Upward and Downward Shock Diagnostics

Why Check Both Directions?

Redington matching is intended to make a portfolio’s surplus locally stable when the valuation yield moves slightly. A local check is therefore not just a calculation for an upward shift: it should also predict the direction of change after a small downward shift. Exact repricing can then confirm whether both results agree with the local prediction.

The comparison is useful because a local estimate and an exact repricing answer different questions. The estimate describes behavior near the base yield; exact repricing measures the effect of a specified finite change. For a sufficiently small parallel shift, a correctly matched portfolio with strictly greater asset convexity should show increased surplus in both directions. The increases need not be equal.

Definition: Suppose fixed asset cash flows $A_t$ and liability cash flows $L_t$ are payable at integer times $t$, and all are valued using the same effective annual yield $i$. The asset value, liability value, and surplus are $$ A(i)=\sum_t A_t(1+i)^{-t},\qquad L(i)=\sum_t L_t(1+i)^{-t},\qquad S(i)=A(i)-L(i). $$ A parallel yield shock $h$ replaces $i$ by $i+h$ for every cash flow. It is valid only when $1+i+h>0$. The change in surplus is $\Delta S(h)=S(i+h)-S(i)$.

Here, cash flows are assumed deterministic, fixed in amount, and unaffected by the yield shock. This tutorial uses annual effective yields and annual cash-flow times. The method does not itself model changing cash flows, a nonparallel movement in yields, or different valuation conventions for assets and liabilities.

The Local Stability Test

For a portfolio to satisfy the usual Redington conditions at a base yield $i_0$, its asset and liability values must agree there, their first derivatives with respect to yield must agree, and the asset second derivative must be strictly greater than the liability second derivative. These are conditions at the base yield, not claims about every possible yield.

Theorem: If $S(i_0)=0$, $S'(i_0)=0$, and $S''(i_0)>0$, then $S(i_0+h)>0$ for every sufficiently small nonzero $h$. Thus the base yield is a strict local minimum of surplus. The result does not guarantee a positive surplus for every finite shock.

The derivatives follow by differentiating the present value of each fixed cash flow. For one payment $C_t$, the first derivative of $C_t(1+i)^{-t}$ is $-tC_t(1+i)^{-t-1}$; differentiating again gives $t(t+1)C_t(1+i)^{-t-2}$. Summing these terms for assets and liabilities gives

S'(i)=-\sum_t t(A_t-L_t)(1+i)^{-t-1}, \qquad S''(i)=\sum_t t(t+1)(A_t-L_t)(1+i)^{-t-2}.

The first derivative condition says that the present-value-weighted cash-flow times agree. When asset and liability present values are equal and positive, this is equivalent to matching their Macaulay durations at the base yield. The second derivative condition compares the corresponding $t(t+1)$-weighted amounts; it requires strictly greater asset convexity in dollar-value terms.

Taylor expansion explains why both shock directions matter. With $h$ measured as a decimal change in the annual yield,

S(i_0+h) = S(i_0)+S'(i_0)h+\frac{1}{2}S''(i_0)h^2+O(h^3).

Under the three Redington conditions, the first two terms are zero, leaving the positive quadratic term $\frac12 S''(i_0)h^2$ as the leading change. Since $h^2$ is positive for either sign of a nonzero shock, the local prediction is positive surplus both above and below $i_0$. The $O(h^3)$ terms explain why the exact increases may differ.

Exact Repricing and a Local Estimate

A convenient way to compute exact shock values is to record each cash flow’s present value at the base yield. If that base present value is $W_t=C_t(1+i_0)^{-t}$, define $x=(1+i_0)/(1+i_0+h)$. Its value after the shock is $W_tx^t$. This changes the arithmetic without changing the cash flow: it is simply exact repricing at the new yield.

Check the base match.
Verify equal present values and equal first derivatives. Do not infer a local minimum from equal present values alone.

Calculate the curvature difference.
Find $S''(i_0)=A''(i_0)-L''(i_0)$. A strict local minimum requires this value to be positive.

Estimate each shock.
For a small positive or negative $h$, calculate $\frac12 S''(i_0)h^2$, using the shock in decimal-rate units.

Reprice exactly.
Evaluate all cash flows at $i_0+h$, calculate the resulting surplus, and compare its sign and size with the local estimate.

Worked Example: A Single Liability and Two Asset Payments

A hypothetical portfolio is valued at an effective annual yield of 10%. It has asset payments of $550,000 at time 1 and $665,500 at time 3, and a liability payment of $1,210,000 at time 2. Check a 1-percentage-point upward and downward parallel shock. Compare exact surplus with the local estimate. The method is to work with base-yield present values; this avoids comparing cash-flow amounts due at different times as though they were already on the same valuation date.

Base values and first derivatives. At 10%, the asset present values are $550{,}000/1.1=500{,}000$ and $665{,}500/1.1^3=500{,}000$. The liability present value is $1{,}210{,}000/1.1^2=1{,}000{,}000$. Thus both sides have value $1,000,000. The asset present-value-weighted time is $500{,}000(1)+500{,}000(3)=2{,}000{,}000$; the liability amount is $1{,}000{,}000(2)=2{,}000{,}000$. The first derivatives match.

Curvature and local estimate. The asset $t(t+1)$-weighted present value is $500{,}000(2)+500{,}000(12)=7{,}000{,}000$. The corresponding liability amount is $1{,}000{,}000(6)=6{,}000{,}000$. Therefore

S''(0.10)=\frac{1{,}000{,}000}{1.1^2} =826{,}446.28.

For either $h=0.01$ or $h=-0.01$, the local estimate is $\frac12(826{,}446.28)(0.01)^2=\$41.32$, rounded to the nearest cent.

Exact upward shock. The new yield is 11%, so $x=1.1/1.11=110/111$. Repricing the base present values gives

\Delta S(0.01) =500{,}000x+500{,}000x^3-1{,}000{,}000x^2 =500{,}000x(x-1)^2 =\frac{55{,}000{,}000}{1{,}367{,}631} =\$40.22.

Exact downward shock. The new yield is 9%, so $x=1.1/1.09=110/109$. Therefore

\Delta S(-0.01) =500{,}000x(x-1)^2 =\frac{55{,}000{,}000}{1{,}295{,}029} =\$42.47.

Both exact surpluses are positive and close to the $41.32 local estimate. They are not equal: higher-order terms affect the upward and downward repricings differently. The base surplus was zero, so these changes are also the post-shock surpluses. This verifies the expected local stability in both directions.

Checking a Stream of Multiple Liabilities

For portfolios with several payment dates, calculate the base present value and time-weighted amounts date by date before repricing. A stream can satisfy the first two matching conditions even though the asset and liability cash flows differ at every date. What matters for this local diagnostic is their aggregate present value, first derivative, and second derivative.

Worked Example: Three Liability Dates and a Half-Point Shock

At a 5% effective annual yield, a hypothetical asset portfolio pays $52,500 at time 1 and $57,881.25 at time 3. Its liabilities are $42,000 at time 1, $22,050 at time 2, and $46,305 at time 3. Calculate the exact changes in surplus after parallel shocks of plus and minus 0.5 percentage points, and compare them with the local prediction. The method is to use the present values at 5% as the weights, which keeps the shock calculation compact.

Base match. The asset present values are $50,000 at times 1 and 3, for a total of $100,000. The liability present values are $40,000, $20,000, and $40,000, also totaling $100,000. The asset first time-weighted amount is $50{,}000(1)+50{,}000(3)=200{,}000$; the liability amount is $40{,}000(1)+20{,}000(2)+40{,}000(3)=200{,}000$. Thus value and first derivative match.

Local prediction. The asset second time-weighted amount is $50{,}000(2)+50{,}000(12)=700{,}000$. The liability amount is $40{,}000(2)+20{,}000(6)+40{,}000(12)=680{,}000$. Hence $S''(0.05)=20{,}000/1.05^2=\$18{,}140.59$. With $|h|=0.005$, the local estimate in either direction is $\frac12(18{,}140.59)(0.005)^2=\$0.23$, rounded to the nearest cent.

Exact repricing. For the upward shock, $x=1.05/1.055=210/211$. The asset-minus-liability surplus is

\Delta S(0.005) =50{,}000x+50{,}000x^3 -\bigl(40{,}000x+20{,}000x^2+40{,}000x^3\bigr) =10{,}000x(x-1)^2 =\frac{2{,}100{,}000}{9{,}393{,}931} =\$0.22.

For the downward shock, $x=1.05/1.045=210/209$. Thus

\Delta S(-0.005) =10{,}000x(x-1)^2 =\frac{2{,}100{,}000}{9{,}129{,}329} =\$0.23.

Each exact result is positive and close to the local prediction. The small difference between the upward and downward values is not a failure of matching; the quadratic estimate gives the leading local behavior, not an exact symmetry rule.

When Convexity Matching Fails

The same two-direction calculation can diagnose a portfolio that matches value and first derivative but fails the strict convexity condition. If $S''(i_0)<0$, the local estimate is negative for both signs of a small shock: the base point is a local maximum of surplus, not a local minimum. If $S''(i_0)=0$, this second-order test alone cannot determine the local behavior.

Worked Example: The Sign of the Curvature Difference

At a 4% effective annual yield, assets have base present value $100,000, all payable at time 2. Liabilities have base present values of $50,000 at time 1 and $50,000 at time 3. Diagnose the effect of plus and minus 0.5-percentage-point shocks. The method is to check all three matching conditions rather than assuming that equal present values imply stability; that assumption would be the common mistake here.

Value and first derivative. Both sides have present value $100,000. Their first time-weighted amounts are $100{,}000(2)=200{,}000$ for assets and $50{,}000(1)+50{,}000(3)=200{,}000$ for liabilities. The first derivatives match.

Curvature. The asset second time-weighted amount is $100{,}000(2)(3)=600{,}000$. The liability amount is $50{,}000(1)(2)+50{,}000(3)(4)=700{,}000$. Thus

S''(0.04)=\frac{-100{,}000}{1.04^2} =-92{,}455.62.

The local prediction for either shock is $\frac12(-92{,}455.62)(0.005)^2=-\$1.16$, rounded to the nearest cent. The sign is negative, so this portfolio does not meet the strict convexity condition for a local surplus minimum.

Exact check. With $x=1.04/(1.04+h)$, the exact surplus is $100{,}000x^2-50{,}000x-50{,}000x^3=-50{,}000x(x-1)^2$. At $h=0.005$, $x=208/209$, giving $-10{,}400{,}000/9{,}129{,}329=-\$1.14$. At $h=-0.005$, $x=208/207$, giving $-10{,}400{,}000/8{,}869{,}743=-\$1.17$. Both are negative, consistent with the local diagnosis.

Shock Size, Units, and Interpretation

A yield shock must be entered in decimal-rate units in the Taylor estimate. A rise of one percentage point is $h=0.01$, not $h=1$; a rise of 25 basis points is $h=0.0025$. Because the leading term contains $h^2$, using 1 instead of 0.01 makes the estimated quadratic effect 10,000 times too large.

For example, the common incorrect calculation for the first portfolio above is $\frac12(826{,}446.28)(1)^2=\$413{,}223.14$ for a one-percentage-point shock. The correction uses $h=0.01$: $\frac12(826{,}446.28)(0.01)^2=\$41.32$. A second trap is to assume that equal upward and downward shocks must produce equal exact surpluses. The Taylor estimate is symmetric to second order, but exact repricing includes higher-order terms and can be asymmetric.

A sound diagnostic reports three things: whether the base value and first derivative match, whether the second derivative difference is strictly positive, and whether exact repricings for both shock signs agree with the local prediction. The conclusion is local: if the shocks are too large, the Taylor estimate may be a poor measure of the exact change. This approach is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Key takeaway: When $S(i_0)=0$, $S'(i_0)=0$, and $S''(i_0)>0$, the leading surplus change is $\frac12S''(i_0)h^2$, which is positive for small shocks in either direction. Confirm the prediction by repricing every dated cash flow at both shifted yields.

Check Your Understanding

For each problem, use decimal-rate units for the shock and compare the exact result with the local diagnostic where requested.

At a 6% effective annual yield, assets have base present values of $400,000 at time 1 and $600,000 at time 3. Liabilities have base present values of $800,000 at time 2 and $200,000 at time 3. Check the value and first-derivative matches, then calculate exact surplus for shocks of plus and minus 1 percentage point. (Answer: The value and first derivative match; exact surplus is $34.61 for the upward shock and $36.63 for the downward shock.)
At a 4% yield, assets have base present value $100,000 at time 2, and liabilities have $50,000 at each of times 1 and 3. Find $S''(0.04)$ and exact surplus for shocks of plus and minus 0.5 percentage points. (Answer: $S''(0.04)=-\$92{,}455.62$; the exact surpluses are $-\$1.14$ and $-\$1.17$, respectively, so the strict convexity condition fails.)
At an 8% yield, assets have base present values of $50,000 at time 1 and $50,000 at time 5; a liability has base present value $100,000 at time 3. Find the local estimate for shocks of either sign with magnitude 0.5 percentage points. (Answer: $S''(0.08)=\$342{,}935.53$; the local estimate is $+\$4.29$ for either shock.)
A portfolio satisfies $S(i_0)=0$ and $S'(i_0)=0$, but its calculated $S''(i_0)$ is zero. Can the strict local-minimum conclusion be made from the second-order test? (Answer: No. The quadratic term is zero, so higher-order behavior must be examined.)
A candidate estimates the local effect of a 25-basis-point yield rise using $h=0.25$. Identify the unit error and state the correct value of $h$. (Answer: The shock was entered in percentage-point units rather than decimal-rate units; $h=0.0025$.)

What You'll Learn