Dependent probabilities and forces of transition

Introduction to Dependent Probabilities

In your previous studies, you likely focused on the Single Decrement Model—where the only way to "leave" a state was through one specific event (usually death). However, in the real world, people leave insurance contracts for many reasons: they might die, they might suffer a critical illness, or they might simply cancel the policy (lapse).

This chapter explores how we model these competing risks. We call these "dependent" probabilities because the likelihood of leaving for one reason depends on the presence of the other reasons. If you die, you can no longer lapse your policy! Understanding the mathematical link between the force of transition (the instantaneous risk) and the dependent probability (the chance of it happening over a year) is the core goal here.

1. The Multiple Decrement Model

The Multiple Decrement Model is a special case of a Multiple-State Markov Model. While a general Markov model allows you to move back and forth between states (like "Healthy" to "Sick" and back to "Healthy"), a decrement model is a "one-way street." Once an individual leaves the "Active" state via a decrement, they cannot return.

Analogy: The Leaky Bucket
Imagine a bucket of water with three different holes at the bottom. Hole 1 is "Death," Hole 2 is "Illness," and Hole 3 is "Lapse." The speed at which water flows out of Hole 1 depends on the size of that hole, but the total amount of water that eventually exits through Hole 1 is limited by how fast water is also escaping through Holes 2 and 3. The holes are "competing" for the same water.

Key Notation

• \(x\): The current age of the individual.
• \(j\): The specific type of decrement (e.g., \(j=1\) for death, \(j=2\) for lapse).
• \(\mu_{x+t}^j\): The force of decrement \(j\) at age \(x+t\). This is the "instantaneous" risk.
• \((aq)_x^j\): The dependent probability that the individual leaves the state due to decrement \(j\) within one year, in the presence of all other decrements.
• \(ap_x\): The probability that the individual stays in the active state for the whole year (survives all decrements).

Quick Review: The "a" in \((aq)_x^j\) stands for "absolute" or "all-causes," signifying we are working within a multi-decrement framework where all causes of exit are active simultaneously.

2. The Relationship Between Force and Probability

To find the probability of a specific event happening, we look at two things: the person must survive all decrements up to time \(t\), and then exit via decrement \(j\) at that exact moment.

The probability of remaining in the active state until time \(t\) is:
\( {}_tp_x^{aa} = \exp\left( -\int_0^t \sum_k \mu_{x+s}^k ds \right) \)

The dependent probability of leaving due to decrement \(j\) over one year is the integral of the probability of being there at time \(t\) multiplied by the force of that specific decrement:
\( (aq)_x^j = \int_0^1 {}_tp_x^{aa} \cdot \mu_{x+t}^j dt \)

Key Takeaway: The total probability of leaving for any reason is simply the sum of the individual dependent probabilities:
\( (aq)_x^{total} = \sum_j (aq)_x^j \)

3. The Constant Force Assumption

In the IFoA CM1 syllabus, a very common and vital assumption is that the forces of transition are constant over single years of age. This means that between age \(x\) and \(x+1\), the risk \(\mu_{x+t}^j\) does not change as \(t\) moves from 0 to 1.

Let's assume there are \(m\) total decrements. If \(\mu_{x+t}^j = \mu^j\) (a constant) for \(0 \le t < 1\):

A. Calculating Dependent Probabilities from Forces

If we know the constant forces \(\mu^1, \mu^2, ..., \mu^m\), the total force is:
\( \mu^{total} = \mu^1 + \mu^2 + ... + \mu^m \)

The probability of staying in the active state for one year is:
\( ap_x = e^{-\mu^{total}} \)

The dependent probability for decrement \(j\) is:
\( (aq)_x^j = \frac{\mu^j}{\mu^{total}} (1 - e^{-\mu^{total}}) \)

Don't worry if this seems tricky! Just remember that the part in the bracket \((1 - e^{-\mu^{total}})\) is the probability of leaving for any reason. We are simply scaling it by the ratio of "force \(j\)" to the "total force."

B. Calculating Forces from Dependent Probabilities

Sometimes the exam gives you the probabilities \((aq)_x^1, (aq)_x^2, ...\) and asks for the forces. First, find the total probability of staying:
\( ap_x = 1 - \sum (aq)_x^j \)

Then, the total force is:
\( \mu^{total} = -\ln(ap_x) \)

Finally, we find the individual force for decrement \(j\):
\( \mu^j = \frac{(aq)_x^j}{\sum (aq)_x^k} \cdot \mu^{total} \)

Common Mistake: Forgetting that \(\sum (aq)_x^k = 1 - ap_x\). Always check that your total probability of leaving plus your probability of staying equals 1.

4. Step-by-Step Example

Question: An insurance policy is subject to two decrements: Death (\(j=1\)) and Lapse (\(j=2\)). Between age 40 and 41, the constant force of death is \(0.02\) and the constant force of lapse is \(0.08\). Calculate the probability that a policyholder aged 40 lapses their policy within the year.

Step 1: Identify the forces.
\( \mu^1 = 0.02 \)
\( \mu^2 = 0.08 \)

Step 2: Calculate the total force.
\( \mu^{total} = 0.02 + 0.08 = 0.10 \)

Step 3: Apply the formula for the lapse probability \((aq)_{40}^2\).
\( (aq)_{40}^2 = \frac{0.08}{0.10} (1 - e^{-0.10}) \)
\( (aq)_{40}^2 = 0.8 \cdot (1 - 0.90484) \)
\( (aq)_{40}^2 = 0.8 \cdot 0.09516 = 0.07613 \)

Interpretation: There was an 8% "instantaneous" risk of lapsing, but because some people died during the year (Decrement 1), only 7.613% actually lived long enough to lapse.

5. Why Does This Matter?

In actuarial modelling, we use these probabilities to project cashflows. If we are calculating the Expected Present Value (EPV) of a death benefit, we must use the dependent probability of death. If we used a single decrement probability (ignoring lapses), we would overstate how many death benefits we expect to pay, because we would be ignoring the fact that many people leave the pool by cancelling their policies before they die.

Key Takeaway Summary:
1. Multiple decrement models are used when there are several competing ways to exit a state.
2. Under the constant force assumption, the probability of a specific exit is proportional to its force.
3. The dependent probability \((aq)_x^j\) will always be smaller than the force \(\mu^j\) (when \(\mu\) is small) because other decrements "steal" potential candidates for exit.

Ready for the next step? This logic forms the foundation for building Multiple Decrement Tables, where we use these probabilities to track a hypothetical cohort of lives (usually starting with \(l_x = 100,000\)).