Conditional Survival Across Multiple Intervals

Why Conditional Survival Uses the Ages After the Condition

A survival probability from birth to a particular age answers a different question from the probability of surviving a further period after reaching that age. For example, someone who has already reached age 51 is not being assessed on survival from birth; the relevant probability concerns the interval after age 51. To calculate it, use either a ratio of survival probabilities or the cumulative force of mortality over precisely that interval. This distinction is central to applying age-varying mortality models and is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $S_0(y)$ be the probability that a life survives from birth to age $y$. For a life that has reached age $x$, the probability of surviving a further $t$ years is $$ {}_tp_x=P(T_x>t)=P(T_0>x+t\mid T_0>x)=\frac{S_0(x+t)}{S_0(x)}, $$ provided $S_0(x)>0$. Here $T_0$ is the age at death measured from birth, and $T_x$ is the future lifetime measured from age $x$.

The denominator matters: the condition is survival to age $x$, so the probability must be divided by $S_0(x)$. A person who has reached age $x$ is already known to have survived the earlier ages. Those earlier survival chances must not remain in the answer.

From Force of Mortality to Conditional Survival

Suppose the lifetime model is continuous and has a force of mortality $\mu_y$ at attained age $y$. On an age interval where the force is defined and integrable, the cumulative hazard from age $x$ to age $x+t$ is the integral of the force over those attained ages. Conditional survival is the exponential of the negative cumulative hazard.

Theorem: Under a continuous lifetime model with an integrable force of mortality over $[x,x+t]$, $$ {}_tp_x=\exp\left(-\int_x^{x+t}\mu_y\,dy\right). $$ Equivalently, if $H(x,x+t)=\int_x^{x+t}\mu_y\,dy$, then ${}_tp_x=e^{-H(x,x+t)}$. This force-based formula assumes the survival model is absolutely continuous on the interval; it does not account for point masses of death not represented by the force.

The force is a rate per unit of age, so the integral accumulates mortality over the elapsed age range. If the force changes at a boundary, split the integral there. A value of the force applying for two years contributes twice as much cumulative hazard as the same value applying for one year.

Set the age interval.
For survival from age $x$ to age $x+t$, the integration limits are the attained ages $x$ and $x+t$.

Split at every change in force.
Write a separate integral for each interval, using the force that applies there.

Add cumulative hazards.
Sum the interval integrals to obtain $H$.

Convert to survival.
Calculate $e^{-H}$, or multiply the exponential survival probabilities for the individual intervals.

Why the Ratio and Integral Formulas Agree

For a continuous survival model, the force at age $y$ satisfies $\mu_y=-\frac{d}{dy}\ln S_0(y)$, wherever the derivative exists. Integrating this identity from $x$ to $x+t$ gives

\int_x^{x+t}\mu_y\,dy =-\int_x^{x+t}\frac{d}{dy}\ln S_0(y)\,dy =\ln S_0(x)-\ln S_0(x+t).

Negating this cumulative hazard and exponentiating yields

\exp\left(-\int_x^{x+t}\mu_y\,dy\right) =\exp\bigl(\ln S_0(x+t)-\ln S_0(x)\bigr) =\frac{S_0(x+t)}{S_0(x)}.

Thus the ratio and force methods are two ways to calculate the same conditional probability when the continuous force model applies. The ratio is useful when survival probabilities are known directly; integration is useful when the force is specified by age.

Worked Examples

Worked Example: Crossing a Change in Constant Force

A hypothetical mortality model has force $\mu_y=0.02$ for $32\le y<33$, and $\mu_y=0.05$ for $33\le y\le35$. Calculate the probability that a life aged 32 survives to age 35. Give the answer to four decimal places.

Method. Integrate the force over each age interval, then exponentiate the negative total cumulative hazard. The first force applies for one year and the second for two years:

$$ H(32,35)=0.02(33-32)+0.05(35-33) =0.02(1)+0.05(2)=0.12. $$

Therefore,

{}_3p_{32}=e^{-0.12}=0.886920\ldots\approx 0.8869.

The result is a conditional probability from age 32, not survival from birth to age 35. The common mistake this method avoids is counting the second interval as only one year: age 33 to age 35 spans two years, so its cumulative hazard is $0.10$, not $0.05$. The total $0.12$ gives a probability between 0 and 1, as required.

Worked Example: A Varying Force Followed by a Constant Force

In a hypothetical model, for $50\le y<52$, let $\mu_y=0.01+0.002(y-50)$. For $y\ge52$, let $\mu_y=0.02$. A life has reached age 51. Find the probability of surviving to age 54, to four decimal places.

Method. Split the cumulative hazard at age 52. In the first interval, use $z=y-50$ as the age offset appearing in the force. From age 51 to 52, this offset runs from 1 to 2:

\begin{aligned} H(51,52) &=\int_1^2(0.01+0.002z)\,dz\\ &=0.01(2-1)+0.001(2^2-1^2)\\ &=0.01+0.003=0.013. \end{aligned}

From age 52 to age 54, the force is $0.02$ for two years, so

$$ H(52,54)=0.02(54-52)=0.04. $$

The total cumulative hazard is $0.013+0.04=0.053$. Hence,

{}_3p_{51}=e^{-0.053}=0.948379\ldots\approx 0.9484.

The answer is conditional on reaching age 51. A frequent error is to integrate the varying-force formula using offsets from 0 to 1. But the specified formula uses $y-50$, so for attained ages 51 to 52 the offset runs from 1 to 2. Integrating from 0 to 1 would give the cumulative hazard for ages 50 to 51, which is not the interval in the question. The calculated probability is close to, but less than, 1, consistent with the small positive cumulative hazard.

Worked Example: Combining Three Intervals

A hypothetical model specifies the force as $0.025$ from age 60 to 62, $0.04$ from age 62 to 63, and $\mu_y=0.03+0.01(y-63)$ from age 63 to 64. Calculate the probability that a life aged 60 survives to age 64, to four decimal places.

Method. Compute a cumulative hazard for each interval. The first force applies for two years:

$$ H(60,62)=0.025(62-60)=0.05. $$

The second applies for one year:

$$ H(62,63)=0.04(63-62)=0.04. $$

For the final interval, set $s=y-63$, which runs from 0 to 1:

\begin{aligned} H(63,64) &=\int_0^1(0.03+0.01s)\,ds\\ &=0.03(1)+0.005(1^2-0^2)=0.035. \end{aligned}

Adding the three contributions gives $H(60,64)=0.05+0.04+0.035=0.125$. Therefore,

{}_4p_{60}=e^{-0.125}=0.882496\ldots\approx 0.8825.

As a check, interval survival probabilities multiply: $e^{-0.05}e^{-0.04}e^{-0.035}=e^{-0.125}$. The common mistake this calculation avoids is applying the final interval’s force outside ages 63 to 64 or treating its changing force as constant at one endpoint. The answer is below 1, as it must be for a positive cumulative hazard.

Products Across Intervals and Common Traps

For any partition $x=a_0<a_1<\cdots<a_k=x+t$, consecutive conditional survival probabilities combine by multiplication:

{}_tp_x=\prod_{j=1}^{k}{}_{a_j-a_{j-1}}p_{a_{j-1}} =\exp\left(-\sum_{j=1}^{k}\int_{a_{j-1}}^{a_j}\mu_y\,dy\right).

The product follows from conditioning successively on reaching each boundary age. The exponential form follows because the exponents add. Both approaches require consistent interval boundaries and the force appropriate to each interval.

A separate trap is confusing conditional survival with survival from birth. For example, suppose $S_0(32)=0.80$ and the force model in the first worked example applies through age 35. Then $S_0(35)=0.80e^{-0.12}=0.709536\ldots$. The conditional probability from age 32 is not $0.7095$; it is

\frac{S_0(35)}{S_0(32)} =\frac{0.80e^{-0.12}}{0.80} =e^{-0.12}\approx0.8869.

The survival-from-birth probability includes the chance of reaching age 32, whereas the conditional probability starts with a life already known to have reached that age.

Key takeaway: For survival from age $x$ to age $x+t$, use $S_0(x+t)/S_0(x)$, or integrate $\mu_y$ from attained age $x$ to $x+t$ and take the negative exponential. When the force changes, split at the change ages and add the interval-specific cumulative hazards.

Check Your Understanding

For each problem, identify the requested conditional interval before calculating. Give each probability to four decimal places.

A hypothetical model has force $0.01$ from age 28 to 29 and $0.03$ from age 29 to 31. Find the probability that a life aged 28 survives to age 31. (Answer: $e^{-0.07}=0.9324$, to four decimals.)
For $45\le y<47$, a hypothetical model has $\mu_y=0.02+0.01(y-45)$; from age 47 to 49 it has force $0.03$. Find the probability of survival from age 45 to age 49. (Answer: $e^{-0.12}=0.8869$, to four decimals.)
A model has constant forces $0.015$, $0.025$, and $0.04$ on the successive intervals from age 70 to 71, 71 to 72, and 72 to 73. Find the probability of survival from age 70 to age 73. (Answer: $e^{-0.08}=0.9231$, to four decimals.)
For $30\le y<32$, let $\mu_y=0.01+0.002(y-30)$; from age 32 to 34, let $\mu_y=0.025$. Find the probability of survival from age 31 to age 34. (Answer: $e^{-0.063}=0.9389$, to four decimals.)

What You'll Learn