Finding a Scenario That Reaches Failure
Reverse stress testing starts with a defined adverse outcome, such as capital resources falling to a required threshold or available liquidity becoming insufficient for payments. The calculation challenge is to identify shock values that reach that outcome. A shock path, an iterative scaling rule, or an optimization criterion can guide the search, but each method answers a different question.
Let \(z=(z_1,\ldots,z_k)\) denote a vector of risk-driver shocks. Define \(H(z)\) as the distance from the failure boundary, with positive values representing headroom, zero representing the boundary, and negative values representing failure. For example, capital headroom can be defined as capital resources after stress minus the required capital threshold. A liquidity model can define headroom as stressed available liquidity minus required payments over a specified horizon.
The function \(H\) must be evaluated using a consistent model: the same valuation date, horizon, capital or liquidity definition, and treatment of interactions for every candidate shock. The earlier material on aggregating losses, liquidity projection, capital resources, and nonlinear effects provides relevant modeling context. The search methods below do not correct a poorly specified failure measure or an unreliable underlying model.
Threshold Search Along a Shock Path
A threshold search reduces a multidimensional search to one dimension by fixing a path \(z(t)\), where \(t\) is a scalar severity parameter. For example, \(z(t)=t b\) scales a specified vector \(b\) of adverse shocks. Define \(h(t)=H(z(t))\). The task is to find a value of \(t\) at which \(h(t)=0\), or, if the model uses a discrete threshold, where the failure condition first holds.
Bisection is useful when a bracket \([a,b]\) has opposite-sign endpoint values, such as \(h(a)>0\) and \(h(b)<0\), and \(h\) is continuous on the interval. The intermediate value theorem then guarantees at least one root in the bracket. Each midpoint evaluation retains a half-interval whose endpoint values still have opposite signs. The bracket width is halved at every iteration, so after \(n\) iterations it is \((b-a)/2^n\).
Continuity and opposite-sign endpoint values guarantee existence, not uniqueness. If \(h\) is strictly monotone on the bracket, the root is unique. A merely nonincreasing or nondecreasing function can be constant at zero over an interval, so it may have multiple roots. If the operational question is the first threshold reached from a safe starting point, a continuous nonincreasing path function with \(h(0)>0\) and a failure endpoint identifies the first boundary as the smallest \(t\) with \(h(t)\leq0\); it does not necessarily identify a unique root over the full bracket.
Worked Example: Bisection for a Capital Threshold
A hypothetical insurer has capital headroom, in millions of dollars, of \(h(t)=12-4t-t^2\) along a specified market-and-credit shock path. Here \(t\geq0\) is the common severity multiplier, and \(h(t)\leq0\) is failure. Find the threshold to within 0.125 severity units using bisection on \([0,3]\).
Method and assumptions. The stated headroom function is continuous. At the endpoints, \(h(0)=12>0\) and \(h(3)=12-12-9=-9<0\), so a root exists. Also, \(h'(t)=-4-2t<0\) for \(0\leq t\leq3\), so the function is strictly decreasing there and the root is unique.
Iteration 1. The midpoint is \(t=1.5\). Then \(h(1.5)=12-6-2.25=3.75>0\), so retain \([1.5,3]\).
Iteration 2. The midpoint is \(t=2.25\). Then \(h(2.25)=12-9-5.0625=-2.0625<0\), so retain \([1.5,2.25]\).
Iteration 3. The midpoint is \(t=1.875\). Then \(h(1.875)=12-7.5-3.515625=0.984375>0\), so retain \([1.875,2.25]\).
Iteration 4. The midpoint is \(t=2.0625\). Then \(h(2.0625)=12-8.25-4.25390625=-0.50390625<0\), so retain \([1.875,2.0625]\). The final bracket width is \(2.0625-1.875=0.1875\), which exceeds the requested 0.125. One more midpoint is needed.
Iteration 5. The midpoint is \(t=1.96875\). Then \(h(1.96875)=12-7.875-3.8759765625=0.2490234375>0\). The bracket becomes \([1.96875,2.0625]\), with width \(0.09375\), below 0.125. Its midpoint is \(2.015625\), so the root is bracketed to the requested tolerance; reporting the bracket as the result avoids implying greater precision than established.
Check and common mistake. Direct substitution gives \(h(2)=12-8-4=0\), consistent with the final bracket. The mistake to avoid is treating a sign change as proof of a unique crossing without checking uniqueness; here strict decrease supplies that check.
Iterative Shock Scaling
When a path is already specified, an iterative scaling method updates its severity based on model evaluations. One option is Newton's method. For a differentiable scalar function \(h(t)\), a Newton update is
The formula follows by approximating \(h(t)\) near \(t_n\) with its tangent, \(h(t_n)+h'(t_n)(t-t_n)\), then setting that approximation to zero and solving for \(t\). This is a local approximation: it can converge quickly near a suitable root, but an update can overshoot, leave the plausible shock domain, or fail when the derivative is zero or the function is poorly behaved. A practical implementation should check every proposed shock and often keep a valid bracket, using a bracketed method if the Newton step is unsafe.
Worked Example: Scaling Two Shocks to a Capital Boundary
In a hypothetical model, the specified shock path is \(z(t)=(t,0.5t)\). The resulting capital loss, in millions of dollars, is \(L(t)=4t+2t^2\), and opening headroom is 16 million dollars. Thus \(h(t)=16-L(t)=16-4t-2t^2\). Starting at \(t_0=1\), use Newton updates to find the failure threshold.
Method and assumptions. The quadratic is differentiable and its derivative is \(h'(t)=-4-4t\), which is nonzero for \(t\geq0\). The model is treated as exact for this calculation, and the shock multiplier must remain nonnegative.
First update. At \(t_0=1\), \(h(1)=16-4-2=10\) and \(h'(1)=-8\). Therefore \(t_1=1-10/(-8)=2.25\). The loss at this scale is \(4(2.25)+2(2.25)^2=9+10.125=19.125\), which exceeds 16; the iterate has crossed the boundary.
Second update. At \(t_1=2.25\), \(h(2.25)=16-9-10.125=-3.125\) and \(h'(2.25)=-13\). Thus \(t_2=2.25-(-3.125)/(-13)=2.009615\) (six decimals). At this scale, \(L(t_2)\) is approximately \(16.1156\), so headroom is approximately \(-0.1156\).
Third update. Using the unrounded Newton value, \(t_3\) is approximately \(2.000015\). The exact boundary is confirmed by \(L(2)=4(2)+2(2)^2=8+8=16\), so \(h(2)=0\). The threshold is \(t=2\), giving shock vector \(z=(2,1)\).
Check and common mistake. Both the loss equation and the original headroom reconcile at \(t=2\). The first update overshot; stopping at \(t_1\) and reporting it as the threshold would be wrong. An iterative result must be checked in the original model, not just accepted because an update was calculated.
For more complicated models, derivatives may be unavailable or unreliable. A finite-difference estimate can approximate the slope, but it adds numerical error. A bracketed update is safer when a valid sign-changing interval is known. In all cases, the stopping rule should reflect decision needs: for example, a maximum headroom residual, a sufficiently narrow severity interval, or a stated shock precision.
Constrained Optimization Across Shock Combinations
A fixed path only searches combinations lying on that path. If several drivers can vary independently, optimization can select a scenario according to an explicit criterion. One common formulation minimizes the squared distance of standardized shocks to the baseline, subject to reaching the failure boundary. Standardization matters: measuring one driver in dollars and another in percentage points would make an unscaled distance arbitrary.
Worked Example: Minimum-Distance Capital Failure Combination
A hypothetical insurer has capital headroom \(H(x,y)=12-4x-3y\), in millions of dollars. The nonnegative shocks \(x\) and \(y\) are standardized units for two risk drivers; each unit has the same stipulated distance cost. Find the shock combination minimizing \(J(x,y)=x^2+y^2\), subject to failure and \(0\leq x,y\leq3\).
Method and assumptions. Failure requires \(4x+3y\geq12\). Since the objective increases with distance from \((0,0)\), the minimum feasible point lies on the boundary \(4x+3y=12\): if the inequality were strict, moving toward the origin along the same ray would reduce the objective and remain feasible until reaching the boundary.
Derivation. Minimize \(x^2+y^2\) subject to \(4x+3y=12\). The Lagrange equations are \(2x=4\lambda\) and \(2y=3\lambda\), so \(x=2\lambda\) and \(y=1.5\lambda\). Substitution into the constraint gives \(8\lambda+4.5\lambda=12\), hence \(\lambda=12/12.5=0.96\). Therefore \(x=1.92\) and \(y=1.44\).
Feasibility and objective check. Both shocks are nonnegative and less than 3. The capital loss is \(4(1.92)+3(1.44)=7.68+4.32=12\), so \(H=0\). The objective is \(1.92^2+1.44^2=3.6864+2.0736=5.76\). The distance is \(\sqrt{5.76}=2.4\).
Why this is the minimum. The objective is strictly convex and the feasible set is convex, so the feasible minimizer is unique. Equivalently, the vector \((1.92,1.44)\) is proportional to the boundary normal \((4,3)\), the closest point on the line; it is feasible under the stated bounds. The result depends on the equal standardized distance costs. Different shock scales or weights can produce a different minimizing combination.
Common mistake. Setting \(x=y\) would give \(7x=12\), or \(x=y=12/7\), a boundary point but not the minimum-distance point. The boundary condition alone does not select the preferred scenario; the objective function does.
Method Choice and Checks
Threshold search is appropriate when a particular path is fixed and a crossing along it is the question. Iterative scaling can reduce model calls near a threshold, but it needs safeguards against overshoot and unstable derivatives. Optimization is appropriate when multiple shock combinations are allowed and a selection criterion has been justified. These methods can be combined: optimization may identify a candidate path, followed by a separate threshold check using the full model.
Define the capital or liquidity metric, its threshold, and the sign convention for headroom.
State the risk drivers, shock bounds, path or permitted combinations, and any dependence or feasibility restrictions used by the model.
Use bisection for a reliable bracket, a safeguarded iterative method for faster local refinement, or optimization when an explicit scenario-selection objective is needed.
Re-evaluate the candidate in the original capital or liquidity model, check all constraints, and report numerical tolerance and any non-uniqueness.
A frequent exam trap is to confuse a boundary found on one path with the least severe failure scenario overall. A path search says only where that path reaches the boundary. Likewise, an optimizer's answer is conditional on its objective, shock scaling, constraints, and model. Report these conditions with the result. A failed search may mean the failure boundary is not reachable within the permitted shock bounds; it does not establish that failure is impossible under every conceivable shock.
The methods in this tutorial are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus. In practice, a computed reverse stress scenario is a model-dependent diagnostic: its usefulness depends on transparent assumptions, reproducible calculations, and a clear distinction between a numerical boundary and a plausible business scenario.
Check Your Understanding
For each problem, state the method or reasoning used and check that the reported scenario meets the specified condition.
- A hypothetical capital headroom function along a fixed path is \(h(t)=9-3t\), for \(0\leq t\leq4\). Find the threshold and verify whether it is unique. (Answer: \(t=3\); it is unique because the function is strictly decreasing.)
- A continuous path function has \(h(0)=5\), \(h(4)=-2\), and is nonincreasing. What does the endpoint information guarantee, and what does it not guarantee about roots? (Answer: At least one root exists; uniqueness is not guaranteed without strict monotonicity or a separate uniqueness condition.)
- A model has headroom \(h(t)=10-2t-t^2\) for \(t\geq0\). Solve for the first failure threshold. (Answer: \(t=\sqrt{11}-1\), approximately \(2.3166\); the positive root is unique because \(h'(t)=-2-2t<0\) for \(t\geq0\).)
- Minimize \(x^2+y^2\) subject to \(3x+4y\geq20\) and \(x,y\geq0\), with no additional upper bounds. Find the minimum-distance failure point and objective value. (Answer: \((x,y)=(2.4,3.2)\), with objective \(16\); it satisfies \(3(2.4)+4(3.2)=20\).)
- A hypothetical loss along a scaled path is \(L(t)=3t+t^2\), and opening headroom is 10. A proposed numerical method returns \(t=2.1\). Is this the threshold? Check the original model and give the exact threshold. (Answer: No; \(L(2.1)=6.3+4.41=10.71\), so the scenario has crossed into failure. The threshold solves \(t^2+3t-10=0\), giving \(t=2\) for \(t\geq0\).)