst after the model changes.
Construct, challenge and retain a reverse-stress certificate
In the workshop enter severity, x and y, then choose the box or shared-budget restriction and the y sensitivity. Construct a box certificate for b=4, save it, switch the joint restriction and observe why the old construction fails. Repair both severity and the scenario point to obtain the budget certificate. Finally increase b to eight and relocate the damaging point; also explore b=6 to demonstrate multiple optimal witnesses. The readout separates candidate loss, set feasibility, worst permitted loss and loss-trigger margin. A certificate requires a feasible point, an actual trigger and the minimum severity. Saving a certificate retains a visit-local comparison of the constructions, not account completion or an investment recommendation.
There are useful boundaries to report. At t=0 only the origin is allowed and loss is zero. If a trigger were zero or negative, the origin would already satisfy it, so the positive-trigger formula must be interpreted with that boundary in mind. If both loss coefficients were zero and K>0, no finite severity could trigger; our activity keeps the x sensitivity positive. For a strict loss greater than fifteen, the same t* is an infimum but is not itself sufficient, because its maximum loss equals fifteen. Finally, expanding a valid shock set cannot reduce its worst loss, but neither the larger set nor a smaller reverse severity predicts how likely failure is. Missing nonlinear exposures, liquidation costs and management constraints can invalidate the valuation or feasibility assumptions. Record those uncertainties instead of calling a successful mathematical certificate a comprehensive real-world resilience assessment.
Worked example
For L=6x+4y and an inclusive 15-chip trigger, certify the smallest severity under each joint shock restriction.
- Box: L≤10t, attained at (t,t); therefore any trigger requires t≥15/10=3/2.
- At t=3/2, (x,y)=(3/2,3/2) is feasible and loses 9+6=15 chips. Wealth becomes 85.
- Shared budget: L≤6(x+y)≤6t, attained at (t,0); hence t≥15/6=5/2.
- At t=5/2, (5/2,0) is feasible and loses 15 chips. The old box point uses budget 3 and is excluded.
- Raise b to eight: budget loss is bounded by 8t, and (0,15/8) attains 15 at severity 15/8.
Minimality needs both an all-scenarios upper bound and a feasible attaining witness; the joint restriction and valuation sensitivity are separate assumptions.
Hands-on: Reverse stress certificate workshop
Construct minimal triggering scenarios under two joint restrictions and relocate the worst direction after a sensitivity change.
Construct a reverse stress certificate
Loss is 6x+by chips; the inclusive trigger is 15. Choose a joint shock restriction, then construct the minimum severity and a feasible triggering point. Save a box certificate, challenge it under the shared budget, then repair it. Change b to 8 and relocate the worst point; at b=6 investigate ties. Decimals and fractions are accepted. Severity and coordinates are limited to 0 through 10 for this workshop.
Joint restriction
0 ≤ x ≤ t and 0 ≤ y ≤ t
Y loss sensitivity
Set box; b 4. Scenario (0, 0); severity 1; required severity 0; feasible. Loss 0 chips; terminal wealth 100; trigger margin -15. Worst permitted loss 10.
Construct a minimal triggering scenario.
Visit-local construction; no account credit. Rounded readouts do not replace the exact bound proof.
Common mistakes
- Adding single-factor worst losses when their combined point violates the joint restriction.
- Calling any breaching point a minimum-severity reverse stress.
- Treating severity as an event probability or a confidence level.
- Keeping the old worst direction after the valuation coefficient crosses six.
- Using an inclusive boundary certificate for a strict greater-than trigger.
Keep these ideas
- State shock units, joint feasibility, valuation and the exact trigger convention.
- Prove Wbox=(6+b)t and Wbudget=max(6,b)t with attaining points.
- For a positive inclusive trigger K, certify t*=K/c through necessity and attainment.
- Separate coefficient sensitivity from set expansion, ties and event probability.
Further reading
- BIS FSI: Stress testing, Executive Summary (2021), usage and limitations ↗
Read usage paragraphs and Limitations: scenario probabilities are not predicted and results depend on judgment, data and models. Supports framing only. All shock sets, proofs, numbers and activities are original; no quotations or video review. Checked 2026-10-25.
- OSFI: Stress Testing Guideline (2009), sections D and E ↗
Read assumptions documentation, interrelations, severity and reverse-stress discussion in sections D and E, especially the specified-outcome working-backward definition. Historical guideline used for conceptual framing, not a current regulatory compliance claim. Our geometric minimal-severity proof and examples are original; no source exercises or videos reused. Checked 2026-10-25.