Converting Between UDD and Constant-Force Rates

Why the Within-Year Assumption Matters

An associated annual decrement probability describes the chance of a specified decrement over a full year in an associated single-decrement model. It does not, by itself, determine how that chance is distributed within the year. To calculate a probability over part of a year, an actuary must state a within-year assumption. Two standard choices are a constant force and uniform distribution of decrements (UDD). They agree on the annual probability but generally give different fractional-year probabilities.

The distinction is important when a valuation or table calculation needs a probability over a fraction of a year. A conversion is valid only for the model whose assumption is being used. In particular, the observed probability of one cause in a multiple-decrement table is not automatically the associated single-decrement probability for that cause.

Definition: Let \(q\) be an associated single-decrement probability of a specified cause during the year from age \(x\) to age \(x+1\). Under the constant-force assumption, the cause has a constant force \(\mu\) throughout that year. Under UDD, the associated exits are uniformly distributed over the year, so the probability of exit by duration \(t\), for \(0\le t\le1\), is \(tq\). Each assumption is a model for the timing of exits within the year; matching the same annual \(q\) does not make the models equivalent at fractional durations.

In multiple-decrement notation, an observed annual probability for cause \(j\) may be written \({}_1q_x^{(j)}\), while the associated single-decrement probability for that cause may be written \({}_1q_x^{\prime(j)}\). This tutorial uses \(q\) for the associated annual probability being converted. First establish which probability the problem supplies. The formulas below apply to \(q\), not automatically to an observed cause-specific probability affected by competing causes.

Converting an Annual Probability to a Constant Force

Suppose the specified decrement follows a constant force \(\mu\) per year during the interval. The probability of no decrement by duration \(t\) is \(e^{-\mu t}\); therefore the probability of decrement by duration \(t\) is \(1-e^{-\mu t}\). At \(t=1\), the annual probability is \(q=1-e^{-\mu}\), which can be rearranged to obtain the equivalent force.

\mu=-\ln(1-q),\qquad 0\le q<1.

The restriction \(q<1\) is necessary for a finite force. If \(q=1\), then \(\ln(1-q)=\ln(0)\) is undefined, and no finite constant force produces probability 1 over a positive period. Probability 1 corresponds only to an infinite force in the limiting sense. If \(q=0\), the equivalent force is \(\mu=0\).

{}_tq_{\text{CF}}=1-e^{-\mu t}=1-(1-q)^t,\qquad 0\le t\le1.

Here the subscript CF labels the fractional-year probability implied by constant force. The annual probability is recovered at \(t=1\). The survival probability is \(e^{-\mu t}=(1-q)^t\), which is useful as a direct check.

Converting an Annual Probability Under UDD

Under UDD, the probability of decrement by duration \(t\) is linear in \(t\) within the year. Since it must equal \(q\) at the end of the year, the fractional-year probability is \(tq\). Its survival probability is \(1-tq\). This assumes UDD for the associated single decrement over the interval; it does not assert that all competing causes in an observed multiple-decrement table have that timing pattern.

{}_tq_{\text{UDD}}=tq,\qquad {}_tp_{\text{UDD}}=1-tq,\qquad 0\le t\le1.

UDD can also be described by a duration-dependent instantaneous force. For \(t<1\) when \(q=1\), or for \(0\le t\le1\) when \(q<1\), differentiating the survival probability gives

\mu_{\text{UDD}}(t) =-\frac{d}{dt}\ln(1-tq) =\frac{q}{1-tq}.

Thus the UDD force is generally not constant: it increases with \(t\) when \(q>0\). When \(q=1\), this expression is finite for \(t<1\) and diverges as \(t\) approaches 1. The annual probability alone therefore does not identify a single force that also reproduces UDD probabilities at every fractional duration.

Key result: Given an associated annual probability \(q\), the equivalent constant force is \(-\ln(1-q)\), when \(0\le q<1\). The fractional-year probability is \(1-(1-q)^t\) under constant force and \(tq\) under UDD. Both reproduce the same annual probability at \(t=1\), but they generally differ for \(0<t<1\).

Why the Conversions Hold

The constant-force formula follows from the survival model. If \(\mu\) is constant, survival for one year is \(e^{-\mu}\), so the probability of decrement during that year is \(1-e^{-\mu}\). Setting this equal to \(q\) gives \(e^{-\mu}=1-q\). Survival to duration \(t\) is then

e^{-\mu t}=(e^{-\mu})^t=(1-q)^t.

Subtracting this survival probability from 1 gives \(1-(1-q)^t\). For UDD, uniform timing means the cumulative probability of decrement grows in direct proportion to elapsed time. At fraction \(t\) of the year, that proportion is \(t\), so the cumulative probability is \(tq\). These arguments also show why the two formulas are different: constant force makes survival exponential in time, whereas UDD makes survival linear in time.

Identify the input.
Confirm that the given annual probability is the associated single-decrement probability for the cause and interval in question.

Choose the within-year assumption.
Use constant force only when the problem specifies or supports a constant force. Use the linear fractional-probability formula only under UDD.

Convert and check.
For constant force, calculate \(\mu=-\ln(1-q)\) and then \(1-e^{-\mu t}\). For UDD, calculate \(tq\). At \(t=1\), each result must return \(q\).

Worked Conversions

Worked Example: Comparing Both Assumptions for One Annual Probability

A hypothetical associated single-decrement table gives an annual probability \(q=0.20\). Calculate the equivalent constant force and the probability of decrement during the first quarter of a year under both constant force and UDD. State the method and check the annual result.

Method. Convert the annual probability to a force for the constant-force calculation; use the linear fractional probability for UDD. The key trap is to treat the two fractional-year formulas as interchangeable.

Constant force. Since \(q=0.20<1\), a finite force exists:

\mu=-\ln(1-0.20)=-\ln(0.80)=0.2231436.

For a quarter-year, \(t=0.25\):

{}_{{0.25}}q_{\text{CF}} =1-e^{-(0.2231436)(0.25)} =1-e^{-0.0557859} =1-(0.80)^{0.25} \approx 0.0542584.

UDD. The fractional-year probability is

{}_{{0.25}}q_{\text{UDD}}=(0.25)(0.20)=0.0500000.

The annual check is \(1-e^{-0.2231436}=0.20\) under constant force and \(1(0.20)=0.20\) under UDD. Rounded to four decimal places, the quarter-year probabilities are \(0.0543\) under constant force and \(0.0500\) under UDD. The difference is expected because the within-year assumptions differ.

Worked Example: Converting a Fractional-Year Probability Back to an Annual Probability

A hypothetical associated decrement study reports a probability of \(0.10\) over half a year. Find the implied annual probability under (a) UDD and (b) constant force. For the constant-force answer, also give the equivalent force. Check each result by recovering the reported half-year probability.

Method. Invert the appropriate fractional-year formula. Do not divide by \(t\) unless UDD is the stated assumption.

(a) UDD. Since \({}_{0.5}q_{\text{UDD}}=0.5q\),

0.10=0.5q \quad\Longrightarrow\quad q=0.20.

The check is \((0.5)(0.20)=0.10\).

(b) Constant force. Half-year survival is \(1-0.10=0.90\). Under constant force, half-year survival is the square root of annual survival, so annual survival is \((0.90)^2=0.81\). Therefore

q=1-0.81=0.19,\qquad \mu=-\ln(0.81)=0.2107210.

The check is \(1-e^{-(0.2107210)(0.5)}=1-\sqrt{0.81}=1-0.90=0.10\). The inferred annual probabilities differ: \(0.20\) under UDD and \(0.19\) under constant force. Rounded to four decimal places, the force is \(0.2107\) per year.

Worked Example: Checking a Boundary Probability

A hypothetical associated single-decrement table gives \(q=1.00\) for a one-year interval. Determine whether a finite constant force can represent this probability, and find the fractional-year probability at \(t=0.60\) under UDD. Explain why the two conclusions do not conflict.

Method. Check the domain of the logarithm before calculating a force, and apply UDD directly to the fractional probability. The trap is to substitute \(q=1\) into the force formula and treat an undefined logarithm as a finite numerical answer.

For constant force, the required expression is \(-\ln(1-q)=-\ln(0)\), which is undefined. No finite constant force gives annual probability 1. In the limiting sense, the probability approaches 1 as the force tends to infinity.

Under UDD, the fractional-year probability is

{}_{{0.60}}q_{\text{UDD}}=(0.60)(1.00)=0.60.

This is consistent with the UDD survival probability \(1-0.60=0.40\) at duration \(0.60\), followed by a linear decline to zero survival at the end of the year. Rounded to four decimal places, the UDD probability is \(0.6000\). The UDD model permits this boundary annual probability; the finite constant-force model does not.

Common Conversion Errors

Using the wrong annual probability. The formulas require an associated single-decrement probability. An observed multiple-decrement probability for a cause may reflect competition from other causes, so identify the quantity before converting it.
Using a linear formula under constant force. Under constant force, use \(1-(1-q)^t\), not \(tq\). The linear expression is the UDD result.
Using a finite-force formula at \(q=1\). The force formula requires \(0\le q<1\). At \(q=1\), the logarithm is undefined and no finite constant force produces the annual probability.
Confusing the equivalent force with the annual probability. The force is \(-\ln(1-q)\), not \(q\). It is measured per unit of time, here per year.
Failing the endpoint check. Set \(t=1\) in the fractional-year result. If it does not return the given annual \(q\), the formula or substitution is wrong.

These conversions matter because a model for annual decrement probabilities may be used in calculations requiring shorter periods. An annual probability alone does not decide which fractional-year pattern is appropriate. State the assumption, apply its formula, check the annual endpoint, and respect the domain of the conversion. This approach is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, state the assumption used and check that the conversion is consistent with the supplied annual or fractional-year probability.

An associated annual probability is \(0.25\). Find the half-year probability under UDD and under constant force. Give probabilities to four decimal places. (Answer: UDD \(0.1250\); constant force \(1-\sqrt{0.75}=0.1340\).)
An associated decrement has constant force \(\mu=0.40\) per year. Find its annual probability and its probability over \(0.30\) years. Give probabilities to four decimal places. (Answer: annual \(1-e^{-0.40}=0.3297\); over \(0.30\) years, \(1-e^{-0.12}=0.1131\).)
A fractional-year probability is \(0.12\) over \(0.40\) years. Find the implied annual probability under UDD and under constant force. Give probabilities to four decimal places. (Answer: UDD \(0.3000\); constant force \(1-(0.88)^{2.5}=0.2735\).)
An associated annual probability is \(0.60\). Find the equivalent constant force and the probability by duration \(0.75\) under constant force. Then find the probability at that duration under UDD. Give probabilities to four decimal places and the force to four decimal places. (Answer: force \(0.9163\); constant force probability \(1-(0.40)^{0.75}=0.4970\); UDD probability \(0.4500\).)
An associated annual probability is \(1.00\). Can a finite constant force reproduce it? What probability by duration \(0.60\) follows under UDD? (Answer: No finite constant force; the force is infinite only in the limiting sense. Under UDD the probability is \(0.6000\).)

What You'll Learn