Sensitivity of Associated Probabilities to Competing Causes

Why Sensitivity Matters

An associated single-decrement probability answers a hypothetical question: what would the probability of exit from one selected cause be if the other causes were removed, under a specified assumption about exits within the year? It is calculated from multiple-decrement data; it is not generally equal to the selected cause’s observed probability. If a competing cause’s observed probability changes, the associated probability can change even when the selected cause’s observed probability is held fixed.

This tutorial examines that sensitivity under two assumptions used in associated-table calculations: uniform distribution of decrements (UDD) and constant force of decrement. The comparison is a mathematical sensitivity analysis: it varies one observed probability while holding the selected cause and any other stated probabilities fixed. It does not, by itself, claim that changing one real-world cause would leave the others unchanged.

Definition: Let \(q_j\) be the observed probability of exit from selected cause \(j\) during a year, and let \(q_k\) be the observed probability of exit from competing cause \(k\). The associated probability \(q_j'\) is the single-decrement probability for cause \(j\) under a specified within-year assumption, with the competing causes removed. Write \(Q=\sum_r q_r\) for the total observed probability of exit from all causes. The observed probabilities must satisfy \(q_r\geq 0\) and \(Q\leq 1\).

Sensitivity Under UDD

Under UDD, exits from each cause are spread uniformly through the year among the lives exposed at its start. For a selected cause \(j\), the associated probability is

q_j'=\frac{q_j}{1-\frac{1}{2}\sum_{k\ne j}q_k}.

The denominator adjusts for the fact that, when competing causes are present, some lives who could have exited from cause \(j\) in the isolated model have already exited from another cause. For sensitivity to one competitor \(k\), hold \(q_j\) and all other competing probabilities fixed. Define \(C=\sum_{r\ne j,k}q_r\). Then

q_j'=\frac{q_j}{1-\frac{1}{2}(q_k+C)} \qquad\text{and}\qquad \frac{\partial q_j'}{\partial q_k} =\frac{q_j}{2\left[1-\frac{1}{2}(q_k+C)\right]^2}.

For \(q_j>0\), this derivative is positive: increasing a competitor’s observed probability increases the selected cause’s associated probability, with the other inputs fixed. The derivative is the local change in \(q_j'\) per unit change in \(q_k\); it is not the change over an arbitrary finite interval. The formula applies when the probabilities remain feasible and the UDD denominator is positive.

Worked Example: UDD Sensitivity to One Competing Cause

A hypothetical decrement table has three causes. The observed probability for selected cause \(j\) is \(q_j=0.12\), and the probability for a second competitor is \(0.08\). Compare the associated probability when the remaining competitor has probability \(0.10\) and \(0.20\). Use UDD, hold \(q_j\) and the second competitor fixed, and calculate the local sensitivity at \(q_k=0.10\).

Method. Apply the UDD formula to each table and differentiate with respect to the changing competing probability. The comparison is feasible: the total observed probabilities are \(0.30\) and \(0.40\), both no greater than 1.

At \(q_k=0.10\), the sum of competing probabilities is \(0.10+0.08=0.18\). Thus the denominator is \(1-\frac{1}{2}(0.18)=0.91\), and \(q_j'=0.12/0.91=0.131868\ldots\).

At \(q_k=0.20\), the competing sum is \(0.20+0.08=0.28\). The denominator is \(1-\frac{1}{2}(0.28)=0.86\), so \(q_j'=0.12/0.86=0.139535\ldots\).

For the local sensitivity at \(q_k=0.10\), use \(\partial q_j'/\partial q_k=q_j/(2D^2)\), where \(D=0.91\). Therefore \(0.12/[2(0.91)^2]=0.072455\ldots\).

Rounded to four decimal places, the associated probabilities are \(0.1319\) and \(0.1395\), and the local sensitivity is \(0.0725\) per unit change in \(q_k\). The finite increase is \(0.139535-0.131868=0.007667\), which is not the same as the local derivative multiplied by \(0.10\), because the derivative changes as \(q_k\) changes.

Check and common mistake avoided. The denominator decreases from \(0.91\) to \(0.86\), so the associated probability must increase when the positive numerator is fixed. Using \(q_j/(1-\frac12Q)\) would incorrectly include the selected cause in the sum of competing probabilities.

Sensitivity Under Constant Force

Under the constant-force assumption, each cause has a constant force during the year. If the total observed probability is \(Q\), the total force is \(-\ln(1-Q)\). The selected cause’s share of that total force is \(q_j/Q\), so its associated single-decrement probability is

q_j'=1-\exp\left(-\frac{q_j}{Q}[-\ln(1-Q)]\right) =1-(1-Q)^{q_j/Q},\qquad 0<Q<1.

Here \(Q\) includes the selected cause and every competitor. To vary one competing probability while holding \(q_j\) and all other observed probabilities fixed, \(Q\) changes at the same rate as that competitor. Define \(h(Q)=-\ln(1-Q)/Q\), so \(q_j'=1-\exp[-q_jh(Q)]\). Differentiating gives

\frac{\partial q_j'}{\partial q_k} =\exp[-q_jh(Q)]q_jh'(Q), \qquad h'(Q)=\frac{\frac{Q}{1-Q}+\ln(1-Q)}{Q^2}.

For \(0<Q<1\), \(h'(Q)>0\). To see the sign, set \(g(Q)=Q/(1-Q)+\ln(1-Q)\). Then \(g(0)=0\) by continuity and \(g'(Q)=Q/(1-Q)^2>0\), so \(g(Q)>0\) for \(Q>0\). The exponential and \(q_j\) are positive when \(q_j>0\); consequently the constant-force sensitivity is positive too. The formula is for an interior value \(0<Q<1\); boundary cases should be handled by limits or directly from the model.

Why the constant-force formula holds: Let \(\mu_r\) be the constant force for cause \(r\), and let \(\mu=\sum_r\mu_r\). The probability of any exit in the year is \(Q=1-e^{-\mu}\), hence \(\mu=-\ln(1-Q)\). The observed probability assigned to cause \(j\) is its force share times the total exit probability: \(q_j=(\mu_j/\mu)Q\). Therefore \(\mu_j=(q_j/Q)[-\ln(1-Q)]\). Removing the competitors leaves force \(\mu_j\) alone, giving \(q_j'=1-e^{-\mu_j}\), which is the stated expression.

Worked Example: Constant-Force Sensitivity With Three Causes

For a hypothetical table, the selected cause has observed probability \(q_j=0.12\), and another competing cause has probability \(0.10\). A third cause has probability \(q_k=0.18\). Under constant force, calculate the associated probability at \(q_k=0.18\), then at \(q_k=0.28\), and find the local sensitivity at the original value.

Method. Recalculate \(Q\) for each case and use the constant-force conversion. The totals are \(0.12+0.10+0.18=0.40\) and \(0.12+0.10+0.28=0.50\), so both are feasible.

Initially, \(Q=0.40\) and \(q_j/Q=0.12/0.40=0.30\). Thus \(q_j'=1-(1-0.40)^{0.30}=1-0.60^{0.30}\). Since \(0.30\ln(0.60)=-0.153248\), \(0.60^{0.30}=e^{-0.153248}=0.8579\) to four decimals. Hence \(q_j'=0.1421\).

After increasing the competitor, \(Q=0.50\), and \(q_j/Q=0.12/0.50=0.24\). Then \(q_j'=1-0.50^{0.24}\). Since \(0.24\ln(0.50)=-0.166355\), \(0.50^{0.24}=e^{-0.166355}=0.8468\) to four decimals. The new associated probability is \(0.1532\).

For the local sensitivity at \(Q=0.40\), \(h(Q)=-\ln(0.60)/0.40=0.510826/0.40=1.277064\). Also, \(h'(Q)=[0.40/0.60+\ln(0.60)]/(0.40)^2 =[0.666667-0.510826]/0.16=0.9740\). The exponential term is \(e^{-0.12(1.277064)}=e^{-0.153248}=0.8579\). Therefore the derivative is \(0.8579(0.12)(0.9740)=0.1003\) per unit change in \(q_k\), to four decimals.

Check and common mistake avoided. The associated probability rises from \(0.1421\) to \(0.1532\), consistent with the positive derivative. A common error is to keep \(Q\) fixed while changing \(q_k\); doing so would ignore that the total observed exit probability changes.

Comparing the Two Sensitivities

Both assumptions give a positive sensitivity when \(q_j\) and the other competing probabilities are fixed. Their mechanisms differ. Under UDD, the competitor changes a denominator directly. Under constant force, it changes the total probability \(Q\), which changes both the inferred total force and the selected cause’s share of that force. Thus the numerical sensitivity depends on the basis; it is not a property of the observed probabilities alone.

Worked Example: Comparing UDD and Constant Force

A hypothetical two-cause table has \(q_j=0.20\) and \(q_k=0.30\). Calculate the associated probability and the local sensitivity to \(q_k\) under each assumption. State which basis gives the larger local sensitivity at these values.

Method. Use each assumption’s own formula; do not transfer the UDD denominator to the constant-force calculation. Here \(Q=0.50\), within the feasible range.

Under UDD, the denominator is \(1-\frac12(0.30)=0.85\). Thus \(q_j'=0.20/0.85=0.235294\ldots\), or \(0.2353\). Its sensitivity is \(0.20/[2(0.85)^2]=0.20/1.445=0.138408\ldots\), or \(0.1384\) per unit.

Under constant force, \(q_j'=1-(1-0.50)^{0.20/0.50}=1-0.50^{0.40}\). Since \(0.40\ln(0.50)=-0.277259\), \(0.50^{0.40}=e^{-0.277259}=0.7579\) to four decimals. Therefore \(q_j'=0.2421\).

For the derivative, \(h(0.50)=0.693147/0.50=1.386294\), and \(h'(0.50)=[1+\ln(0.50)]/(0.50)^2 =[1-0.693147]/0.25=1.2274\). The exponential term is \(e^{-0.20(1.386294)}=0.7579\). Thus the sensitivity is \(0.7579(0.20)(1.2274)=0.1861\) per unit, to four decimals.

At these values, constant force gives the larger local sensitivity: \(0.1861>0.1384\). This is a comparison at the specified point, not a universal ranking for every probability set. As a check, both associated probabilities exceed the observed \(q_j=0.20\), as expected when removing the competitor under either assumption.

Interpretation and Exam Traps

A sensitivity is meaningful only after the quantities held fixed are stated. In the calculations above, the selected observed probability stays fixed while a competing observed probability changes. A different question—for example, changing all cause probabilities proportionally—would require a different derivative. Also, a finite change should be calculated by evaluating the associated probability at both endpoints; a derivative is a local rate, not an exact finite-change rule.

Do not confuse observed and associated probabilities. The selected observed \(q_j\) is an input; the associated \(q_j'\) is the isolated-cause result.
Use the right total. In the UDD denominator, sum only the competing causes. In the constant-force formula, \(Q\) is the sum over all causes, including \(j\).
Check feasibility. Observed probabilities must be nonnegative and sum to at most 1. When varying one probability, verify this at both endpoints.
Do not interpret a positive derivative as a causal claim. It describes the specified mathematical variation with other observed probabilities fixed.

These sensitivity calculations are useful when assessing how an associated table depends on competing-cause experience or on a chosen assumption about decrements within the year. The direction of change agrees under UDD and constant force, but the size of the change can differ. Clear statements of the basis, the varied probability, and the held-fixed quantities make the result interpretable and aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, state the assumption and keep track of which probabilities are held fixed.

Under UDD, \(q_j=0.09\) and the two competing probabilities are \(0.12\) and \(0.06\). Find the associated probability for cause \(j\), to four decimals. (Answer: \(0.0989\))
Under UDD, \(q_j=0.15\), one competitor is fixed at \(0.10\), and another increases from \(0.10\) to \(0.20\). Find the associated probability before and after the increase, to four decimals. (Answer: \(0.1667\) before; \(0.1765\) after)
Under constant force, \(q_j=0.08\) and the total observed probability is \(Q=0.30\). Find \(q_j'\), to four decimals. (Answer: \(1-0.70^{0.08/0.30}=0.0907\))
Under constant force, \(q_j=0.10\) and \(Q=0.40\). Find the local sensitivity of \(q_j'\) to a competing probability when all other observed probabilities are held fixed, to four decimals. (Answer: \(0.0857\) per unit)
Under UDD, \(q_j=0.20\) and \(q_k=0.30\), with only these two causes. Find both the associated probability and the local sensitivity to \(q_k\), to four decimals. (Answer: \(q_j'=0.2353\); sensitivity \(=0.1384\) per unit)

What You'll Learn