Small-Interval Approximation Using Force

Why a Local Force Gives a Short-Interval Approximation

The force of mortality describes an instantaneous rate, while a death probability describes an event over an interval. For a sufficiently short interval, the force at its beginning provides a useful first-order estimate of the conditional probability of death during that interval. The qualification “sufficiently short” matters: the force may change across the interval, and a local approximation can be inaccurate if the interval is too long.

Let a life be alive at age \(x+t\), and let \(h>0\) be an interval measured in the same time units as the force. The quantity \({}_h q_{x+t}\) is the probability that the life dies within the next \(h\) time units, conditional on being alive at age \(x+t\). The force at the start of the interval is \(\mu_{x+t}\).

Definition: If the force is continuous near age \(x+t\), then as \(h\) decreases to zero, the conditional probability of death over the next \(h\) time units satisfies \({}_h q_{x+t}=\mu_{x+t}h+o(h)\). Equivalently, \({}_h q_{x+t}/h\) tends to \(\mu_{x+t}\). The practical first-order approximation is \({}_h q_{x+t}\approx\mu_{x+t}h\).

Here \(o(h)\) denotes a term whose ratio to \(h\) tends to zero as \(h\) tends to zero. The formula is a limit statement: it says what happens for very short intervals. It does not assert that the approximation is exact for a particular nonzero interval. Since a probability cannot exceed one, the product \(\mu_{x+t}h\) should also be interpreted as an approximation, not as a probability formula for arbitrary \(h\).

Derivation from Conditional Survival

The exact conditional probability is determined by the force over the whole interval, not just by its starting value. Suppose the force is nonnegative and locally integrable over the interval. Define its cumulative value from the start of the interval through length \(h\) by \(M(h)=\int_0^h\mu_{x+t+s}\,ds\). Under the survival model, conditional survival through that interval is \(\exp(-M(h))\). Therefore, conditional death during it is one minus that survival probability.

{}_h q_{x+t}=1-\exp\left(-\int_0^h\mu_{x+t+s}\,ds\right)=1-e^{-M(h)}.

This exact expression explains why the starting force can be used when \(h\) is small. If the force is continuous at the starting age, its average over an interval that shrinks to zero tends to its value at the start. Thus \(M(h)=\mu_{x+t}h+o(h)\). Also, for \(z\) near zero, \(1-e^{-z}=z+o(z)\). Because a continuous force is bounded near the starting age, \(M(h)=O(h)\), so the remaining terms are of smaller order than \(h\). Combining these facts gives the stated first-order result.

\frac{{}_h q_{x+t}}{h} = \frac{1-e^{-M(h)}}{M(h)} \frac{M(h)}{h} \longrightarrow 1\cdot\mu_{x+t}=\mu_{x+t},

where the first factor is understood by its limit of 1 when \(M(h)\) is positive and tends to zero. This ratio argument also handles a starting force of zero: the probability divided by \(h\) still tends to zero.

If the force is differentiable at the start, the next-order terms make the sources of error more visible. Writing \(\mu_0=\mu_{x+t}\) and \(\mu'_0\) for the derivative of force with respect to elapsed time at the start, expansion of the integral and exponential gives

M(h)=\mu_0h+\frac{\mu'_0}{2}h^2+o(h^2), \qquad {}_h q_{x+t} =\mu_0h+\frac{\mu'_0-\mu_0^2}{2}h^2+o(h^2).

The approximation omits the terms of order \(h^2\) and smaller. The force’s change across the interval contributes through \(\mu'_0\); the conversion from cumulative force to probability contributes through \(-\mu_0^2\).

Using the Approximation Carefully

To apply the approximation, identify the starting age, read the force at that age, and express the interval in the force’s time units. Multiply the starting force by the interval length. If the force is specified throughout the interval, the exact expression \(1-e^{-M(h)}\) can be used to assess the approximation; if only the starting force is known, an exact probability generally cannot be recovered from that information alone.

Locate the interval’s start.
Use the force at age \(x+t\), not a force at the interval’s endpoint or an unconditional probability from a different starting age.

Match the time units.
If force is stated per year, express \(h\) in years. A force per year multiplied by a number of days is dimensionally inconsistent.

Calculate the first-order estimate.
Multiply \(\mu_{x+t}\) by \(h\), then report the product as an approximate conditional probability.

Check whether the approximation is plausible.
It is most useful when \(h\) is short and the force changes little over the interval. If the force is known across the interval, compare with the exact integrated-force probability.

The approximation is conditional on survival to the beginning of the interval. It is not, by itself, the probability that a life initially aged \(x\) survives to \(x+t\) and then dies in the next \(h\) time units. That unconditional probability would also include the probability of reaching age \(x+t\).

Worked Example: A Mildly Increasing Force Over a Short Interval

A hypothetical life is alive at age 60. Over the next \(0.2\) years, suppose the force at elapsed time \(s\) is \(\mu_{60+s}=0.025+0.010s\) per year, where \(s\) is measured in years. Find the first-order approximation and the exact conditional death probability, then compare them.

Method. Use the starting force for the approximation. For the exact probability, integrate the specified force over the interval and apply \(1-e^{-M(h)}\). This avoids treating a changing force as constant when calculating the exact probability.

The starting force is \(\mu_{60}=0.025\) per year, and the interval is \(h=0.2\) years. The first-order estimate is

{}_h q_{60}\approx 0.025(0.2)=0.005.

For the exact value, integrate from \(s=0\) to \(s=0.2\):

M(0.2)=\int_0^{0.2}(0.025+0.010s)\,ds =0.025(0.2)+\frac{0.010(0.2)^2}{2} =0.005+0.0002=0.0052.

Therefore,

{}_{0.2}q_{60}=1-e^{-0.0052}\approx 0.0051865.

The approximation is \(0.0050000\), and the exact probability is about \(0.0051865\). The signed error, approximation minus exact value, is about \(-0.0001865\). The estimate is slightly low because the force increases during the interval. The comparison is consistent with a short-interval first-order approximation, but it also shows that the exact probability cannot be found from the starting force alone.

Worked Example: A Changing Force and the Size of the Error

A hypothetical policyholder is alive at age 50. Over the next \(0.10\) years, let the force at elapsed time \(s\) be \(\mu_{50+s}=0.040+0.080s\) per year, with \(s\) in years. Calculate the starting-force approximation, the exact probability, and the signed and relative errors.

Method. First multiply the force at the start by the interval length. Then integrate the entire force function for the exact result. Define signed error as approximation minus exact probability, and divide by the exact probability to obtain relative error.

The starting force is \(0.040\) per year, so the approximation is

{}_{0.10}q_{50}\approx 0.040(0.10)=0.0040000.

The cumulative force over the interval is

M(0.10)=\int_0^{0.10}(0.040+0.080s)\,ds =0.040(0.10)+\frac{0.080(0.10)^2}{2} =0.004+0.0004=0.0044.

Thus the exact conditional death probability is

{}_{0.10}q_{50}=1-e^{-0.0044}\approx 0.004390334.

The signed error is

0.0040000-0.004390334=-0.000390334\approx -0.0003903.

The relative error, using the exact probability as the denominator, is

\frac{-0.000390334}{0.004390334}\times 100\% \approx -8.891\%.

The negative sign means the approximation is below the exact probability. The force rises from \(0.040\) to \(0.048\) per year over the interval, so using only its starting value omits the increase. The error is about \(8.891\%\) of the exact probability, even though the absolute difference is less than \(0.0004\).

Worked Example: Zero Force at the Starting Age

A hypothetical life is alive at age 40. Suppose the force during the next \(0.20\) years is \(\mu_{40+s}=0.50s\) per year, where \(s\) is measured in years. Find the first-order estimate and the exact conditional death probability. Explain what the comparison shows.

Method. Evaluate the force at the starting age before applying the local approximation. Then integrate the force to determine the exact probability. This avoids assuming that a zero first-order estimate means the actual probability is zero.

At the start, \(s=0\), so \(\mu_{40}=0\) per year. The first-order estimate is therefore

{}_{0.20}q_{40}\approx \mu_{40}(0.20)=0.

The cumulative force over the interval is positive:

M(0.20)=\int_0^{0.20}0.50s\,ds =\frac{0.50(0.20)^2}{2}=0.010.

The exact conditional probability is

{}_{0.20}q_{40}=1-e^{-0.010}\approx 0.0099502.

Thus, for this particular interval, the first-order estimate is zero while the exact probability is about \(0.0099502\). The first-order limit remains valid: as \(h\) tends to zero, \(M(h)=0.25h^2\), so

\frac{{}_h q_{40}}{h} =\frac{1-e^{-0.25h^2}}{h} \longrightarrow 0=\mu_{40}.

But the probability is not asymptotically equivalent to \(\mu_{40}h\), since that product is identically zero. The first-order approximation describes the probability’s rate relative to \(h\); when the starting force is zero, it need not provide a useful relative estimate for a fixed short interval.

Common Exam Traps and Practical Checks

Confusing a limit with an exact formula. The result \({}_h q_{x+t}=\mu_{x+t}h+o(h)\) does not say that the death probability equals the product for a finite interval.
Using the wrong time unit. If force is per year, \(h\) must be in years. For example, 30 days is \(30/365\) years when a 365-day year is specified.
Ignoring force variation. The approximation uses the starting force. If the force rises or falls appreciably, it may differ from the exact probability based on the full interval.
Claiming relative equivalence when the starting force is zero. The correct general conclusion is \({}_h q_{x+t}/h\to\mu_{x+t}\). The statement \({}_h q_{x+t}\sim\mu_{x+t}h\) requires a positive starting force; if that force is zero, the product cannot serve as a relative benchmark.
Dropping the conditional interpretation. The probability is conditional on being alive at the interval’s start. Do not multiply or divide by a survival probability unless the question asks for an unconditional probability from an earlier age.

Key takeaway: Over a short interval, conditional death probability is approximately the starting force multiplied by the interval length, with matching time units. This is a first-order approximation. It is most informative when the force is continuous and positive near the start, changes little across the interval, and the interval is short.

Check Your Understanding

Use the starting force for each first-order estimate. When an exact probability is requested, integrate the stated force over the interval first. All elapsed times are in years.

A hypothetical life is alive at age 52, where the force is \(0.030\) per year. Approximate the conditional probability of death during the next month, using \(1/12\) year for one month. (Answer: \(0.0025\), or \(0.2500\%\))
For a hypothetical life alive at age 63, suppose \(\mu_{63+s}=0.020+0.040s\) per year for \(0\le s\le0.50\). Find the starting-force approximation and the exact conditional death probability over the half-year; give probabilities to 4 decimals. (Answer: approximation \(0.0100\); exact probability \(1-e^{-0.015}\approx0.0149\))
A hypothetical life is alive at age 45, with force \(\mu_{45+s}=0.012+0.020s\) per year for the next \(0.20\) years. Calculate the first-order estimate and the exact conditional probability to 6 decimals. (Answer: approximation \(0.002400\); exact probability \(1-e^{-0.0028}\approx0.002796\))
For a hypothetical life alive at age 38, let \(\mu_{38+s}=0.30s\) per year during the next \(0.10\) years. Find the first-order estimate and exact probability. State why the first-order result is not a relative approximation here. (Answer: approximation \(0\); exact probability \(1-e^{-0.0015}\approx0.001499\); the starting force is zero)
A hypothetical life is alive at age 70, where the force is \(0.12\) per year. Estimate the conditional probability of death in the next 30 days, using a 365-day year, and give the result to 4 decimals. (Answer: \(0.12(30/365)\approx0.0099\))

What You'll Learn