Curtate Future Lifetime Random Variable

Studying for Exam LTAM, Part 1.8

When somebody asks you how old you are, how do you answer? Are you like most people? Do you give your age from your previous birthday? Or do you get extra-precise and say something like “I am 21 years, 5 months, and 1 week old”?

If you answer the way that most people do, then you already know a little bit about curate survival random variables. You have also made use of the “greatest integer” or “floor” function without even knowing it. Essentially this is a function that takes an arbitrary real number input $x$ and “rounds down” to the nearest integer less than or equal to $x$ .

In textbooks, this function is written either as $f(x)=\lfloor x\rfloor$ or $f(x)=[[x]]$ . A sampling of its values includes $f(3.1)=f(3.9)=f(3.9999)=3$ and $f(4)=4,$ while $f(-3.1)=f(-3.9)=f(-3.9999)=f(-4)=-4$ . The graph of $f$ is shown below. Open circles are included at points that are not on the graph of $f$ .

The graph of the greatest integer (floor) function $f(x)=\lfloor x\rfloor=[[x]]$ . Note that the open circles are at points which are **not** on the graph.

The graph of the greatest integer (floor) function $f(x)=\lfloor x\rfloor=[[x]]$ . Note that the open circles are at points which are **not** on the graph.

The Discrete Curtate Lifetime Random Variable

We can apply the greatest integer function to a continuous survival random variable $T_{0}$ to get a discrete random variable $K_{0}=\lfloor T_{0}\rfloor$ . If $T_{0}$ is the lifetime of a newborn, then $K_{0}$ will be the rounded-down age at death. For example, if the observed value of $T_{0}$ is 78.7591 years (about 75 years, 9 months, and 3 days), then the observed value of $K_{0}$ will be 78 years — that person did not make it to their 79th birthday.

Likewise, if we assume a newborn lives to age $x>0$ and has remaining lifetime $T_{x}=T_{0}-x$ , then $K_{x}=\lfloor T_{x}\rfloor$ is the remaining lifetime, rounded down to the nearest integer. The random variable $T_{x}$ is discrete. In general, discrete random variables take on either a finite number or a countably infinite number of values. For this situation, it can take on only non-negative integer values $0,1,2,3,\ldots.$

Whereas probabilities for continuous random variables are found by integrating probability density functions (PDFs), probabilities for discrete random variables are found by summing values of probability mass functions (PMFs).

The probability mass function for $K_{x}$ can be written symbolically as $P[K_{x}=k]$ , where $k=0,1,2,3,\ldots$ . This is a function of $k$ . It is often written as $p(k)$ or $f(k)$ .

Computing the Probability Mass Function

A question now naturally arises: how should the PMF be computed? Since $K_{x}$ is defined in terms of $T_{x}$ , the answer should be related to the distribution functions of $T_{x}$ .

The fact that $K_{x}=\lfloor T_{x}\rfloor$ means that $K_{x}=k$ is equivalent to $k\leq T_{x}<k+1$ . Therefore, $P[K_{x}=k]=P[k\leq T_{x}<k+1]$ . But, since $T_{x}$ is a continuous random variable, we can say that $P[k\leq T_{x}<k+1]=P[k<T_{x}\leq k+1]=P[T_{x}\leq k+1]-P[T_{x}\leq k]$ . This is the same as $F_{x}(k+1)-F_{x}(k)$ , where $F_{x}(t)=\,_{t}q_{x}$ is the cumulative distribution function (CDF) of $T_{x}$ . On the other hand, if $\,_{t}p_{x}=1-\,_{t}q_{x}$ is the survival function (SF) of $T_{x}$ , this also equals $\,_{k}p_{x}-\,_{k+1}p_{x}$ .

To summarize, we have just found that the difference $\,_{k}p_{x}-\,_{k+1}p_{x}$ is the PMF of $K_{x}$ . That is, we can say that $P[K_{x}=k]=\,_{k}p_{x}-\,_{k+1}p_{x}$ for $k=0,1,2,3,\ldots$ .

Alternatively, the general multiplication rule implies that $\,_{k+1}p_{x}=P[T_{x}>k+1]=P[T_{x}>k]\cdot P[T_{x+k}>1|T_{x}>k]=\,_{k}p_{x}\cdot \,_{1}p_{x+k}$ . Hence, we can also write the PMF of $K_{x}$ as $P[K_{x}=k]=\,_{k}p_{x}-\,_{k}p_{x}\cdot \,_{1}p_{x+k}=\,_{k}p_{x}\cdot (1-\,_{1}p_{x+k})=\,_{k}p_{x}\cdot \,_{1}q_{x+k}$ for $k=0,1,2,3,\ldots$ .

We should note that a pre-subscript equal to “1” is usually omitted. In other words, we usually write $p_{x+k}$ in place of $\,_{1}p_{x+k}$ and $q_{x+k}$ in place of $\,_{1}q_{x+k}$ . Hence, it is common to write the PMF as $P[K_{x}=k]=\,_{k}p_{x}-\,_{k+1}p_{x}=\,_{k}p_{x}\cdot q_{x+k}.$

This has an intuitive interpretation. If we assume that time is measured in years, the chances of $(x)$ living between $k$ and $k+1$ whole years is the probability that $(x)$ lives at least $k$ years times the probability that, once $(x)$ attains age $x+k$ , he or she dies in the next year.

Computations and Graphs for Specific Survival Models

Let us now consider the distribution of $K_{x}$ for the various specific survival models we have considered so far: 1) uniform (De Moivre’s Law), 2) exponential (constant force), 3) triangular, and 4) Gompertz-Makeham.

Uniform Lifetime (De Moivre’s Law)

The SF of $T_{x}$ is $\,_{t}p_{x}=1-\frac{t}{\omega-x}$ and the CDF of $T_{x}$ is $\,_{t}q_{x}=\frac{t}{\omega-x}$ for $0\leq x<\omega$ and $0\leq t\leq \omega-x$ .

Hence, the PMF of $K_{x}$ is $\,_{k}p_{x}-\,_{k+1}p_{x}=\frac{(k+1)-k}{\omega-x}=\frac{1}{\omega-x}$ for $k=0,1,2,\ldots,\lfloor \omega-x\rfloor-1$ . And, for $k=\lfloor \omega-x\rfloor,$ the PMF value is $\,_{\lfloor \omega-x\rfloor}p_{x}=1-\frac{\lfloor\omega-x\rfloor}{\omega-x}$ since $\,_{\lfloor \omega-x\rfloor+1}p_{x}=0$ . Note that if $\omega-x$ is an integer, then $\,_{\lfloor \omega-x\rfloor}p_{x}=0$ . Also note that $\,_{\lfloor\omega-x\rfloor}p_{x}=\frac{\omega-x-\lfloor\omega-x\rfloor}{\omega-x}<\frac{1}{\omega-x}$ since $\omega-x-\lfloor\omega-x\rfloor<1$ (think about this inequality for specific examples if that helps).

As a double-check on this, we note that if $k=0,1,2,\ldots,\lfloor \omega-x\rfloor-1$ , then $\,_{k}p_{x}\cdot q_{x+k}=\left(1-\frac{k}{\omega-x}\right)\cdot \frac{1}{\omega-x-k}=\frac{\omega-x-k}{(\omega-x)(\omega-x-k)}=\frac{1}{\omega-x}$ , as seen above. And if $k=\lfloor\omega-x\rfloor$ , then $q_{x+k}=1$ so $\,_{k}p_{x}\cdot q_{x+k}=\,_{\lfloor\omega-x\rfloor}p_{x}=\frac{\omega-x-\lfloor\omega-x\rfloor}{\omega-x}<\frac{1}{\omega-x}$ as seen above as well.

An animation of the graph of this PMF is shown below. In the animation, $x$ ranges from 0 to 40 while $\omega$ ranges from 60 to 100. This discrete distribution is “almost uniform” except for the very last probability when $k=\lfloor \omega-x\rfloor$ (except in the case where $\omega-x$ is exactly an integer, in which case the discrete distribution is exactly uniform).

The probability mass function (PMF) for $K_{x}=\lfloor T_{x}\rfloor$ when $T_{x}$ has a (continuous) uniform distribution over $[0,\omega-x]$ . This discrete PMF is (usually) ‘almost uniform’. It is exactly uniform when $\omega-x$ is an integer. In the animation $x$ ranges from 0 to 40 while $\omega$ ranges from 60 to 100.

The probability mass function (PMF) for $K_{x}=\lfloor T_{x}\rfloor$ when $T_{x}$ has a (continuous) uniform distribution over $[0,\omega-x]$ . This discrete PMF is (usually) ‘almost uniform’. It is exactly uniform when $\omega-x$ is an integer. In the animation $x$ ranges from 0 to 40 while $\omega$ ranges from 60 to 100.

Exponential Lifetime (Constant Force)

The SF of $T_{x}$ is $\,_{t}p_{x}=e^{-\lambda t}$ and the CDF of $T_{x}$ is $\,_{t}q_{x}=1-e^{-\lambda t}$ for $\lambda>0, x\geq 0$ and $t\geq 0$ . (Recall that this distribution is memoryless so that these formulas do not actually depend on $x$ .)

Therefore, the PMF of $K_{x}$ is $\,_{k}p_{x}-\,_{k+1}p_{x}=e^{-\lambda k}-e^{-\lambda (k+1)}=(1-e^{-\lambda})e^{-\lambda k}=\frac{e^{\lambda}-1}{e^{\lambda}}e^{-\lambda k}$ for $k=0,1,2,3,\ldots$ .

Double-checking this with the other formula gives $\,_{k}p_{x}\cdot q_{x+k}=e^{-\lambda k}\left(1-e^{-\lambda\cdot 1}\right)=(1-e^{-\lambda})e^{-\lambda k}$ for $k=0,1,2,3,\ldots$ .

An animation of this PMF does not vary with $x$ . Hence, we make an animation showing what happens as $\lambda$ varies instead. This means we are really looking at a bunch of different exponential distributions rather than the family of distributions for $K_{x}$ at some particular value of $\lambda$ .

In the animation, $\lambda$ varies from 0.1 to 10. Note that when $\lambda$ increases to around 5 or so, “almost all” the probability is at $k=0$ . In other words, there is almost no chance of surviving at least one more year in that situation. In fact, if $\lambda=5$ , then $P[K_{x}=0]=\,_{0}p_{x}-\,_{1}p_{x}=q_{x}=1-e^{-5}\approx 0.993262$ .

The probability mass function (PMF) for $K_{x}=\lfloor T_{x}\rfloor$ when $T_{x}$ has a (continuous) exponential distribution over $[0,\infty)$ . This discrete PMF is constant in $x$ . The animation instead shows what happens as the parameter $\lambda$ varies from 0.1 to 10. Note that the mean of the corresponding continuous distribution is $\frac{1}{\lambda}$ . Also note that once $\lambda$ gets up to about 5 or so, there is almost no chance of surviving at least 1 more year.

It is also interesting to note here that the values of $P[K_{x}=k]$ do not increase or decrease monotonically in $\lambda$ . They go up and then down.

For example, if $k=1$ , then $P[K_{x}=1]=(1-e^{-\lambda})e^{-\lambda}=e^{-\lambda}-e^{-2\lambda}$ . Taking the derivative of this with respect to $\lambda$ gives $\frac{d}{d\lambda}(e^{-\lambda}-e^{-2\lambda})=-e^{-\lambda}+2e^{-2\lambda}=e^{-\lambda}(2e^{-\lambda}-1)$ . This is equal to zero when $e^{-\lambda}=\frac{1}{2}\Leftrightarrow \lambda=\ln(2)\approx 0.693$ .

It turns out that the graph of $P[K_{x}=1]=(1-e^{-\lambda})e^{-\lambda}=e^{-\lambda}-e^{-2\lambda}$ , as a function of $\lambda$ , has a maximum value of $0.25$ at $\lambda=\ln(2)\approx 0.693$ . This is the value of $\lambda$ where the “second spike” (at $k=1$ ) has a maximum height (of $0.25$ ) in the animation above.

Triangular Lifetime

For this example, the formulas are piecewise and very complicated (see “Triangular Survival Models”). We will be content just to look at the animated graph of the PMF below. We see that it has a triangular shape to it, just like in the continuous case. In the animation $x$ varies from 0 to 20, $d$ varies from 40 to 60, and $\omega$ varies from 80 to 100.

The PMF of $K_{x}$ when $T_{x}$ has a triangular distribution. In the animation $x$ varies from 0 to 20, $d$ varies from 40 to 60, and $\omega$ varies from 80 to 100.

The PMF of $K_{x}$ when $T_{x}$ has a triangular distribution. In the animation $x$ varies from 0 to 20, $d$ varies from 40 to 60, and $\omega$ varies from 80 to 100.

Gompertz-Makeham Lifetime

In the Gompertz-Makeham model, the SF of $T_{x}$ is $\,_{t}p_{x}=\exp\left(\frac{Bc^{x}}{\ln(c)}\left(1-c^{t}\right)-At\right)$ for $t\geq 0$ .

Therefore, the PMF $\,_{k}p_{x}\cdot q_{x+k}=P[K_{x}=k]$ of $K_{x}$ is $\exp\left(\frac{Bc^{x}}{\ln(c)}\left(1-c^{k}\right)-Ak\right)-\exp\left(\frac{Bc^{x}}{\ln(c)}\left(1-c^{k+1}\right)-A(k+1)\right)$ . This formula can be rewritten in various ways. But we will not bother doing so and just look at our animation.

The PMF of $K_{x}$ when $T_{x}$ has a Gompertz-Makeham distribution. In the animation $A$ varies from .0001 to .01, $B$ varies from .0003 to .001, $c$ varies from 1.07 to 1.12, and $x$ varies from 0 to 60. It is perhaps most interesting to note the effect that $A$ has on the height of the graph for small values of $k$ : when $A$ is relatively large, there is a relatively high chance of dying young.

The PMF of $K_{x}$ when $T_{x}$ has a Gompertz-Makeham distribution. In the animation $A$ varies from .0001 to .01, $B$ varies from .0003 to .001, $c$ varies from 1.07 to 1.12, and $x$ varies from 0 to 60. It is perhaps most interesting to note the effect that $A$ has on the height of the graph for small values of $k$ : when $A$ is relatively large, there is a relatively high chance of dying young.

In the next post of this series on studying for Exam LTAM, we will explore the mean and standard deviation of $K_{x}$ for these different models.

Next: Curtate Expectation of Life