Minimizing Surplus Risk Across Rate Scenarios

Why Optimize Across Rate Scenarios?

Duration matching is designed to control the first-order effect of a small, parallel change in yields. A portfolio can satisfy that condition and still experience a material asset-liability mismatch under a nonparallel curve movement, or when the rate change is large enough for curvature to matter. Scenario-based optimization addresses a different question: which feasible portfolio has the smallest specified surplus mismatch across a chosen set of rate outcomes?

The result depends on the scenarios, their probabilities, the measure of mismatch, and the allowed portfolio weights. It is not a claim that the chosen scenarios exhaust all possible interest-rate movements. The method is aligned with the SOA Exam FM (Financial Mathematics) syllabus, especially the interpretation of duration matching and surplus stability.

Definition: Let $S_0=A_0-L_0$ be the asset-liability surplus at the valuation date. For scenario $j$, let $A_j$ and $L_j$ be the revalued asset and liability amounts, and define the surplus change $\Delta S_j=(A_j-L_j)-S_0$. If the portfolio is exactly funded initially, $S_0=0$, so $\Delta S_j=A_j-L_j$. A scenario-based objective assigns a nonnegative probability $p_j$ to each scenario, with $\sum_j p_j=1$, and minimizes a stated measure built from the scenario surplus changes.

Set Up the Objective Before Optimizing

A common choice is the mean squared surplus change, also called a probability-weighted squared mismatch from the initial surplus:

Q=\sum_{j=1}^{m}p_j(\Delta S_j)^2.

This objective penalizes both variation and a systematic shift away from the initial surplus. If instead the goal is to minimize centered scenario variance, the objective is

\operatorname{Var}(\Delta S)=\sum_{j=1}^{m}p_j(\Delta S_j-\overline{\Delta S})^2, \qquad \overline{\Delta S}=\sum_{j=1}^{m}p_j\Delta S_j.

These are not interchangeable when the expected surplus change is nonzero: $Q=\operatorname{Var}(\Delta S)+(\overline{\Delta S})^2$. The first measure targets zero change from the initial surplus. The second measures dispersion around the scenario mean, even if that mean differs from zero. State which one is being minimized; otherwise an answer may be numerically correct for the wrong objective.

Suppose asset $k$ has weight $w_k$, with $w_k\geq 0$ and $\sum_k w_k=1$. If $r_{kj}$ is the change in value of asset $k$ under scenario $j$, and $\ell_j$ is the liability change, then

\Delta S_j=\sum_k w_k r_{kj}-\ell_j.

With fixed liability values and linearly weighted asset holdings, each $\Delta S_j$ is linear in the weights. Therefore $Q$ is a quadratic function of those weights. It is convex because it is a sum of squared linear expressions with nonnegative probabilities. A minimum exists on a closed feasible set such as the nonnegative weight simplex. A constrained minimum can occur at a boundary, so an unconstrained first-order solution alone is not enough.

Why the Quadratic Objective Has This Form

For each scenario, the amount by which the revalued assets exceed or fall short of the revalued liabilities is $\Delta S_j$. Squaring gives a nonnegative penalty, with larger mismatches receiving disproportionately larger penalties. Taking the probability-weighted average gives the expected squared penalty. In vector form, if $\boldsymbol{\Delta S}$ is the scenario mismatch vector and $P$ is the diagonal matrix of scenario probabilities, then

Q=\boldsymbol{\Delta S}^{\,T}P\boldsymbol{\Delta S}.

This expression also explains why a scenario with zero probability has no effect on this objective, while assigning more probability to a scenario increases its influence. The probabilities should reflect the decision-maker’s stated scenario weighting, not be selected afterward to favor a desired allocation.

Specify the valuation basis.
Record the initial asset and liability values and the rate assumptions used to revalue them.

Define scenarios and probabilities.
Include the relevant parallel and nonparallel changes, and state their probabilities or weights.

Express each surplus change in the portfolio weights.
Subtract the scenario liability value from the scenario asset value, then subtract the initial surplus.

Minimize the stated objective over feasible weights.
Check nonnegativity, budget constraints, and any other constraints that are part of the problem.

Revalue and verify.
Report the resulting scenario changes and objective value, and compare them with a relevant benchmark such as duration matching.

Worked Scenario Optimizations

Worked Example: Checking a Boundary Portfolio Across Three Scenarios

A hypothetical insurer has an initially funded liability portfolio. The baseline curve is flat at a 4% effective annual yield. Three equally weighted scenarios are considered: a parallel increase of 1 percentage point, a parallel decrease of 1 percentage point, and a nonparallel twist in which short yields rise by 1 percentage point while long yields fall by 1 percentage point. A valuation model gives the following scenario surplus changes, in thousands of dollars, for each asset when held as the entire portfolio:

Asset A: $(3,0,-3)$; Asset B: $(3,1,-4)$; Asset C: $(2,0,-2)$. Let $x$ and $y$ be the weights in A and B, so the weight in C is $1-x-y$, with $x\geq0$, $y\geq0$, and $x+y\leq1$. Minimize the equally weighted mean squared surplus change.

Method. Form the three scenario changes as the weighted sum of the asset vectors, then minimize the sum of their squares. All three vectors sum to zero, so every feasible portfolio has zero mean change across the equally likely scenarios. Here the mean squared change is also the centered variance.

The portfolio vector is

\begin{aligned} \boldsymbol{\Delta S} &=x(3,0,-3)+y(3,1,-4)+(1-x-y)(2,0,-2)\\ &=(2+x+y,\ y,\ -2-x-2y). \end{aligned}

At the feasible corner $x=0,\ y=0$, the portfolio is entirely in C, giving scenario changes $(2,0,-2)$. Its objective value is

Q(0,0)=\frac{2^2+0^2+(-2)^2}{3}=\frac{8}{3}.

To verify that this is the global minimum, rewrite the vector as $(2,0,-2)+x(1,0,-1)+y(1,1,-2)$. The dot products of $(2,0,-2)$ with the two directions are $4$ and $6$, respectively. Therefore

3Q=8+8x+12y+\left\|x(1,0,-1)+y(1,1,-2)\right\|^2\geq 8

for every feasible $x,y\geq0$. Thus the minimum is $Q=8/3$, at $x=0,\ y=0$, with weight 1 in C. For comparison, at $x=0,\ y=0.5$, the scenario vector is $(2.5,0.5,-3)$, not $(2.5,0.5,-1.5)$, and its objective is $(2.5^2+0.5^2+(-3)^2)/3=15.5/3=31/6$, which is larger than $8/3$.

Check. The selected weights are nonnegative and sum to 1. The scenario changes are $2,0,-2$, whose mean is 0 and whose variance is $(4+0+4)/3=8/3$, matching the objective.

Common mistake avoided. A boundary solution cannot be ruled out by looking only for an interior stationary point. Also, each scenario component must be computed from every asset weight; a sign or multiplication error in one component can reverse the apparent ranking.

Worked Example: Minimizing a Weighted Portfolio Mismatch

A hypothetical fund is initially matched to its liabilities at a 5% effective annual yield. For three equally likely rate scenarios—a parallel rise of 0.75 percentage point, a parallel fall of 0.75 percentage point, and a nonparallel twist—the model gives the following surplus changes, in thousands of dollars, if the portfolio is entirely in a particular asset: Asset A has $(4,-2,6)$, and Asset B has $(-2,2,0)$. The portfolio puts weight $x$ in A and $1-x$ in B, where $0\leq x\leq1$. Minimize the mean squared change from the initial surplus.

Method. Write each scenario mismatch as a linear function of $x$, then minimize the sum of squared mismatches. Since all scenarios have equal probability, dividing the sum by 3 does not affect the minimizing weight.

The scenario vector is

\begin{aligned} \boldsymbol{\Delta S} &=x(4,-2,6)+(1-x)(-2,2,0)\\ &=(-2+6x,\ 2-4x,\ 6x). \end{aligned}

Thus $3Q(x)=(-2+6x)^2+(2-4x)^2+(6x)^2$. Differentiate:

\begin{aligned} \frac{d(3Q)}{dx} &=12(-2+6x)-8(2-4x)+72x\\ &=-40+176x. \end{aligned}

Setting the derivative to zero gives $x=40/176=5/22$, which is feasible. The second derivative is $176>0$, so this is the minimum. The B weight is $1-x=17/22$. Substitution gives the three changes:

\left(-\frac{7}{11},\ \frac{12}{11},\ \frac{15}{11}\right) \text{ thousand dollars}.

The sum of their squares is $(49+144+225)/121=418/121=38/11$. Therefore $Q=(38/11)/3=38/33$, or approximately $1.1515$ thousand-dollars squared. The check is that the three changes sum to $20/11$, so their mean is not zero; the objective is specifically the mean squared change from zero, not centered variance.

Common mistake avoided. Minimizing centered variance would subtract the nonzero scenario mean before squaring and solve a different problem. Here the requested target is no change from the initial surplus, so the uncentered squared changes are required.

Worked Example: Scenario Optimization Versus Duration Matching

A hypothetical liability portfolio has a value of \$100,000 and modified duration 6. Two initially available asset portfolios each also have value \$100,000. Asset A has modified duration 3 and Asset B has modified duration 9. Let $x$ be the fraction invested in A and $1-x$ the fraction in B. A duration match requires

3x+9(1-x)=6, \qquad x=\frac12.

Now use a hypothetical scenario valuation model with three equally likely outcomes: a parallel yield increase of 1 percentage point, a parallel yield decrease of 1 percentage point, and a nonparallel twist. In units of \$100, the surplus changes for the mixed portfolio are supplied by the model as follows:

Parallel increase: $2x-1$.
Parallel decrease: $-(2x-1)$.
Nonparallel twist: $4x-1$.

The first two expressions are opposite, consistent with the parallel first-order mismatch being controlled when the duration-matching weight $x=1/2$ is used. The twist has a different exposure, so duration matching alone does not force its mismatch to zero.

Method. Minimize the equal-probability mean squared mismatch across all three modelled scenarios. The sum of squares is

\begin{aligned} 3Q(x) &=(2x-1)^2+[-(2x-1)]^2+(4x-1)^2\\ &=2(2x-1)^2+(4x-1)^2. \end{aligned}

Differentiating gives $\frac{d(3Q)}{dx}=8(2x-1)+8(4x-1)=48x-16$. The stationary point is $x=1/3$. The second derivative is $48>0$, and $1/3$ is feasible, so it is the minimum. Its scenario changes are $(-1/3,\ 1/3,\ 1/3)$ in units of \$100, and

Q\left(\frac13\right)=\frac{1}{3}\left(\frac19+\frac19+\frac19\right)=\frac19.

At the duration-matched weight $x=1/2$, the scenario changes are $(0,0,1)$, giving $Q(1/2)=1/3$. The scenario-optimal allocation reduces this specified objective by $1/3-1/9=2/9$, but it no longer exactly matches duration: its weighted asset duration is $3(1/3)+9(2/3)=7$, rather than 6.

Check. The scenario-optimal weights are nonnegative and sum to 1. The three squared changes at $x=1/3$ are each $1/9$, confirming the reported objective. This is a comparison under the stated scenario set and objective, not a general claim that duration matching should be abandoned.

Common mistake avoided. A duration-matched portfolio is not necessarily the scenario-risk minimizer. Conversely, a scenario minimizer should not be described as duration matched unless its weighted duration actually equals the liability duration.

Interpreting and Communicating the Result

A scenario-optimal weight is conditional on the model inputs. If scenario probabilities change, if new scenarios are included, or if the objective changes from mean squared mismatch to centered variance, the optimal weights can change. A useful result therefore reports more than the weights: include the scenario definitions, probabilities, surplus changes, objective measure, and feasibility constraints.

Scenario revaluations may be obtained from a full repricing model. For small parallel changes, duration and convexity can provide an approximation, while key-rate sensitivities help describe nonparallel changes. Such approximations should be identified as approximations; they need not equal the results of repricing the cash flows under a changed yield curve. The optimization is only as representative as the revaluation inputs.

A practical comparison should also distinguish risk reduction from perfect matching. An allocation may lower the chosen objective while leaving nonzero mismatches in individual scenarios. It may also increase exposures outside the selected scenario set. The scenario-based result is most informative when presented alongside a benchmark such as duration matching and when the remaining scenario residuals are examined directly.

Key takeaway: Build scenario-specific asset-liability surplus changes first, state whether the objective is mean squared mismatch or centered variance, and then optimize over feasible weights. Verify the result by revaluing every scenario and checking boundary constraints. Duration matching controls a particular first-order exposure; it does not guarantee the smallest mismatch over a broader scenario set.

Check Your Understanding

For each problem, use the stated objective and constraints. Amounts are scenario surplus changes unless otherwise indicated.

Two equally likely scenarios have mismatch changes $3x-1$ and $x+1$, where $0\leq x\leq1$. Find the weight minimizing mean squared mismatch. (Answer: $x=1/5$.)
An asset with modified duration 3 and another with modified duration 9 are used to match a liability duration of 6. The portfolio weights are nonnegative and sum to 1. Find the duration-matching weight in the first asset. (Answer: $1/2$.)
A portfolio has three equally likely scenario changes $(1,-1,2)$, in thousands of dollars. Calculate its mean squared change from zero. (Answer: $2$ thousand-dollars squared.)
For equally likely scenarios, an allocation has changes $(2,0,-2)$, in thousands of dollars. Find its centered variance. (Answer: $8/3$ thousand-dollars squared.)
A duration-matched portfolio has scenario changes $(0,0,1)$, in units of \$100, across three equally likely scenarios. A feasible alternative has changes $(-1/3,1/3,1/3)$. Which has the smaller mean squared mismatch, and by what amount? (Answer: The alternative; its objective is $1/9$ rather than $1/3$, a reduction of $2/9$ in squared \$100 units.)

What You'll Learn