Conditional Density after Survival to a Given Time

Why Condition on Survival?

A lifetime model describes the time until an event such as death. Once a person or insured has already survived for a specified duration, questions about their future lifetime must be answered conditional on that survival. For example, the probability of surviving another five years is not generally the same at every duration already survived.

The key calculation is a change of origin: if the original future lifetime is \(T_x\), then after survival to duration \(s\), remaining lifetime is \(T_x-s\). Its density is not generally the original density evaluated at the remaining time. The original density must be evaluated at the total lifetime \(s+t\), then divided by the probability of reaching \(s\).

Definition: Let \(T_x\) be the future lifetime of a life aged \(x\), with continuous density \(f_x(t)\) and survival function \(S_x(t)=P(T_x>t)\). In standard life-contingency notation, \(S_x(t)={}_tp_x\). Given \(S_x(s)>0\), define the remaining lifetime conditional on survival to duration \(s\) by \(R_{x,s}=T_x-s\mid T_x>s\). For \(t\geq 0\), its density is \(g_{x,s}(t)=f_x(s+t)/S_x(s)\).

The condition \(S_x(s)>0\) is necessary: conditioning on survival to a duration with zero probability is not defined. The density formula assumes a continuous lifetime distribution with density \(f_x\). If a model has point masses, a density alone does not describe all its probability.

Deriving the Conditional Density

The derivation starts with the conditional distribution of remaining lifetime. For \(t\geq0\), the event that remaining lifetime exceeds \(t\), given survival to \(s\), is the event that total lifetime exceeds \(s+t\), given that it exceeds \(s\). Since surviving to \(s+t\) implies surviving to \(s\), conditional probability gives:

P(R_{x,s}>t) =P(T_x>s+t\mid T_x>s) =\frac{S_x(s+t)}{S_x(s)}.

Therefore the conditional distribution function is one minus this conditional survival probability. Differentiating it with respect to \(t\) gives the density:

G_{x,s}(t)=1-\frac{S_x(s+t)}{S_x(s)}, \qquad g_{x,s}(t)=\frac{-S'_x(s+t)}{S_x(s)} =\frac{f_x(s+t)}{S_x(s)}.

The denominator normalizes the density. Indeed, integrating over all possible remaining lifetimes gives:

\int_0^\infty g_{x,s}(t)\,dt =\frac{1}{S_x(s)}\int_0^\infty f_x(s+t)\,dt =\frac{1}{S_x(s)}\int_s^\infty f_x(u)\,du =\frac{S_x(s)}{S_x(s)} =1.

The substitution \(u=s+t\) in the second equality changes the integration variable from remaining lifetime \(t\) to total lifetime \(u\). The last equality uses the defining relationship between density and survival: the probability of a lifetime exceeding \(s\) is the density integrated from \(s\) onward.

Theorem: For \(0\leq a<b\), the probability that conditional remaining lifetime lies between \(a\) and \(b\) is \(P(a<R_{x,s}\leq b)=\int_a^b f_x(s+t)/S_x(s)\,dt=[S_x(s+a)-S_x(s+b)]/S_x(s)\). For a continuous lifetime distribution, including or excluding either endpoint does not change the probability.

This interval formula is often faster than integrating the conditional density directly. It also makes the time shift explicit: the endpoints for the original lifetime are \(s+a\) and \(s+b\), not \(a\) and \(b\).

Using the Formula

Worked Example: Remaining Lifetime Under Constant Force

A hypothetical lifetime model has a constant force of mortality \(\mu_x(t)=0.04\) per year at every duration \(t\geq0\). A life has already survived 10 years. Find the probability that remaining lifetime is between 5 and 12 years.

Method and assumptions. Under a constant force of \(0.04\) per year, the survival function is \(S_x(t)=e^{-0.04t}\), and the density is \(f_x(t)=0.04e^{-0.04t}\). The model is continuous, and \(S_x(10)>0\). No interest assumption is needed because this is a lifetime probability, not a present-value calculation.

The conditional interval probability is the difference between conditional survival at 5 and at 12 years of remaining lifetime:

P(5<R_{x,10}\leq12) =\frac{S_x(10+5)-S_x(10+12)}{S_x(10)} =\frac{e^{-0.04(15)}-e^{-0.04(22)}}{e^{-0.04(10)}}.

Evaluate the exponents: \(0.04(15)=0.60\), \(0.04(22)=0.88\), and \(0.04(10)=0.40\). Dividing each term by \(e^{-0.40}\) gives:

e^{-0.60+0.40}-e^{-0.88+0.40} =e^{-0.20}-e^{-0.48} =0.8187308-0.6187834 =0.1999474.

Rounded to four decimal places, the probability is 0.1999. As a check, the conditional density is \(g_{x,10}(t)=0.04e^{-0.04t}\), which integrates from 5 to 12 to \(e^{-0.20}-e^{-0.48}\), the same result. The equality of this density with the original density reflects the memoryless property of the constant-force model; it is not true for every lifetime model. The common mistake to avoid is using that simplification automatically when the force is not constant.

Worked Example: An Age-Varying Force

Consider a hypothetical lifetime model with force of mortality \(\mu_x(t)=0.01+0.002t\) per year at duration \(t\), for \(t\geq0\). Given survival to duration \(s=5\), calculate the probability that remaining lifetime is between 2 and 4 years.

Method and assumptions. The force is nonnegative for \(t\geq0\). Integrating it from 0 to \(t\) gives cumulative force \(H(t)=0.01t+0.001t^2\); hence \(S_x(t)=e^{-H(t)}\). The density is \(f_x(t)=\mu_x(t)S_x(t)\). The model is continuous, and the conditioning probability \(S_x(5)\) is positive.

First evaluate the cumulative force at the three total-lifetime durations required:

$$ H(5)=0.01(5)+0.001(5^2)=0.050+0.025=0.075, $$

H(7)=0.01(7)+0.001(7^2)=0.070+0.049=0.119, \qquad H(9)=0.01(9)+0.001(9^2)=0.090+0.081=0.171.

The interval probability uses total durations \(5+2=7\) and \(5+4=9\). Thus:

P(2<R_{x,5}\leq4) =\frac{S_x(7)-S_x(9)}{S_x(5)} =e^{-[H(7)-H(5)]}-e^{-[H(9)-H(5)]} =e^{-0.044}-e^{-0.096}.

Numerically, \(e^{-0.044}=0.9569540\) and \(e^{-0.096}=0.9084640\), so the difference is \(0.0484900\). Rounded to four decimal places, the answer is 0.0485.

The conditional density provides a check. Substituting \(s+t=5+t\) into the original density and dividing by \(S_x(5)\) gives \(g_{x,5}(t)=(0.020+0.002t)e^{-0.020t-0.001t^2}\). Its integral from 2 to 4 equals \(e^{-0.044}-e^{-0.096}\). A common error is to calculate \(S_x(7)-S_x(9)\) and stop: that is an unconditional probability, and it must be divided by \(S_x(5)\).

Worked Example: A Survival Function With a Heavy Tail

A hypothetical lifetime model has survival function \(S_x(t)=(1+t/20)^{-2}\) for \(t\geq0\), with durations measured in years. Given survival to year 20, find the probability that remaining lifetime is between 10 and 30 years.

Method and assumptions. The given function decreases continuously from 1 toward 0, so it defines a continuous lifetime model. Its density is \(f_x(t)=-S'_x(t)=0.1(1+t/20)^{-3}\). The requested interval for remaining lifetime corresponds to total lifetime from 30 to 50 years.

Evaluate the survival function at the conditioning duration and the two total-lifetime endpoints:

S_x(20)=(1+20/20)^{-2}=2^{-2}=\frac14,

S_x(30)=(1+30/20)^{-2}=(2.5)^{-2}=0.16, \qquad S_x(50)=(1+50/20)^{-2}=(3.5)^{-2}=\frac{4}{49}\approx0.0816327.

Therefore:

P(10<R_{x,20}\leq30) =\frac{0.16-4/49}{1/4} =4\left(\frac{4}{25}-\frac{4}{49}\right) =\frac{384}{1225} \approx0.3134694.

Rounded to four decimal places, the probability is 0.3135. As a check, conditional survival at 10 years is \(S_x(30)/S_x(20)=0.64\), and at 30 years it is \(S_x(50)/S_x(20)=16/49\approx0.3265306\). Their difference is \(0.64-0.3265306=0.3134694\). The method avoids the trap of treating the survival value at 10 years as conditional survival: after survival to 20, the correct ratio is \(S_x(20+10)/S_x(20)\).

What to Check in an Exam Calculation

The conditional-density formula is compact, but it contains two distinct operations: shift the argument by the elapsed duration \(s\), then normalize by survival to \(s\). A reliable calculation keeps the conditioning duration, the remaining-lifetime interval, and the corresponding total-lifetime interval separate.

Identify the model.
Write down \(S_x(t)\) or \(f_x(t)\), the time units, and any parameters. If starting from a force, integrate it to obtain the survival function.

State the conditioning event.
Survival to duration \(s\) means \(T_x>s\), and the denominator is \(S_x(s)\). Confirm that this probability is positive.

Shift the interval.
An interval \(a<R_{x,s}\leq b\) corresponds to total lifetime \(s+a<T_x\leq s+b\).

Normalize and check.
Use \([S_x(s+a)-S_x(s+b)]/S_x(s)\), or integrate \(f_x(s+t)/S_x(s)\) from \(a\) to \(b\). Check that the result is between 0 and 1.

A frequent trap is to substitute \(t\) for \(s+t\) in the density, effectively calculating with the original lifetime as though the clock had restarted. Another is to shift the numerator endpoints correctly but omit the denominator. The conditional density \(f_x(s+t)/S_x(s)\) prevents both errors: its shifted argument represents total lifetime, while its denominator accounts for the fact that only lives surviving to \(s\) are in the conditioned group.

Check Your Understanding

For each problem, keep remaining-lifetime time distinct from total-lifetime time. Probabilities are rounded to four decimal places where needed.

A hypothetical model has constant force of mortality \(0.03\) per year. Given survival to year 8, calculate the probability that remaining lifetime is between 4 and 9 years. (Answer: \(0.1235\))
A lifetime model has \(S_x(t)=e^{-0.005t^2}\) for \(t\geq0\). Given survival to year 10, find the probability of remaining lifetime between 3 and 6 years. (Answer: \(0.2498\))
A model has \(S_x(t)=(1+t/10)^{-1}\) for \(t\geq0\). Given survival to year 10, calculate the probability that remaining lifetime is between 5 and 15 years. (Answer: \(0.2286\))
A continuous lifetime model has \(S_x(t)=(1-t/20)^2\) for \(0\leq t\leq20\), and \(S_x(t)=0\) for \(t>20\). Given survival to year 5, find the probability that remaining lifetime is between 5 and 10 years. (Answer: \(1/3\), or \(0.3333\))

What You'll Learn