Asymmetric and Nonlinear Responses to Stress

Why Symmetric Shocks Can Have Asymmetric Effects

A stress is often described as an increase or decrease of the same size in an assumption. That symmetry applies to the input, not necessarily to the result. If a reserve, profit measure, or asset share is a curved function of the assumption, moving the assumption up by $h$ and down by $h$ can produce changes with different magnitudes. Floors, caps, and other thresholds can make the difference more pronounced.

The first step is to specify what “equal-sized” means. An absolute shock of $h$ changes an assumption $\theta$ to $\theta+h$ or $\theta-h$. It is not generally the same as applying equal percentage changes to $\theta$. The second step is to identify the function being stressed: for example, a prospective reserve $R(\theta)$, or an accumulated asset share $AS_t(\theta)$. The direction and size of the output change depend on that function and on the model assumptions.

Definition: For a baseline assumption $\theta$ and an absolute shock $h>0$, the upward and downward changes in an output $f$ are $f(\theta+h)-f(\theta)$ and $f(\theta-h)-f(\theta)$. The shocks are symmetric in the assumption. The output changes are equal and opposite only when the function is locally linear over the interval, or when a particular model happens to give that result.

This tutorial uses simple hypothetical models to isolate three causes of asymmetry: curvature in survival probabilities, compounding of changing expenses, and a guarantee threshold. These are diagnostic examples, not universal predictions about the direction of a stress result.

Curvature and the Size of a Response

For a twice-differentiable output $f(\theta)$, a Taylor expansion around the baseline gives an explanation for the difference between symmetric shocks. For sufficiently small $h$, the first-order terms have opposite signs, while the second-order terms have the same sign:

f(\theta+h)-f(\theta)\approx f'(\theta)h+\frac{1}{2}f''(\theta)h^2,\qquad f(\theta-h)-f(\theta)\approx -f'(\theta)h+\frac{1}{2}f''(\theta)h^2.

Adding the changes cancels the first-order terms. Thus, to second order, their sum is $f''(\theta)h^2$. If $f''(\theta)>0$, the output is locally convex: the sum of the two changes is positive. If $f''(\theta)<0$, it is locally concave and the sum is negative. This is a local diagnostic; for a large shock, the curvature can vary across the interval, so direct recalculation is preferable.

Worked Example: A Two-Year Survival-Contingent Benefit

A hypothetical contract pays \$20,000 at the end of year 2 if the insured survives both years. The annual conditional mortality probability is the same in each year, $q$, and the interest rate is zero. Assume the conditional mortality probabilities are unchanged by duration and that survival in year 2 is conditional on survival to that year. Compare the reserve at $q=0.20$ with reserves under equal absolute shocks of 0.05.

Method. Under these assumptions, the probability of surviving both years is $(1-q)^2$. The time-0 reserve is therefore $R(q)=20{,}000(1-q)^2$. There are no other benefits, expenses, or premiums in this calculation.

Baseline. The annual survival probability is $1-0.20=0.80$, so the two-year survival probability is $0.80^2=0.64$. The reserve is $20{,}000(0.64)=\$12{,}800$.

Upward mortality shock. The shocked mortality probability is $0.20+0.05=0.25$. The annual survival probability is $0.75$, and the two-year survival probability is $0.75^2=0.5625$. The reserve is $20{,}000(0.5625)=\$11{,}250$. Its change from baseline is $\$11{,}250-\$12{,}800=-\$1{,}550$.

Downward mortality shock. The shocked mortality probability is $0.20-0.05=0.15$. The annual survival probability is $0.85$, and the two-year survival probability is $0.85^2=0.7225$. The reserve is $20{,}000(0.7225)=\$14{,}450$. Its change from baseline is $\$14{,}450-\$12{,}800=+\$1{,}650$.

The changes are not equal and opposite: their magnitudes differ by \$100. To see why, expand the reserve: $R(q)=20{,}000(1-2q+q^2)$, so $R''(q)=40{,}000>0$. In this quadratic model, the second-order contribution to either shocked reserve is $20{,}000(0.05)^2=\$50$. It adds \$50 to each of the first-order approximations, making the downward-reserve increase \$50 larger and the upward-reserve decrease \$50 smaller than their corresponding linear estimates. The direct calculations confirm the result.

Common mistake avoided. Do not treat an increase in mortality as if it reduced a survival probability by the same dollar amount as a matching mortality decrease raises it. Survival over two years is the product $(1-q)^2$, not a linear function of $q$.

Compounding Can Make Asset-Share Changes Unequal

An asset share accumulates net cash flows under a specified timing and investment assumption. Even if each year’s expense is defined by a simple trend, the resulting asset share can be nonlinear in the trend parameter: later expenses depend on powers of that parameter, and cash flows at different times accumulate for different periods.

Model definition: In the example below, $AS_t$ is a hypothetical accumulated balance after year $t$. A year’s premium less expense is added to the prior balance, and the resulting balance earns the stated annual effective return. This explicit recurrence fixes the cash-flow timing; changing the timing can change the result.

Worked Example: Expense Trend and a Three-Year Asset Share

A hypothetical policy receives a premium of \$400 in each of three years. Expenses are \$60, $60(1+g)$, and $60(1+g)^2$, respectively. There are no other cash flows. For this calculation, each year’s net cash flow is added to the existing balance before the balance earns 4% for that year:

AS_0=0,\qquad AS_t=(AS_{t-1}+400-E_t)(1.04).

Calculate the year-3 asset share for $g=0$, $g=0.10$, and $g=-0.10$. Report dollar amounts to the nearest cent.

Baseline, $g=0$. The expenses are \$60 in all three years. Year 1 gives $AS_1=(0+400-60)(1.04)=340(1.04)=\$353.60$. Year 2 gives $AS_2=(353.60+400-60)(1.04)=693.60(1.04)=\$721.344$. Year 3 gives $AS_3=(721.344+400-60)(1.04)=1{,}061.344(1.04)=\$1{,}103.79776$, or \$1,103.80.

Upward expense-trend shock, $g=0.10$. The expenses are \$60, \$66, and \$72.60. Year 1 remains $AS_1=\$353.60$. Year 2 gives $AS_2=(353.60+400-66)(1.04)=687.60(1.04)=\$715.104$. Year 3 gives $AS_3=(715.104+400-72.60)(1.04)=1{,}042.504(1.04)=\$1{,}084.20416$, or \$1,084.20. Relative to the unrounded baseline, the change is $\$1{,}084.20416-\$1{,}103.79776=-\$19.59360$, or a decrease of \$19.59.

Downward expense-trend shock, $g=-0.10$. The expenses are \$60, \$54, and \$48.60. Year 1 gives $AS_1=\$353.60$. Year 2 gives $AS_2=(353.60+400-54)(1.04)=699.60(1.04)=\$727.584$. Year 3 gives $AS_3=(727.584+400-48.60)(1.04)=1{,}078.984(1.04)=\$1{,}122.14336$, or \$1,122.14. The change from the unrounded baseline is $\$1{,}122.14336-\$1{,}103.79776=+\$18.34560$, or an increase of \$18.35.

The equal and opposite changes in $g$ produce a \$19.59 decrease and an \$18.35 increase. This results from the different expense amounts in years 2 and 3 and from the accumulation timing. A useful check is to retain the unrounded balances through the recurrence and round only the final amounts.

Common mistake avoided. Do not calculate the final-year expense as $60+2g$, or apply the trend only once to all years. The stated pattern is $60(1+g)^{t-1}$: for $g=-0.10$, the third-year expense is $60(0.90)^2=\$48.60$.

Thresholds, Guarantees, and Direction Changes

A guarantee may create a kink in a financial result. It is useful to separate the signed gap between a guarantee and available assets from the amount actually payable under a floor. The signed gap can be negative when assets exceed the guarantee; a payment defined as a shortfall cannot be negative.

Worked Example: A One-Year Guarantee Near Its Threshold

A hypothetical one-year contract has \$2,000 invested and a guarantee amount of \$2,100. The deterministic baseline annual return is 4%. Compare equal absolute return shocks of six percentage points. Define the signed funding gap as $D=G-A$, where $G$ is the guarantee and $A$ is the accumulated asset amount. If the contract pays a top-up, its amount is $C=\max(D,0)$. There are no fees, taxes, or other cash flows.

Baseline. The return is 4%, so $A=2{,}000(1.04)=\$2{,}080$. The signed gap is $D=2{,}100-2{,}080=\$20$, and the top-up is $C=\max(20,0)=\$20$.

Upward return shock. The return is $4\%+6\%=10\%$. Assets are $A=2{,}000(1.10)=\$2{,}200$. The signed gap is $D=2{,}100-2{,}200=-\$100$, so the top-up is $C=\max(-100,0)=\$0$. The signed gap changes by $-\$100-\$20=-\$120$; the top-up changes by $-\$20$.

Downward return shock. The return is $4\%-6\%=-2\%$. Assets are $A=2{,}000(0.98)=\$1{,}960$. The signed gap is $D=2{,}100-1{,}960=\$140$, and the top-up is $C=\$140$. The signed gap changes by $+\$120$, and the top-up changes by $+\$120$.

The signed-gap changes are equal and opposite, but the payable changes are not: the upward shock reduces the payment by \$20, while the downward shock increases it by \$120. The floor at zero explains the asymmetry. A negative signed gap of $-\$100$ is a surplus relative to the guarantee, not a negative payment.

Common mistake avoided. Do not report the signed gap and the guarantee payment as if they were the same quantity. Keep $D=G-A$ signed, then apply the positive-part rule only when calculating $C$.

A Practical Diagnostic for Stress Responses

When equal upward and downward stresses give unequal results, the difference is not automatically an error. It is a prompt to check the model. A disciplined calculation separates the assumption change from the mechanics that convert it into an output.

Define the shock.
State the baseline, the shocked values, and whether the shock is an absolute change or a percentage change. Verify that both directions use the same stated convention.

Write the output function.
Identify the reserve, profit measure, or asset-share recurrence being recalculated. State cash-flow timing, interest, and any benefit or expense conditions before substituting values.

Check for curvature and thresholds.
Products of probabilities, powers of a trend factor, and accumulated cash flows can create curvature. Floors, caps, and guarantee triggers can change the calculation rule across a threshold.

Recalculate both scenarios and compare.
Use the same baseline and timing in each scenario, retain sufficient precision, and calculate each change from the unrounded baseline. Interpret the sign in the context of the defined output.

The same discipline applies when the output is profit. A stress can affect expected cash flows and the value assigned to future obligations; those effects need not move one-for-one. Do not infer that a reserve increase must equal a profit decrease without specifying the profit definition, valuation basis, and timing. For this topic, the key diagnosis is whether the mapping from stressed assumption to the reported output is linear, curved, or interrupted by a threshold.

Key takeaway: Equal-sized shocks describe the inputs, not the outputs. Explain an unequal response by writing down the relevant function or recurrence, checking its curvature and thresholds, and recomputing both directions under identical assumptions and timing.

Check Your Understanding

For each problem, state the relevant model and compare the two stressed outcomes with the baseline.

A hypothetical benefit of \$10,000 is payable at the end of year 3 only if the insured survives all three years. The same conditional annual mortality probability $q$ applies in each year, interest is zero, and there are no other cash flows. At $q=0.10$, calculate the reserve and the changes under absolute shocks of $+0.02$ and $-0.02$. (Answer: baseline \$7,290.00; at $q=0.12$, \$6,814.72, a decrease of \$475.28; at $q=0.08$, \$7,786.88, an increase of \$496.88.)
A hypothetical two-year policy receives \$500 at each year and has expenses of \$100 in year 1 and $100(1+g)$ in year 2. There are no other cash flows. Use $AS_0=0$ and $AS_t=(AS_{t-1}+500-E_t)(1.05)$. Calculate the year-2 asset share at $g=0$ and at $g=0.10$ and $g=-0.10$. (Answer: baseline \$861.00; at $g=0.10$, \$850.50, a decrease of \$10.50; at $g=-0.10$, \$871.50, an increase of \$10.50.)
A one-year guarantee is \$1,500 against \$1,400 invested, with deterministic baseline return 5%. Define the signed gap as guarantee less accumulated assets, and the payable top-up as the positive part of that gap. Compare the baseline with equal absolute return shocks of $+5$ and $-5$ percentage points. (Answer: baseline signed gap and top-up \$30; at 10%, signed gap $-\$40$ and top-up \$0; at 0%, signed gap and top-up \$100.)
A hypothetical benefit of \$8,000 is payable at the end of two years if the insured survives both years. The same conditional annual mortality probability $q$ applies in each year, with zero interest and no other cash flows. At $q=0.25$, calculate the reserve changes under equal absolute shocks of $+0.05$ and $-0.05$. (Answer: baseline \$4,500; at $q=0.30$, \$3,920, a decrease of \$580; at $q=0.20$, \$5,120, an increase of \$620.)
A hypothetical benefit with initial amount \$1,000 increases by the same annual factor $1+j$ for two years, so its year-2 amount is $1{,}000(1+j)^2$. Ignore discounting and all other cash flows. At $j=0.05$, compare the amount with equal absolute shocks of $+0.10$ and $-0.10$. (Answer: baseline \$1,102.50; at $j=0.15$, \$1,322.50, an increase of \$220.00; at $j=-0.05$, \$902.50, a decrease of \$200.00.)

What You'll Learn