Curtate Future Lifetime Moments from Survival Probabilities

Why Survival Probabilities Give Curtate Lifetime Moments

A life table or a sequence of annual survival probabilities describes how many years a life aged $x$ is expected to complete. Those probabilities can be used to find moments of curtate future lifetime without first constructing every probability of death in an exact year. The key is to recognize that a survival probability is a tail probability: it tells us the chance that curtate future lifetime is at least a given integer.

This tutorial focuses on the mean and second moment of curtate future lifetime, and on the variance obtained from them. The formulas use integer-year survival probabilities only. They do not require an assumption about the timing of death within a year.

Definition: For a life aged $x$, the curtate future lifetime $K_x$ is the number of complete years lived after age $x$. If $T_x$ is complete future lifetime, then $K_x=\lfloor T_x\rfloor$. Thus $K_x$ is a nonnegative integer-valued random variable. Its integer-year survival probabilities are ${}_kp_x=P(T_x\geq k)=P(K_x\geq k)$, for integers $k\geq 0$, with ${}_0p_x=1$.

For example, if death occurs 2.4 years after age $x$, then $K_x=2$: two full years were completed. The event $K_x\geq 2$ is survival for at least two years, and its probability is ${}_2p_x$. In general, ${}_kp_x$ is not the probability that death occurs during year $k$; it is the probability of surviving through $k$ years.

Tail-Sum Formulas

For any nonnegative integer-valued random variable $K$, the event $K\geq k$ contributes one unit to $K$ for each integer $k$ from 1 through $K$. Therefore $K$ can be written as a sum of indicator variables:

K=\sum_{k=1}^{\infty} \mathbf{1}_{\{K\geq k\}}.

Taking expectations gives the tail-sum formula for the mean. To obtain the second moment, note that $K^2=1+3+5+\cdots +(2K-1)$, because the sum of the first $K$ positive odd integers is $K^2$. The $k$th odd integer is included exactly when $K\geq k$.

Theorem: If $K_x$ is nonnegative and integer-valued, then $$ E[K_x]=\sum_{k=1}^{\infty}{}_kp_x \qquad\text{and}\qquad E[K_x^2]=\sum_{k=1}^{\infty}(2k-1){}_kp_x. $$ The sums may be finite when survival ends by a finite duration. For an unbounded lifetime, the formulas apply as nonnegative series; the corresponding moment is finite when its series converges.

The variance follows from the standard identity $\operatorname{Var}(K_x)=E[K_x^2]-(E[K_x])^2$. These formulas are useful because they use the survival probabilities directly, rather than requiring a separate calculation of every exact-year probability of death.

Why the Tail-Sum Formulas Hold

The justification is an application of indicator variables and linearity of expectation. For the mean, the indicator $\mathbf{1}_{\{K_x\geq k\}}$ equals 1 if at least $k$ years are completed and 0 otherwise, so its expectation is $P(K_x\geq k)={}_kp_x$. Summing the indicators counts exactly the number of completed years.

E[K_x] =E\!\left[\sum_{k=1}^{\infty}\mathbf{1}_{\{K_x\geq k\}}\right] =\sum_{k=1}^{\infty}P(K_x\geq k) =\sum_{k=1}^{\infty}{}_kp_x.

For the second moment, the same count uses the odd numbers instead of ones. If $K_x=n$, then the indicators are 1 for $k=1,\ldots,n$, and the weighted sum is $1+3+\cdots +(2n-1)=n^2$. Taking expectations therefore gives the second tail-sum formula:

E[K_x^2] =E\!\left[\sum_{k=1}^{\infty}(2k-1)\mathbf{1}_{\{K_x\geq k\}}\right] =\sum_{k=1}^{\infty}(2k-1){}_kp_x.

For a finite maximum lifetime, these are finite sums and linearity of expectation is immediate. For an unbounded lifetime, the nonnegative summands allow the corresponding infinite sums to be taken as limits of finite partial sums. A finite mean or second moment still requires convergence of its series.

Calculating Moments from a Finite Survival Table

Suppose the available survival probabilities are ${}_1p_x,\ldots,{}_mp_x$, and ${}_{m+1}p_x=0$. Then $K_x$ cannot exceed $m$, and the sums stop at $m$. A useful calculation layout is to list each $k$, its survival probability, and the second-moment weight $2k-1$.

Worked Example: Moments from a Hypothetical Annual Survival Table

A hypothetical mortality model for a life aged $x$ gives ${}_1p_x=0.80$, ${}_2p_x=0.60$, ${}_3p_x=0.30$, ${}_4p_x=0.10$, and ${}_5p_x=0$. Find $E[K_x]$, $E[K_x^2]$, and $\operatorname{Var}(K_x)$. No interest rate or within-year mortality assumption is needed, because this calculation concerns only the integer-valued lifetime.

Method. Apply the two tail-sum formulas. Since ${}_5p_x=0$, the nonzero sums end at $k=4$.

Mean. Add the four nonzero survival probabilities:

E[K_x]=0.80+0.60+0.30+0.10=1.80\text{ years}.

Second moment. The weights for $k=1,2,3,4$ are $1,3,5,7$, respectively:

E[K_x^2] =1(0.80)+3(0.60)+5(0.30)+7(0.10) =0.80+1.80+1.50+0.70 =4.80\text{ years}^2.

Variance. Square the mean and subtract it from the second moment:

\operatorname{Var}(K_x)=4.80-(1.80)^2=4.80-3.24=1.56\text{ years}^2.

Check. The implied probabilities of $K_x=0,1,2,3,4$ are $0.20,0.20,0.30,0.20,0.10$, obtained by differencing successive survival probabilities. They sum to 1. Their weighted mean is $0(0.20)+1(0.20)+2(0.30)+3(0.20)+4(0.10)=1.80$, and their weighted second moment is $0+1(0.20)+4(0.30)+9(0.20)+16(0.10)=4.80$. This confirms both moments. The tail-sum method avoids mistakenly treating survival probabilities as exact-year probabilities.

Using Life-Table Counts

A life table gives the number $\ell_y$ of lives surviving to exact age $y$. For a life aged $x$, the integer-year survival probability is ${}_kp_x=\ell_{x+k}/\ell_x$, provided $\ell_x>0$. The denominator matters: life-table counts are not themselves probabilities, so each future count must be divided by the starting count.

Worked Example: Converting Life-Table Counts to Moments

A hypothetical life table has $\ell_{60}=1000$, $\ell_{61}=900$, $\ell_{62}=720$, $\ell_{63}=360$, $\ell_{64}=120$, and $\ell_{65}=0$. For a life aged 60, calculate the mean and variance of $K_{60}$. The relevant model is the life-table survival model; no interest or fractional-year assumption is required.

Step 1: Convert counts to survival probabilities. Divide each count by $\ell_{60}=1000$:

{}_1p_{60}=0.90,\quad {}_2p_{60}=0.72,\quad {}_3p_{60}=0.36,\quad {}_4p_{60}=0.12,\quad {}_5p_{60}=0.

Step 2: Find the mean. Add the four nonzero probabilities:

E[K_{60}]=0.90+0.72+0.36+0.12=2.10\text{ years}.

Step 3: Find the second moment. Apply weights $1,3,5,7$:

E[K_{60}^2] =0.90+3(0.72)+5(0.36)+7(0.12) =0.90+2.16+1.80+0.84 =5.70\text{ years}^2.

Step 4: Find the variance.

\operatorname{Var}(K_{60})=5.70-(2.10)^2=5.70-4.41=1.29\text{ years}^2.

Check. The implied probabilities of $K_{60}=0,1,2,3,4$ are $0.10,0.18,0.36,0.24,0.12$. They sum to 1. Their mean is $0.18+0.72+0.72+0.48=2.10$; their second moment is $0.18+1.44+2.16+1.92=5.70$. The results agree with the tail sums.

Infinite Survival Sequences

Some models give nonzero survival probabilities at every finite duration. In that case, the tail-sum formulas remain useful, but the series must be evaluated or shown to converge. A common tractable structure is a constant conditional annual survival probability after a specified duration. This creates a geometric tail in ${}_kp_x$.

For $0\leq r<1$, the geometric series and its first-moment counterpart are

\sum_{j=0}^{\infty}r^j=\frac{1}{1-r}, \qquad \sum_{j=0}^{\infty}jr^j=\frac{r}{(1-r)^2}.

Consequently, a tail with survival probabilities proportional to $r^j$ can be summed for both moments. The condition $r<1$ ensures convergence; if the annual survival probability were 1 indefinitely, the lifetime would not have finite moments under that tail model.

Worked Example: A Geometric Survival Tail

A hypothetical life aged $x$ has ${}_1p_x=0.90$ and ${}_2p_x=0.72$. From year 2 onward, the conditional probability of surviving each additional year is $r=0.75$, independent of the number of years already survived beyond year 2. Find the mean and variance of $K_x$. This is an annual survival model with a geometric tail; there is no interest assumption.

Step 1: Write the tail probabilities. For $k\geq2$, each additional year multiplies survival by $0.75$:

{}_kp_x=0.72(0.75)^{k-2},\qquad k\geq2.

Step 2: Calculate the mean. Separate the first-year term, then sum the geometric tail:

\begin{aligned} E[K_x] &=0.90+\sum_{k=2}^{\infty}0.72(0.75)^{k-2}\\ &=0.90+0.72\sum_{j=0}^{\infty}(0.75)^j\\ &=0.90+\frac{0.72}{1-0.75} =0.90+2.88=3.78\text{ years}. \end{aligned}

Step 3: Calculate the second moment. Put $j=k-2$, so the weight $2k-1$ becomes $2j+3$:

\begin{aligned} E[K_x^2] &=0.90+\sum_{k=2}^{\infty}(2k-1)0.72(0.75)^{k-2}\\ &=0.90+0.72\sum_{j=0}^{\infty}(2j+3)(0.75)^j\\ &=0.90+0.72\left( 2\frac{0.75}{(1-0.75)^2} +3\frac{1}{1-0.75}\right)\\ &=0.90+0.72(24+12) =0.90+25.92=26.82\text{ years}^2. \end{aligned}

Step 4: Calculate the variance.

\operatorname{Var}(K_x)=26.82-(3.78)^2=26.82-14.2884=12.5316\text{ years}^2.

Check. The tail begins with ${}_2p_x=0.72$, and its first terms are $0.72,0.54,0.405,\ldots$. The geometric sum of this tail is $0.72/(1-0.75)=2.88$, matching its contribution to the mean. The weighted tail sum is 25.92, so adding the year-1 contributions gives the stated second moment. The common indexing trap is to use $0.72(0.75)^{k}$ instead of $0.72(0.75)^{k-2}$, which incorrectly shifts the starting probability away from ${}_2p_x=0.72$.

Common Mistake: Squaring the Tail Index

A tempting but incorrect approach is to multiply ${}_kp_x$ by $k^2$ and add. That calculation treats each survival probability as though it were the probability of the exact value $K_x=k$. It is not: ${}_kp_x$ is the probability of all values $K_x\geq k$. The correct second-moment weight is $2k-1$, because each possible lifetime $n$ accumulates the first $n$ odd integers to make $n^2$.

Common mistake and correction: For the finite table in the first example, the incorrect calculation would be $$ \sum_{k=1}^{4}k^2{}_kp_x =0.80+4(0.60)+9(0.30)+16(0.10) =0.80+2.40+2.70+1.60=7.50. $$ This is not $E[K_x^2]$. The correct tail-sum calculation is $$ \sum_{k=1}^{4}(2k-1){}_kp_x =0.80+1.80+1.50+0.70=4.80. $$

Another frequent error is to omit the variance subtraction. The tail-sum formula gives $E[K_x^2]$, not the variance itself. Always calculate $E[K_x]$ and $E[K_x^2]$ first, then use $\operatorname{Var}(K_x)=E[K_x^2]-(E[K_x])^2$.

Key takeaway: For curtate future lifetime, add the integer-year survival probabilities to obtain the mean, and weight them by successive odd integers to obtain the second moment. Convert life-table counts to probabilities first, and use the variance identity only after both moments are calculated.

Check Your Understanding

For each problem, use integer-year survival probabilities and show the tail-sum weights before calculating. Answers are provided for checking.

A hypothetical life has ${}_1p_x=0.85$, ${}_2p_x=0.65$, ${}_3p_x=0.25$, and ${}_4p_x=0$. Find $E[K_x]$, $E[K_x^2]$, and $\operatorname{Var}(K_x)$. (Answer: $1.7500$ years; $4.0500$ years squared; variance $0.9875$ years squared.)
A hypothetical life table gives $\ell_x=800$, $\ell_{x+1}=600$, $\ell_{x+2}=300$, and $\ell_{x+3}=0$. Find the mean and variance of $K_x$. (Answer: mean $1.1250$ years; variance $0.6094$ years squared.)
A hypothetical life has ${}_1p_x=0.80$, ${}_2p_x=0.60$, and conditional annual survival probability $0.50$ for each additional year after year 2. Find $E[K_x]$, $E[K_x^2]$, and $\operatorname{Var}(K_x)$. (Answer: $2.0000$ years; $6.8000$ years squared; variance $2.8000$ years squared.)
A hypothetical model has ${}_1p_x=0.90$, ${}_2p_x=0.60$, ${}_3p_x=0.30$, and ${}_4p_x=0$. Find the probability that $K_x=2$, then calculate $E[K_x]$ and $E[K_x^2]$. (Answer: $P(K_x=2)=0.30$; mean $1.8000$ years; second moment $4.2000$ years squared.)

What You'll Learn