The Effect of Removing One Competing Cause

What Changes When a Competitor Is Removed?

In a multiple-decrement setting, an individual can leave through several causes. Removing one cause from the model gives the selected cause more opportunity to occur, but the size of that change depends on the assumption used to describe decrement timing. Under UDD, the relevant quantities are associated single-decrement probabilities. Under constant force, they are cause-specific forces.

It is important to separate two questions. The associated single-decrement probability for cause $j$ is the probability of that decrement when all other causes are eliminated. The revised probability for cause $j$ after removing only cause $r$ allows any other causes to continue competing. These are equal only when no other competing causes remain.

Definition: Let $q_j$ be the observed probability of decrement $j$ during a one-year interval in a multiple-decrement model. Let $q_j^{*}$ denote its associated single-decrement probability, with all other causes removed. Let $q_j^{(-r)}$ denote the probability of decrement $j$ when cause $r$ is removed but the other causes remain. For a removal calculation, assume the underlying behavior of each retained cause is unchanged; only the specified competitor is removed.

The assumptions matter. The UDD calculation holds each cause’s associated probability fixed when a competitor is removed. The constant-force calculation holds each retained cause’s force fixed. Neither method says that the original observed probability $q_j$ should simply be carried over unchanged.

UDD: Recover First, Then Recalculate

For one year under UDD, let $q_j$ be the observed probability of cause $j$, and let the other causes have observed probabilities $q_k$. The associated probability is

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

This conversion uses the other causes’ observed probabilities in the original multiple-decrement table. After recovering the associated probabilities, removing cause $r$ means forming a new competing-decrement table using the associated probabilities of the causes that remain. If more than one cause remains, their revised observed probabilities must satisfy the UDD conversion equations together.

UDD recalculation: For the retained causes, hold each $q_j^{*}$ fixed and solve $$ q_j^{(-r)}=q_j^{*}\left(1-\frac{1}{2}\sum_{\substack{k\ne j\\k\ne r}}q_k^{(-r)}\right). $$ If removing $r$ leaves only cause $j$, the sum is empty and $q_j^{(-r)}=q_j^{*}$.

The equation follows from the UDD associated-probability identity applied to the revised set of causes: the observed probability for cause $j$ is its associated probability multiplied by the UDD adjustment for the competing decrements that remain. Thus, when multiple retained causes remain, their revised probabilities are linked; substituting the original observed probabilities in place of the revised ones generally does not solve the new system.

Worked Example: Removing the Only Competitor Under UDD

In a hypothetical annual decrement table, the observed probabilities are $q_A=0.12$ and $q_B=0.08$. Assume UDD. Find the probability of decrement $A$ after cause $B$ is removed. State the associated probability and check the result.

Method. Convert the observed probability for $A$ to its associated single-decrement probability. Since $B$ is the only competitor, that associated probability is also the revised probability after $B$ is removed.

The UDD conversion gives

q_A^{*}=\frac{0.12}{1-\frac{1}{2}(0.08)} =\frac{0.12}{0.96} =0.125.

Once $B$ is removed, no competitor remains. Therefore,

q_A^{(-B)}=q_A^{*}=0.125.

The result is a probability of $0.125$, or 12.5%. As a check, substituting the original probabilities into the conversion identity returns $0.12$: $0.125(1-0.08/2)=0.125(0.96)=0.12$. The common mistake is to retain $0.12$ as the revised probability; that is the original observed probability while $B$ is still competing, not the probability after $B$ is removed.

Worked Example: Removing One Cause While Two Causes Remain Under UDD

A hypothetical table has three causes, $A$, $B$, and $C$, with observed one-year probabilities $q_A=q_B=q_C=0.10$. Assume UDD. Remove cause $C$ and calculate the revised probability of $A$, while $B$ remains. Also compare it with the associated probability of $A$.

Step 1: Recover the associated probabilities. For cause $A$, the other observed probabilities sum to $0.10+0.10=0.20$. Hence

q_A^{*}=\frac{0.10}{1-\frac{1}{2}(0.20)} =\frac{0.10}{0.90} =\frac{1}{9}.

The same calculation applies to $B$: $q_B^{*}=1/9$. This is the basis to use after removing $C$.

Step 2: Set up the revised UDD equations. The revised probabilities of $A$ and $B$ are equal by symmetry; call each $x$. With the other retained cause still competing,

x=\frac{1}{9}\left(1-\frac{x}{2}\right).

Step 3: Solve and check. Multiplying by 9 gives $9x=1-x/2$, so $9.5x=1$, and therefore $x=2/19=0.105263\ldots$. To four decimal places, $q_A^{(-C)}=0.1053$. Substitution checks the result: $(1/9)(1-(2/19)/2)=(1/9)(18/19)=2/19$.

The associated probability $q_A^{*}=1/9=0.1111$ is larger than the revised probability $0.1053$, because $B$ still competes in the revised table. The common mistake is to call $1/9$ the probability after removing only $C$; it is the probability after removing every competitor of $A$.

Constant Force: Remove the Force, Not the Original Probability

Under the constant-force assumption over a one-year interval, each cause $j$ has a constant force $\mu_j$, and the total force is $\mu=\sum_j\mu_j$. The observed probability of cause $j$ is

q_j=\frac{\mu_j}{\mu}\left(1-e^{-\mu}\right).

If the observed probabilities are given, first calculate $q=\sum_j q_j$. The total force is $\mu=-\ln(1-q)$, and cause $j$'s force is $\mu_j=(q_j/q)\mu$. When cause $r$ is removed, retain the forces of the other causes, set $\mu^{(-r)}=\sum_{j\ne r}\mu_j$, and recalculate each remaining probability using that new total force.

Constant-force recalculation: If the cause-specific forces remain unchanged when cause $r$ is removed, then for any retained cause $j$, $$ q_j^{(-r)}=\frac{\mu_j}{\mu^{(-r)}}\left(1-e^{-\mu^{(-r)}}\right), \qquad \mu^{(-r)}=\sum_{k\ne r}\mu_k. $$ If all competitors of $j$ are removed, its associated single-decrement probability is $1-e^{-\mu_j}$.

The formula is justified by survival under constant total force. With total force $\mu^{(-r)}$, the probability of remaining in the initial state through time $t$ is $e^{-\mu^{(-r)}t}$. Cause $j$'s instantaneous decrement probability over a small interval of length $dt$ is $\mu_j e^{-\mu^{(-r)}t}dt$. Integrating over the year gives

\int_0^1\mu_j e^{-\mu^{(-r)}t}\,dt =\frac{\mu_j}{\mu^{(-r)}}\left(1-e^{-\mu^{(-r)}}\right).

The cause-specific force is unchanged, but the total force is lower after a competitor is removed. The revised cause probability therefore needs to be calculated with the new total force.

Worked Example: Removing a Competitor Under Constant Force

A hypothetical annual model has constant cause-specific forces $\mu_A=0.10$, $\mu_B=0.20$, and $\mu_C=0.10$, all per year. Find the original observed probability of $A$, then find its probability after $C$ is removed while $A$ and $B$ remain. Also calculate the associated single-decrement probability of $A$. Give probabilities to four decimal places.

Step 1: Calculate the original observed probabilities. The original total force is $0.10+0.20+0.10=0.40$, so $1-e^{-0.40}=0.32967995$. The force shares are $0.10/0.40=0.25$ for $A$, $0.20/0.40=0.50$ for $B$, and $0.10/0.40=0.25$ for $C$. Thus

q_A=0.25(0.32967995)=0.0824200,\quad q_B=0.50(0.32967995)=0.1648400,\quad q_C=0.25(0.32967995)=0.0824200.

To four decimal places, these are $0.0824$, $0.1648$, and $0.0824$.

Step 2: Remove cause $C$. The remaining total force is $\mu^{(-C)}=0.10+0.20=0.30$, and $1-e^{-0.30}=0.25918178$. The force share of $A$ is now $0.10/0.30=1/3$, so

q_A^{(-C)}=\frac{1}{3}(0.25918178)=0.08639393.

Therefore, $q_A^{(-C)}=0.0864$ to four decimal places. As a check, $q_A^{(-C)}+q_B^{(-C)}=0.25918178$, the total probability of any decrement under the remaining force $0.30$.

Step 3: Find the associated probability. If $A$ were the only cause, its probability would be

q_A^{*}=1-e^{-0.10}=0.09516258,

or $0.0952$ to four decimal places. This exceeds $0.0864$ because $B$ still competes after $C$ is removed. The common mistake is to use $q_A=0.0824$ as the revised probability. That original probability was calculated with all three forces contributing to the competition; the correct revised value uses the remaining total force $0.30$.

Choosing the Right Calculation

The two assumptions share the same modelling idea: retain the selected cause’s underlying basis and recalculate the probability in the new set of competing causes. Their bases differ. Under UDD, retain associated probabilities and solve the revised equations. Under constant force, retain cause-specific forces and use the new total force. Do not mix the UDD conversion with constant-force allocation.

Question	UDD approach	Constant-force approach
What is held fixed when a competitor is removed?	The associated single-decrement probabilities of retained causes	The cause-specific forces of retained causes
What must be recalculated?	The revised multiple-decrement probabilities, jointly if several causes remain	The total force and each retained cause’s probability
When does the revised probability equal the associated probability?	When no competing cause remains	When no competing cause remains

A useful exam check is to state explicitly which quantity the question requests. If it asks for the probability after removing one specified cause, retain any other competitors. If it asks for an associated single-decrement probability, remove all competing causes. This distinction is central to calculations aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Key takeaway: Removing a competitor generally increases the selected cause’s probability, but does not usually make it equal to its associated single-decrement probability. Under UDD, recover the associated probabilities and solve for the revised table. Under constant force, retain cause-specific forces and recalculate using the reduced total force.

Check Your Understanding

For each question, assume the stated model and give probabilities to four decimal places unless an exact value is requested.

Under UDD, two causes have observed annual probabilities $q_A=0.09$ and $q_B=0.06$. Find the probability of $A$ after $B$ is removed. (Answer: $0.0928$)
Under UDD, three causes have observed annual probabilities $0.10$, $0.10$, and $0.10$. Remove the third cause. Find the revised probability of the first cause while the second remains. (Answer: $2/19=0.1053$)
Under constant force, causes $A$ and $B$ have forces $0.08$ and $0.12$ per year. Find the probability of $A$ when $B$ is removed. (Answer: $1-e^{-0.08}=0.0769$)
Under constant force, the original observed probabilities are $q_A=0.05$ and $q_B=0.15$. Infer the cause-specific force for $A$, then find the probability of $A$ after $B$ is removed. (Answer: $\mu_A=0.0558$ per year; revised probability $0.0543$)

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