Triangular Survival Models - Infinity is Really Big

Studying for Exam LTAM, Part 1.5

*An example of a triangular distribution probability density function (PDF).*

An important principle in mathematical modeling is to start simple. For example, if you have a situation where a linear function matches the trend in your data, go ahead and use a linear function rather than, for example, a quadratic function.

If a simple model is not accurate enough, then you will want to use a more complicated model.

For example, a linear function is definitely not accurate enough as a representation of the distance fallen by an object under the influence of the Earth’s gravity. A quadratic function must be used in this situation.

So far in our discussions about continuous survival random variables (in this series on studying for actuarial exam LTAM), we have focused on two relatively simple models.

Uniform Distribution (De Moivre’s Law): the probability density function (PDF) is piecewise constant (and constant over the time of life).
Constant Force of Mortality (Exponential Distribution): the force of mortality is constant, resulting in a PDF and a survival function (SF) that are exponential decay functions.

In this post, we will discuss another simple model that has the potential to be somewhat more accurate for human lifetimes: triangular distributions (also see https://learnandteachstatistics.files.wordpress.com/2013/07/notes-on-triangle-distributions.pdf). These are continuous random variables whose PDFs have graphs that look like the triangle in the figure above. They have the potential to be somewhat accurate models of human lifetimes because they can model the following facts:

Not many people die at very young or very old ages (where the values of the PDF are low).
Most commonly, people die in the age-range of 50 to 90 years (where the values of the PDF are large — near the high “peak of the tent”).

One aspect of human mortality that such models do not account for is relatively high infant mortality. Very young humans (< 2 months old) are more susceptible to death than, for example, young humans between the ages of 1 and 10 years old.

Formula for the PDF in Terms of Two Parameters

Let $T_{0}$ be a continuous survival random variable with a triangular distribution. For some “limiting age” $\omega>0$ , this means we are assuming that the PDF $f_{0}(t)$ takes on positive values for $t\in (0,\omega)$ . It also means that the graph of $f_{0}$ looks like the top part (two legs) of a triangle whose base is on the horizontal axis (see the image at the start of this post).

We assume that there is some number $d$ where $f_{0}$ has a maximum value (technically this would be called the “mode” of the continuous distribution). It will be most accurate for human lifetimes to assume that $0<d<\omega$ , though the cases where $d=0$ or $d=\omega$ could be separately considered.

In order to be a PDF, the area under this graph, which is the area of the triangle, must be equal to 1. If $h$ is the height of the triangle (maximum value of $f_{0}$ ), then $1=\frac{1}{2}\omega h$ so that $h=\frac{2}{\omega}$ .

This means the slope of the first (left-most) piece of this graph is $\frac{2/\omega}{d}=\frac{2}{\omega d}$ . And the slope of the second (right-most) piece is $\frac{-2/\omega}{\omega-d}=\frac{2}{\omega(d-\omega)}.$

Therefore, over the interval $[0,\omega],$ the formula for $f_{0}$ can be written in piecewise form as $f_{0}(t)=\begin{cases}\frac{2}{\omega d}t & \mbox{if } 0\leq t\leq d \\ \frac{2}{\omega(d-\omega)}(t-\omega) & \mbox{if } d<t\leq \omega \end{cases}.$

In this formula, $\omega$ and $d$ would be labeled as “parameters”, while $t$ is the true “variable”. A better notation for this (family of) function(s) would be $f_{0,\omega,d}(t)$ . However, we will avoid such complication in our writing. If you want to use a program such as Mathematica to deal with this in its full generality, however, you will need to take the facts that $\omega$ and $d$ are arbitrary into account.

In the case of the graph at the top of this post, $\omega=120$ and $d=80$ , so the formula becomes $f_{0}(t)=\begin{cases}\frac{t}{4800} & \mbox{if } 0\leq t\leq 80 \\ -\frac{t-120}{2400} & \mbox{if } 80<t\leq 120\end{cases}$ .

Survival Function(s)

The survival function (SF) for $T_{0}$ is the complement of the cumulative distribution function (CDF): $S_{0}(t)=1-F_{0}(t)=P[T_{0}>t].$ If $0\leq t\leq d$ , we get $S_{0}(t)=1-\frac{t^{2}}{\omega d}$ .

If $d<t\leq \omega$ , the formula is $S_{0}(t)=1-\left(\frac{d}{\omega}+\displaystyle\int_{d}^{t}\frac{2}{\omega(d-\omega)}(s-\omega)\, ds\right)$ . (Think about this!)

After a bit of tricky algebra, when $d<t\leq \omega$ , you should check that this simplifies to $S_{0}(t)=\frac{(t-\omega)^{2}}{\omega(\omega-d)}$ .

Hence, $S_{0}(t)=\begin{cases}1-\frac{t^{2}}{\omega d} & \mbox{if } 0\leq t\leq d \\ \frac{(t-\omega)^{2}}{\omega(\omega-d)} & \mbox{if } d<t\leq \omega\end{cases}$ .

When $d=80$ and $\omega=120$ , the graph of $S_{0}(t)$ is as shown below. Note the inflection point at $t=d=80.$ Actually, this is a point where the function is not twice-differentiable, since $f_{0}(t)$ fails to be differentiable at $t=d=80$ (and $S_{0}'(t)=-f_{0}(t)$ ).

*A triangular distribution survival function (SF) with $d=80$ and $\omega=120$ .*

*A triangular distribution survival function (SF) with $d=80$ and $\omega=120$ .*

Assuming survival to age $x>0$ , let $T_{x}=T_{0}-x$ be the corresponding remaining-life random variable.

In the case where $0\leq x\leq d$ , the corresponding survival function for $T_{x}$ is $\,_{t}p_{x}=\frac{S_{0}(x+t)}{S_{0}(x)}=\begin{cases}\frac{\omega d-(x+t)^{2}}{\omega d-x^{2}} & \mbox{if } 0\leq t\leq d-x \\ \frac{d(x+t-\omega)^{2}}{(\omega-d)(\omega d-x^{2})} & \mbox{if } d-x<t\leq \omega-x \end{cases}$ .

In the case where $0<d<x\leq \omega$ , the corresponding survival function for $T_{x}$ is $\,_{t}p_{x}=\frac{S_{0}(x+t)}{S_{0}(x)}=\left(\frac{x+t-\omega}{x-\omega}\right)^{2}$ for $0\leq t\leq \omega-x$ .

The following animation shows the how the graph of $\,_{t}p_{x}$ changes as $x$ increases from 0 to 60, still in the case where $d=80$ and $\omega=120$ .

Note that, for any fixed $x>0$ , the graph of $\,_{t}p_{x}$ is obtained from the graph of $S_{0}(t)$ by a horizontal shift to the left by $x$ units combined with a vertical stretch by a factor of $\frac{1}{S_{0}(x)}$ . The resulting graph is, in a sense, “similar” to a piece of the original, at least with respect to these two transformations.

The graph of the conditional survival function $\,_{t}p_{x}=\frac{S_{0}(x+t)}{S_{0}(x)}$ as $x$ increases from 0 to 60, in the case where $d=80$ and $\omega=120$ .

The graph of the conditional survival function $\,_{t}p_{x}=\frac{S_{0}(x+t)}{S_{0}(x)}$ as $x$ increases from 0 to 60, in the case where $d=80$ and $\omega=120$ .

Force(s) of Mortality (Hazard Rate Function(s))

The force of mortality (FM) for a newborn is $\mu_{t}=-\frac{S_{0}'(t)}{S_{0}(t)}$ . You should take the time to check that this definition gives $\mu_{t}=\begin{cases} \frac{2t}{\omega d-t^{2}} & \mbox{if } 0<t<d \\ \frac{2}{\omega-t} & \mbox{if } d<t<\omega \end{cases}$ . Do not worry about the fact that $\mu_{t}$ is undefined at $t=0,d,\omega$ .

The FM for an individual $(x)$ of attained age $x$ is then taken to be $\mu_{x+t}=\begin{cases} \frac{2(x+t)}{\omega d-(x+t)^{2}} & \mbox{if } 0<t<d-x \\ \frac{2}{\omega-x-t} & \mbox{if } d-x<t<\omega-x \end{cases}$ .

The following animation shows the how the graph of $\mu_{x+t}$ changes as $x$ increases from 0 to 60, in the case where $d=80$ and $\omega=120$ .

Note that there is a vertical asymptote at $t=120-x$ and a point of non-differentiability at the location of the dot $t=80-x$ . In fact, as mentioned above, $\mu_{t}$ is undefined at $t=d=80$ , so $\mu_{x+t}$ is undefined at $t=80-x$ . The dot really represents a “hole” in the graph.

*The graph of the conditional force of mortality $\mu_{x+t}=-\frac{S_{0}'(x+t)}{S_{0}(x+t)}$ as $x$ increases from 0 to 60, in the case where $d=80$ and $\omega=120$ .*

*The graph of the conditional force of mortality $\mu_{x+t}=-\frac{S_{0}'(x+t)}{S_{0}(x+t)}$ as $x$ increases from 0 to 60, in the case where $d=80$ and $\omega=120$ .*

Complete Expectation of Life

The complete expectation of life is, for this example with limiting age $\omega>0$ , equal to $\stackrel{\circ}e_{x}=E[T_{x}]=\displaystyle\int_{0}^{\omega-x}\,_{t}p_{x}\, dt.$

When $d=80$ and $\omega=120$ , I will leave it to you as an exercise to determine a formula for $\stackrel{\circ}e_{x}$ (separate it into cases where $0\leq t\leq 80-x$ versus $80-x<t\leq 120-x$ ).

Below is a static graph of $\stackrel{\circ}e_{x}$ when $d=80$ and $\omega=120$ . I have also plotted the graph of $x+\stackrel{\circ}e_{x}$ .

It turns out that this second function is always an increasing function of $x$ , no matter what the situation is (no matter what continuous survival model we are considering). You may also want to see if you can figure out why this is true.

*Graphs of $\stackrel{\circ}e_{x}$ and $x+\stackrel{\circ}e_{x}$ when $d=80$ and $\omega=120$ . The graph of $x+\stackrel{\circ}e_{x}$ is always increasing, no matter what the situation is.*

*Graphs of $\stackrel{\circ}e_{x}$ and $x+\stackrel{\circ}e_{x}$ when $d=80$ and $\omega=120$ . The graph of $x+\stackrel{\circ}e_{x}$ is always increasing, no matter what the situation is.*

You should make sure you understand the real-life meaning of the complete expectation of life. In this model, it turns out that $\stackrel{\circ}e_{0}\approx 66.7$ . This is the expected lifetime of a newborn baby. Note that it is less than the mode $d=80$ . That should not be too surprising when you realize the distribution is left-skewed (you might also see if you can determine the median lifetime to be about 69.3).

As another example, for a person who is currently age 80, the model predicts an expected remaining lifetime of $\stackrel{\circ}e_{80}\approx 13.3$ .

You might also wonder whether anything special is happening for this graph at $x=d=80$ . For example: is $\stackrel{\circ}e_{x}$ differentiable there, or not? I do not know the answer off-hand. You can send me a note if you think you figure it out.

One final animation is worth doing. What happens to the graph of $\stackrel{\circ}e_{x}$ as, for example, the mode $d$ increases?

From this animation, we see that the mean remaining lifetimes are getting larger as the mode $d$ increases, but they go down at a faster rate. The derivative $\frac{d}{dx}\left(\stackrel{\circ}e_{x}\right)$ becomes more and more negative as $d$ increases. This should make intuitive sense.

As a final challenge, you might want to see if you can determine how $\frac{d}{dx}\left(\stackrel{\circ}e_{x}\right)$ depends on $d$ .

The graph of $\stackrel{\circ}e_{x}$ as $d$ increases from 60 to 100. Note that the graph gets higher and steeper. The mean remaining lifetimes are getting larger as the mode increases, but they go down at a faster rate.

Mathematica Code

Below, for your reference, are pictures of the Mathematica code for the three animations above.

Mathematica code for the animation of the graph of $\,_{t}p_{x}$ when $d=80$ and $\omega=120$ as $x$ increases from 0 to 60.

*Mathematica code for the animation of the graph of $\mu_{x+t}$ when $d=80$ and $\omega=120$ as $x$ increases from 0 to 60.*

*Mathematica code for the animation of the graph of $\mu_{x+t}$ when $d=80$ and $\omega=120$ as $x$ increases from 0 to 60.*