Why Conversion Consistency Matters
An associated single-decrement table describes one cause as if the other causes were absent. Its one-year decrement probability is therefore not, in general, the probability of that cause in the original multiple-decrement model. Conversion consistency is the requirement that, after converting all associated tables using the same within-year assumption, the reconstructed cause-specific probabilities add to the total probability of exit.
This is a useful check on a conversion as well as a result in its own right. It catches errors such as adding associated probabilities directly, applying a conversion for the wrong within-year assumption, or using a cause-specific probability where a total decrement probability belongs. The methods below are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
The prime on \(q'^{(j)}\) distinguishes an associated single-decrement probability from the multiple-decrement probability \(q^{(j)}\). The two are not generally equal: in the multiple-decrement model, a life can leave from another cause before being exposed to cause \(j\) for the full year. Conversion formulas account for this competition. We consider two alternative assumptions about the timing of exits within the year: uniform distribution of decrements (UDD) and constant forces of decrement.
Consistency Under UDD
Under UDD, decrements are spread uniformly through the year. Write \(Q=\sum_j q^{(j)}\). If \(l\) lives are present at the start of the year, the expected number still present at time \(t\), for \(0\le t\le1\), is \(l(1-tQ)\). In particular, the average number exposed to risk over the year is \(l(1-Q/2)\). The cause-\(j\) decrement count is \(lq^{(j)}\), so its central rate is \(m^{(j)}=q^{(j)}/(1-Q/2)\).
In the associated single-decrement table for cause \(j\), UDD gives an average exposure of \(l'(1-q'^{(j)}/2)\) from a starting number \(l'\), and decrement count \(l'q'^{(j)}\). Thus its central rate is \(q'^{(j)}/(1-q'^{(j)}/2)\). Equating these rates gives the standard UDD conversion:
The denominator in the second form contains the other causes, not cause \(j\) itself. To prove the consistency result in the reverse direction, define \(r_j=q'^{(j)}/(1-q'^{(j)}/2)\) and \(R=\sum_j r_j\). Rearranging the rate equation gives \(q^{(j)}=(1-Q/2)r_j\). Summing over causes yields \(Q=(1-Q/2)R\), so:
This proves the key point: reconstructing every cause with the same UDD conversion produces cause-specific probabilities that sum exactly to the total decrement probability. The factors \(1+R/2\) are shared across causes; calculating a different total for each cause would break that common scaling.
Worked Example: Reconstructing Two UDD Causes
A hypothetical membership group has two competing exit causes. Its associated UDD tables give \(q'^{(1)}=2/11\) and \(q'^{(2)}=2/7\). Reconstruct the multiple-decrement probabilities, find the total exit probability, and check the result.
Method. Convert each associated probability to \(r_j=q'^{(j)}/(1-q'^{(j)}/2)\), then apply the shared UDD scaling factor \(1+R/2\).
For cause 1, \(r_1=(2/11)/(1-1/11)=(2/11)/(10/11)=1/5\). For cause 2, \(r_2=(2/7)/(1-1/7)=(2/7)/(6/7)=1/3\). Therefore \(R=1/5+1/3=8/15\), and the common denominator is \(1+R/2=1+4/15=19/15\).
The reconstructed probabilities are \(q^{(1)}=(1/5)/(19/15)=3/19\) and \(q^{(2)}=(1/3)/(19/15)=5/19\). Thus \(Q=3/19+5/19=8/19\), or \(0.4211\) to four decimal places. The probability of remaining in the group is \(1-Q=11/19\).
Check. Converting back, cause 1 gives \((3/19)/(1-(5/19)/2)=(3/19)/(33/38)=2/11\). Cause 2 gives \((5/19)/(1-(3/19)/2)=(5/19)/(35/38)=2/7\). Both associated inputs are recovered. Common mistake avoided: adding \(2/11+2/7\) gives the sum of the associated probabilities, not the multiple-decrement total.
Worked Example: Three UDD Causes and the Shared Scaling Factor
A hypothetical benefit program has three competing exit causes. The associated UDD probabilities are \(2/21\), \(2/11\), and \(6/23\). Find each multiple-decrement probability and verify closure, meaning that the cause-specific probabilities add to the total.
For the three causes, respectively, \(r_1=(2/21)/(1-1/21)=1/10\), \(r_2=(2/11)/(1-1/11)=1/5\), and \(r_3=(6/23)/(1-3/23)=3/10\). Hence \(R=1/10+1/5+3/10=3/5\), and the common scaling denominator is \(1+R/2=1+3/10=13/10\).
It follows that \(q^{(1)}=(1/10)/(13/10)=1/13\), \(q^{(2)}=(1/5)/(13/10)=2/13\), and \(q^{(3)}=(3/10)/(13/10)=3/13\). Their sum is \(Q=(1+2+3)/13=6/13\), or \(0.4615\) to four decimal places. The corresponding stay probability is \(7/13\), and \(6/13+7/13=1\).
As a cause-by-cause check, for cause 3 the other two probabilities sum to \(3/13\). Its associated probability is therefore \((3/13)/(1-(3/13)/2)=(3/13)/(23/26)=6/23\), as given. The same conversion applies to the other causes. Common mistake avoided: applying a separate scale based only on each cause would not preserve a common total \(Q\).
Consistency Under Constant Forces
For the alternative constant-force assumption, suppose the force for each cause \(j\) is a nonnegative constant \(\mu_j\) throughout the year. In its associated single-decrement table, \(q'^{(j)}=1-e^{-\mu_j}\), so \(\mu_j=-\ln(1-q'^{(j)})\). When all causes operate together, the total force is \(M=\sum_j\mu_j\). The chance of any exit during the year is \(Q=1-e^{-M}\).
The probability of exit from cause \(j\) is the integral, over the year, of the probability of surviving all causes to time \(t\) multiplied by the cause-\(j\) force. Since total-force survival to time \(t\) is \(e^{-Mt}\), this gives:
If \(M=0\), all the cause forces are zero, so \(Q=0\) and every \(q^{(j)}=0\); the result still holds without using the expression that divides by \(M\). This derivation shows why adding the cause-specific probabilities works: each cause receives the same total-exit factor \(1-e^{-M}\), apportioned in proportion to its force.
Worked Example: Constant-Force Conversion for Two Causes
For a hypothetical one-year model, the associated single-decrement probabilities correspond to constant forces \(\mu_1=0.10\) and \(\mu_2=0.20\). Find the associated probabilities, the multiple-decrement probabilities, and the total exit probability.
Method. Add the forces to get \(M=0.10+0.20=0.30\). The total exit probability is \(Q=1-e^{-0.30}=0.2591818\), which is \(0.2592\) to four decimal places.
The cause shares are \(\mu_1/M=0.10/0.30=1/3\) and \(\mu_2/M=0.20/0.30=2/3\). Therefore \(q^{(1)}=(1/3)(0.2591818)=0.0863939\), or \(0.0864\), and \(q^{(2)}=(2/3)(0.2591818)=0.1727879\), or \(0.1728\). Before rounding, their sum is \(0.0863939+0.1727879=0.2591818=Q\). The associated probabilities themselves are \(q'^{(1)}=1-e^{-0.10}=0.0952\) and \(q'^{(2)}=1-e^{-0.20}=0.1813\), to four decimal places.
Check and common mistake. The cause probabilities add to the total. Adding the associated probabilities instead gives \(0.0952+0.1813=0.2765\) using rounded values, which is not the multiple-decrement total. Associated probabilities describe hypothetical single-cause models, not simultaneous exits.
Choosing and Applying the Assumption
The UDD and constant-force conversions are alternative models of timing within the year. Consistency is guaranteed when all causes are converted on the same basis; it does not justify using a UDD conversion for one cause and a constant-force conversion for another. The two assumptions generally produce different reconstructed values, even when they start from similar-looking associated probabilities.
A reliable calculation follows a short sequence:
Check whether each input is an associated single-decrement probability or a multiple-decrement probability.
Use UDD for every cause or constant forces for every cause, as specified by the problem.
Under UDD, calculate every \(r_j\) and use the shared factor. Under constant forces, recover each force, sum them, and allocate the total exit probability.
Sum the reconstructed cause probabilities and compare with the total decrement probability calculated under the same assumption.
A frequent exam trap is to recognize a familiar conversion but omit the “other causes” part of its denominator. Under UDD, the forward conversion is \(q'^{(j)}=q^{(j)}/[1-(Q-q^{(j)})/2]\), not \(q^{(j)}/(1-Q/2)\). The latter incorrectly includes cause \(j\)'s own decrements in the adjustment for competing causes. Another trap is to round intermediate values: keep full precision through the shared scaling calculation and round only the final reported probabilities.
Check Your Understanding
Use the stated within-year assumption throughout each problem. Round probabilities to four decimal places where a decimal answer is requested.
- Under UDD, two associated tables have probabilities \(1/4\) and \(1/5\). Calculate \(r_1\), \(r_2\), and the reconstructed total decrement probability. (Answer: \(r_1=2/7,\ r_2=2/9,\ Q=32/79\approx0.4051\))
- Under UDD, associated probabilities are \(2/9\) and \(2/11\). Find each multiple-decrement probability and their total. (Answer: \(q^{(1)}=10/49,\ q^{(2)}=8/49,\ Q=18/49\approx0.3673\))
- Under constant forces, associated probabilities are \(1-e^{-0.04}\) and \(1-e^{-0.11}\). Find the total decrement probability. (Answer: \(1-e^{-0.15}\approx0.1393\))
- Under UDD, three associated probabilities are \(0.10,\ 0.20,\ 0.30\). Convert each to \(r_j\), then find the total decrement probability. (Answer: \(R\approx0.6804,\ Q=R/(1+R/2)\approx0.5077\))
- Two associated tables imply constant forces \(0.12\) and \(0.18\). A calculation adds their associated probabilities instead of converting the forces. State the correct total and explain the error. (Answer: \(Q=1-e^{-0.30}\approx0.2592\); adding the associated probabilities does not give the multiple-decrement total.)