Complete Future Lifetime Moments from Survival Functions

Why Use the Survival Function for Lifetime Moments?

The complete future lifetime $T_x$ measures the exact duration, in years, from age $x$ until death. Unlike the curtate future lifetime, it includes the fractional part of the final year. Its distribution can be described by the survival function ${}_tp_x=P(T_x>t)$, the probability that a life aged $x$ survives more than $t$ years. When that function is known, its integrals give moments of $T_x$ directly; a density is not required.

These formulas are useful when survival is specified continuously, such as under a constant force of mortality, or by a model with different forces over different durations. They also make clear that a lifetime can have a finite mean but an infinite second moment. Checking whether the necessary integrals converge is therefore part of calculating a moment, not an optional technical detail.

Definition: The complete future lifetime $T_x$ is a nonnegative random variable measured in years. Its survival function is ${}_tp_x=P(T_x>t)$, for $t\geq 0$, with ${}_0p_x=1$ for a life known to be alive at age $x$. The $r$th raw moment, when finite, is $E[T_x^r]$. In particular, $E[T_x]$ is the mean and $\operatorname{Var}(T_x)=E[T_x^2]-E[T_x]^2$.

Tail-Integral Formulas

For a nonnegative lifetime, each realized value $T_x$ can be represented by the total length of the interval from zero up to that value. For the second moment, the corresponding representation weights each duration $t$ by $2t$. Taking expectations of these representations gives the tail-integral formulas.

Theorem: For any nonnegative $T_x$, with infinite values allowed for the integrals, $$ E[T_x]=\int_0^\infty {}_tp_x\,dt,\qquad E[T_x^2]=2\int_0^\infty t\,{}_tp_x\,dt. $$ More generally, for $r>0$, $E[T_x^r]=r\int_0^\infty t^{r-1}{}_tp_x\,dt$. Each moment is finite exactly when its corresponding integral is finite.

Here is a justification of the first two formulas. For a fixed nonnegative number $z$, the area under the indicator of $z>t$ is $z$. Likewise, integrating $2t$ over that interval gives $z^2$:

z=\int_0^\infty \mathbf{1}_{\{z>t\}}\,dt,\qquad z^2=\int_0^\infty 2t\,\mathbf{1}_{\{z>t\}}\,dt.

Set $z=T_x$ and take expectations. Since the integrands are nonnegative, expectation and integration can be interchanged, even if an answer is infinite. The expected indicator is $P(T_x>t)={}_tp_x$, which yields the stated formulas. This argument does not require $T_x$ to have a density, and isolated points where the survival function changes do not affect the integrals.

The formulas also provide a direct test for finiteness. The mean is finite if the survival function is integrable over all durations. The second moment is finite if the survival function, after multiplication by $t$, is integrable. A tail that decreases slowly may pass the first test but fail the second.

Applying the Formulas

For a survival model given in pieces, split the integrals at the points where the formula changes. On an unbounded interval, a change of variable can often reduce the integral to standard exponential or power integrals. Keep exact expressions until the final rounding step; summing rounded interval contributions can change the last reported digit.

Worked Example: Moments Under a Changing Force of Mortality

A hypothetical life aged $x$ has constant force of mortality $0.10$ per year for the first two years of future lifetime and $0.20$ per year thereafter. Find $E[T_x]$, $E[T_x^2]$, and $\operatorname{Var}(T_x)$. Give the final values to four decimal places. The method is to integrate the survival function separately over each interval; this avoids treating the force after year two as if it applied from time zero.

The survival function must be continuous at $t=2$. It is

{}_tp_x= \begin{cases} e^{-0.10t}, & 0\leq t\leq 2,\\ e^{-0.20}e^{-0.20(t-2)}, & t>2. \end{cases}

For the mean, the first interval contributes

\int_0^2 e^{-0.10t}\,dt =\frac{1-e^{-0.20}}{0.10} =1.8126924692\ldots.

The remaining contribution is

e^{-0.20}\int_0^\infty e^{-0.20s}\,ds =\frac{e^{-0.20}}{0.20} =4.0936537654\ldots,

where $s=t-2$. Adding the unrounded expressions gives

E[T_x]=\frac{1-e^{-0.20}}{0.10}+\frac{e^{-0.20}}{0.20} =5.9063462346\ldots\text{ years} \approx 5.9063\text{ years}.

For the second moment, use $2\int_0^\infty t\,{}_tp_x\,dt$. The first interval contributes

2\int_0^2 t e^{-0.10t}\,dt =200\left[1-1.2e^{-0.20}\right] =3.5046192613\ldots.

For the tail, write $t=2+s$, so its contribution is

2e^{-0.20}\int_0^\infty (2+s)e^{-0.20s}\,ds =2e^{-0.20}\left(\frac{2}{0.20}+\frac{1}{0.20^2}\right) =70e^{-0.20} =57.3111527155\ldots.

Thus $E[T_x^2]=60.8157719767\ldots\text{ years}^2$. Squaring the unrounded mean gives $E[T_x]^2=34.8849258\ldots$, so

\operatorname{Var}(T_x) =60.8157719767\ldots-34.8849258\ldots =25.9308461\ldots\text{ years}^2 \approx 25.9308\text{ years}^2.

As a check, the survival function starts at one, decreases continuously at $t=2$, and has an exponentially decreasing tail, so both moments are finite. The mean is also greater than two years, consistent with the substantial survival beyond the first two years.

Worked Example: A Power-Tail Survival Model

Suppose a hypothetical lifetime has survival function ${}_tp_x=(1+t/2)^{-3}$ for $t\geq0$, with $t$ measured in years. Find its mean, second moment, variance, and probability of surviving more than four years. The method is substitution in the tail integrals, followed by a separate substitution into the survival function for the probability. It avoids assuming that a heavy-tailed model necessarily has infinite moments.

Let $u=1+t/2$, so $dt=2\,du$ and $t=2(u-1)$. For the mean,

E[T_x]=2\int_1^\infty u^{-3}\,du =2\left[\frac{u^{-2}}{-2}\right]_1^\infty =1\text{ year}.

For the second moment, the integrand includes the additional factor $t$:

\begin{aligned} E[T_x^2] &=2\int_0^\infty t(1+t/2)^{-3}\,dt\\ &=8\int_1^\infty (u-1)u^{-3}\,du\\ &=8\int_1^\infty (u^{-2}-u^{-3})\,du\\ &=8\left(1-\frac12\right) =4\text{ years}^2. \end{aligned}

Therefore $\operatorname{Var}(T_x)=4-1^2=3\text{ years}^2$. At four years,

P(T_x>4)={}_4p_x=(1+4/2)^{-3}=\frac{1}{27}=0.0370.

The model's power tail still decreases quickly enough for both of the displayed moment integrals to converge. The direct integral checks finiteness as well as producing the values.

Worked Example: Moments From a Bounded Survival Curve

A hypothetical survival model is ${}_tp_x=(1-t/4)^2$ for $0\leq t\leq4$, and zero for $t>4$. Find the mean, second moment, variance, and probability of surviving more than two years. The method is to integrate only over the interval where survival is positive, rather than extending the polynomial beyond its stated range.

The mean is

E[T_x]=\int_0^4(1-t/4)^2\,dt =4\int_0^1(1-u)^2\,du =4\left(\frac13\right) =\frac43\text{ years},

where $u=t/4$. For the second moment, the same substitution gives

\begin{aligned} E[T_x^2] &=2\int_0^4 t(1-t/4)^2\,dt\\ &=32\int_0^1u(1-u)^2\,du\\ &=32\left(\frac12-\frac23+\frac14\right) =\frac83\text{ years}^2. \end{aligned}

Consequently,

\operatorname{Var}(T_x)=\frac83-\left(\frac43\right)^2 =\frac{24}{9}-\frac{16}{9} =\frac89\text{ years}^2.

The survival probability at two years is ${}_2p_x=(1-2/4)^2=1/4=0.2500$. As a check, the mean $4/3$ is less than the maximum possible lifetime of four years, and the calculated variance is positive.

Moment Finiteness and Common Traps

For a power-tail model with ${}_tp_x=(1+t/b)^{-\alpha}$, where $b>0$ and $\alpha>0$, the mean integral is finite only when $\alpha>1$, and the second-moment integral is finite only when $\alpha>2$. These conditions follow from the large-$t$ behavior: the mean integrand is proportional to $t^{-\alpha}$, while the second-moment integrand is proportional to $t^{1-\alpha}$. A finite mean therefore does not guarantee a finite variance.

Common trap: The second-moment formula has a factor of $2$. For example, if ${}_tp_x=e^{-0.25t}$, omitting that factor incorrectly gives $1/(0.25)^2=16$. The correct calculation is $E[T_x^2]=2\int_0^\infty t e^{-0.25t}\,dt=2/(0.25)^2=32$. The mean is $1/0.25=4$, so the variance is $32-4^2=16$, not $32$.

Another frequent error is to use $E[T_x^2]$ as though it were the variance. The second moment is a raw moment about zero; variance measures squared deviations from the mean and must be calculated by subtracting $E[T_x]^2$. A negative result signals an arithmetic or formula error, since variance cannot be negative.

The tail-integral approach matters whenever complete future lifetime is specified through survival probabilities or a continuous survival model. It allows moments to be calculated without first finding a density, accommodates piecewise models by splitting the integral, and exposes whether an answer exists before numerical evaluation. This topic is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each model, use the survival-function integrals and retain units in your answers.

A hypothetical life has ${}_tp_x=e^{-0.25t}$. Find the mean, second moment, variance, and ${}_3p_x$. (Answer: mean $=4$ years; second moment $=32$ years squared; variance $=16$ years squared; ${}_3p_x=e^{-0.75}=0.4724$.)
A survival model is ${}_tp_x=(1+t/3)^{-4}$ for $t\geq0$. Find the mean, second moment, variance, and ${}_3p_x$. (Answer: mean $=1$ year; second moment $=3$ years squared; variance $=2$ years squared; ${}_3p_x=1/16=0.0625$.)
A hypothetical life has force $0.20$ per year for the first year and $0.40$ per year thereafter. Write its survival function and find its mean. (Answer: ${}_tp_x=e^{-0.20t}$ for $0\leq t\leq1$, and $e^{-0.20}e^{-0.40(t-1)}$ for $t>1$; mean $=(1-e^{-0.20})/0.20+e^{-0.20}/0.40=2.9532$ years.)
A model has ${}_tp_x=(1-t/5)^3$ for $0\leq t\leq5$ and zero afterward. Find the mean, second moment, variance, and probability of surviving more than two years. (Answer: mean $=1.25$ years; second moment $=2.5$ years squared; variance $=0.9375$ years squared; probability $=0.2160$.)
A survival function is ${}_tp_x=(1+t)^{-1.8}$ for $t\geq0$. Determine whether the mean and second moment are finite; calculate the mean if it exists. (Answer: mean $=1.25$ years and finite; second moment is infinite, so the variance is infinite.)

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