Curtate Expectation of Life - Infinity is Really Big

Studying for Exam LTAM, Part 1.9

Image by Clker-Free-Vector-Images from Pixabay

Means of random variables can be thought of as centers of mass. If the random variable is continuous, think of the region under the graph of its probability density function (PDF) as a thin sheet of metal with constant density. The mean will be at the center of mass (balance point) in the horizontal direction.

If the random variable is discrete, you can think of its probability mass function (PMF) either in terms of a graph or in terms of point masses located at the possible values of the variable. Once again, the mean will be at the center of mass in the horizontal direction.

We saw in the previous article “Curtate Future Lifetime Random Variable” that if $T_{x}$ is the continuous random variable representing remaining life of an individual $(x)$ at age $x\geq 0$ , then there is a corresponding discrete random variable $K_{x}=\lfloor T_{x}\rfloor$ representing the remaining life rounded down to the nearest whole number of years. This random variable is called the curtate future lifetime random variable. The word “curtate” basically means “shortened”.

We also saw that if $\,_{t}p_{x}$ is the survival function (SF) of $T_{x}$ and $\,_{t}q_{x}=1-\,_{t}p_{x}$ is its cumulative distribution function (CDF), then the PMF of $K_{x}$ is $P[K_{x}=k]=\,_{k}p_{x}-\,_{k+1}p_{x}=\,_{k}p_{x}\cdot q_{x+k}$ for $k=0,1,2,3,\ldots$ . We note that this is also written symbolically as $\,_{k}\lvert q_{x}$ , which is the notation for the probability that $(x)$ will die between time $k$ and $k+1$ .

The Mean of a Curtate Future Lifetime Random Variable

In general, the mean (expected value) of a discrete random variable $X$ with PMF $p(x)$ is $\mu=E[X]=\displaystyle\sum xp(x)$ , where the sum ranges over all values of $x$ where $p(x)>0$ (which is, by definition, either a finite or countably infinite set of $x$ values).

If you know the physical definition of the center of mass of “particles” (point masses) along a number line, you should see the connection: the values of $X$ are the locations of the particles and the corresponding probabilities are the masses.

Therefore, the mean of the curtate future lifetime $K_{x}$ is $E[K_{x}]=\displaystyle\sum_{k=0}^{\infty}k\left(\,_{k}p_{x}-\,_{k+1}p_{x}\right)$ . Actuaries denote this mean by $e_{x}$ and call it the curtate expectation of life. Note that we can obviously delete the first term in this sum when $k=0$ if we want. Also note that this makes sense for any value of $x\geq 0$ , not just integer values of $x$ .

A generic partial sum for this infinite series is $\displaystyle\sum_{k=0}^{n}k\left(\,_{k}p_{x}-\,_{k+1}p_{x}\right)=(\,_{1}p_{x}-\,_{2}p_{x})+(2\,_{2}p_{x}-2\,_{3}p_{x})+(3\,_{3}p_{x}-3\,_{4}p_{x})+\cdots$

$\cdots+((n-1)\,_{n-1}p_{x}-(n-1)\,_{n}p_{x})+(n\,_{n}p_{x}-n\,_{n+1}p_{x})$

This sum (partially) “telescopes” to become $\,_{1}p_{x}+\,_{2}p_{x}+\,_{3}p_{x}+\cdots+\,_{n-1}p_{x}+\,_{n}p_{x}$ . The infinite series therefore telescopes as well and we can write $e_{x}=\displaystyle\sum_{k=1}^{\infty}\,_{k}p_{x}$ . We remark that this formula can also be derived using summation-by-parts. This is something you might want to check on your own.

Note that this sum begins at $k=1$ . Contrast this with the integral beginning at 0 for the complete expectation of life $\stackrel{\circ}e_{x}=\displaystyle\int_{0}^{\infty}\,_{t}p_{x}\, dt$ .

Computations and Graphs for Specific Survival Models

As in the previous post, we now do computations and make graphs of $e_{x}$ for our specific survival models 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 $\,_{t}p_{x}=1-\frac{t}{\omega-x}$ for $0\leq t\leq \omega-x$ . Hence $e_{x}=\displaystyle\sum_{k=1}^{\lfloor \omega-x\rfloor}\left(1-\frac{k}{\omega-x}\right)=\lfloor \omega-x\rfloor-\frac{\lfloor \omega-x\rfloor(\lfloor\omega-x\rfloor+1)}{2(\omega-x)}$ , where we have used the formula for the sum of the first $\lfloor \omega-x\rfloor$ positive integers. If $\omega-x$ is an integer, this simplifies to $e_{x}=\omega-x-\frac{\omega-x+1}{2}=\frac{\omega-x}{2}-\frac{1}{2}$ .

An animated graph of $e_{x}$ is shown below as $\omega$ varies from 80 to 100. In spite of its formula involving the floor function, it is a continuous function. It turns out to be just barely below the graph of $\frac{\omega-x}{2}-\frac{1}{2}$ for all values of $x$ (and equal to it for a discrete set of values of $x$ ).

The curtate expectation of life $e_{x}$ when $T_{x}$ has a uniform distribution over $[0,\omega]$ . The value of $\omega$ varies from 80 to 100 here. This graph is continuous and is just barely below the graph of $\frac{\omega-x}{2}-\frac{1}{2}$ .

The curtate expectation of life $e_{x}$ when $T_{x}$ has a uniform distribution over $[0,\omega]$ . The value of $\omega$ varies from 80 to 100 here. This graph is continuous and is just barely below the graph of $\frac{\omega-x}{2}-\frac{1}{2}$ .

As a check (but not a proof) of the continuity of this function, let us consider an example with $\omega=100$ . We note that, for example, $e_{50}=\frac{100-50}{2}-\frac{1}{2}=24.5$ . To be continuous at $x=50$ , we require that the one-sided limits $\displaystyle\lim_{x\rightarrow 50-}e_{x}=\displaystyle\lim_{x\rightarrow 50+}e_{x}=24.5$ as well.

For the left-hand limit, we can consider $49<x<50$ so that $\lfloor \omega-x\rfloor=\lfloor 100-x\rfloor=50$ . The limit is then $\displaystyle\lim_{x\rightarrow 50-}\left(50-\frac{50\cdot 51}{2(100-x)}\right)=50-\frac{51}{2}=50-25.5=24.5$ .

For the right-hand limit, we can consider $50<x<51$ so that $\lfloor \omega-x\rfloor=\lfloor 100-x\rfloor=49$ . The limit is then $\displaystyle\lim_{x\rightarrow 50+}\left(49-\frac{49\cdot 50}{2(100-x)}\right)=49-\frac{49}{2}=49-24.5=24.5$ .

Recall also that the complete expectation of life in this situation is $\stackrel{\circ}e_{x}=\frac{\omega-x}{2}$ . Therefore, $e_{x}\approx \stackrel{\circ}e_{x}-\frac{1}{2}$ . This should make intuitive sense. In fact, this approximation is typically pretty good no matter what model we consider.

Exponential Lifetime (Constant Force)

The SF of $T_{x}$ is $\,_{t}p_{x}=e^{-\lambda t}$ for $\lambda>0$ and $t\geq 0$ . Hence $e_{x}=\displaystyle\sum_{k=1}^{\infty}e^{-\lambda k}$ (don’t get confused by the different uses of the symbol $e$ !). This is a geometric series with first term $e^{-\lambda}$ and common ratio $e^{-\lambda}$ . Since $|e^{-\lambda}|<1$ when $\lambda>0$ , we can say that $e_{x}=\frac{e^{-\lambda}}{1-e^{-\lambda}}=\frac{1}{e^{\lambda}-1}$ .

In other words, $e_{x}$ is a constant function for this example. This should not be surprising because exponential distributions are memoryless. An animated graph of $e_{x}$ would show a horizontal line moving downward toward zero (at a slower and slower rate) as $\lambda>0$ increases.

Recall that, for this model, the complete expectation of life is $\stackrel{\circ}e_{x}=\frac{1}{\lambda}$ . Therefore, $\stackrel{\circ}e_{x}-e_{x}=\frac{e^{\lambda}-1-\lambda}{\lambda(e^{\lambda}-1)}$ . Using the Taylor series $e^{\lambda}=1+\lambda+\frac{\lambda^{2}}{2!}+\frac{\lambda^{3}}{3!}+\cdots$ , we see that when $\lambda>0$ is small, $\stackrel{\circ}e_{x}-e_{x}\approx \frac{\lambda^{2}/2}{\lambda\cdot \lambda}=\frac{1}{2}$ . This is consistent with the approximation $e_{x}\approx \stackrel{\circ}e_{x}-\frac{1}{2}$ (more commonly written $\stackrel{\circ}e_{x}\approx e_{x}+\frac{1}{2}$ ) from the previous example.

On the other hand, this approximation gets less accurate as $\lambda>0$ gets large. Since both $\stackrel{\circ}e_{x}\rightarrow 0$ and $e_{x}\rightarrow 0$ as $\lambda\rightarrow \infty$ , it follows that $\stackrel{\circ}e_{x}-e_{x}\rightarrow 0$ as $\lambda\rightarrow \infty$ as well.

Triangular Lifetime

As in the previous post, formulas for this model are so complicated they are probably not worth writing down. But, we can once again make an animated graph.

As you should expect by now, the graph is not that much different than the graph of $\stackrel{\circ}e_{x}$ found in the post “Triangular Survival Models”. In fact, we continue see the approximation $\stackrel{\circ}e_{x}\approx e_{x}+\frac{1}{2}$ hold.

The curtate expectation of life $e_{x}$ when $T_{x}$ has a triangular distribution over $[0,\omega]$ with mode at $x=d$ . The value of $d$ varies from 30 to 50 and the value of $\omega$ varies from 80 to 100.

The curtate expectation of life $e_{x}$ when $T_{x}$ has a triangular distribution over $[0,\omega]$ with mode at $x=d$ . The value of $d$ varies from 30 to 50 and the value of $\omega$ 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 $t\geq 0$ .

We will once again be content with an animated graph.

The curtate expectation of life $e_{x}$ when $T_{x}$ has a Gompertz-Makeham distribution. The value of $A$ varies from .0001 to .01, the value of $B$ varies from .0003 to .001, and the value of $c$ varies from 1.07 to 1.12.

The curtate expectation of life $e_{x}$ when $T_{x}$ has a Gompertz-Makeham distribution. The value of $A$ varies from .0001 to .01, the value of $B$ varies from .0003 to .001, and the value of $c$ varies from 1.07 to 1.12.

Second Moment, Variance, and Standard Deviation

The second moment of a discrete random variable $X$ with PMF $p(x)$ is defined to be $E[X^{2}]=\displaystyle\sum x^{2}p(x)$ . Once again, the sum is over all values of $x$ where $p(x)>0$ .

For $K_{x}$ , this can be written as $E[K_{x}^{2}]=\displaystyle\sum_{k=0}^{\infty}k^{2}\left(\,_{k}p_{x}-\,_{k+1}p_{x}\right)$ . By thinking about partial sums again, you should check that this can be rewritten as $E[K_{x}^{2}]=2\left(\displaystyle\sum_{k=1}^{\infty}k\,_{k}p_{x}\right)-e_{x}$ . Because of subtracting $e_{x}$ , this has a different form than the corresponding second moment for $T_{x}$ , which is $E[T_{x}^{2}]=2\displaystyle\int_{0}^{\infty}t\,_{t}p_{x}\, dt$ .

The variance of $K_{x}$ is then $\mbox{Var}(K_{x})=E[K_{x}^{2}]-(e_{x})^{2}$ and the standard deviation is $\sigma_{x}=\sqrt{\mbox{Var}(K_{x})}$ .

We could also attempt to interpret these visually as we have done in previous posts. The interpretations would be similar. In particular, we would once again observe the validity of the rule of thumb that: almost all the probability is within 2 standard deviations of the mean $e_{x}\pm 2\sigma_{x}$ .

Next: Transformed Power Function Survival Models

2 Replies to “Curtate Expectation of Life”

Katie says:

May 3, 2019 at 3:04 pm

This is such an awesome post!! I LOVE your blog. It is so interesting, helpful, and well put together!! Thank you for all the work you put into your YouTube channel and blog so that others can learn 🙂 I can’t wait to see the future content you create!!
1. Bill says:
  
  May 5, 2019 at 8:13 pm
  
  Thanks Katie. I appreciate that you are interested in it find it helpful. Make sure to let others know about it!

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