Constant Force of Mortality (Exponential Distribution)

Studying for Exam LTAM, Part 1.4

Survival random variables apply to other situations besides the lifetimes of people. A common example in textbooks is the lifetime of a light bulb.

Modern-day light bulb technology seems to be much better than it was even thirty years ago. I think this is especially true with regard to car headlights. My experience back in the 1980’s was that a headlight on a car would go out every 6-12 months or so. It was very common to see vehicles with only one headlight working.

In fact, one time I went to work for a 4-10 pm shift and when I started my car to head home, I realize to my horror that both headlights were out. Fortunately, my high-beams still worked, but of course other people flashed their high beams back at me the whole way home. It was hard to overcome the instinct to keep saying “sorry” to them in spite of their not being able to hear me.

Light bulbs do seem to fail at somewhat random times. How has the random lifetimes of light bulbs been mathematically modeled with probability? While it is not guaranteed to be the most accurate model, the one with a constant force of mortality (hazard rate function) is simple and has been used by engineers. This is also referred to as an exponential distribution, for reasons we will see below.

Review of Survival Random Variables

Let $T_{0}$ be a continuous survival random variable representing the age-at-failure (death) of a “newborn”. In this case, the “newborn” is a newly made light bulb — or, at least, a light bulb that has just been turned on for the first time. Let $S_{0}(t)=P[T_{0}>t]$ be the survival function of $T_{0}.$ The corresponding probability density function is $f_{0}(t)=-S_{0}'(t)$ .

If we assume survival to age $x>0,$ we can define $T_{x}=T_{0}-x$ to be the remaining-lifetime random variable. The corresponding survival function for $T_{x}$ is then a conditional probability $\,_{t}p_{x}=P[T_{x}>t|T_{0}>x]=P[T_{0}>x+t|T_{0}>x]=\frac{S_{0}(x+t)}{S_{0}(x)}$ and the PDF is $f_{x}'(t)=-\frac{d}{dt}\left(\,_{t}p_{x}\right).$

The complete expectation of life is the mean of $T_{x}$ , which we denote by $\stackrel{\circ}e_{x}$ . Under certain reasonable assumptions, it can be shown that $\stackrel{\circ}e_{x}=\displaystyle\int_{0}^{\infty}tf_{x}(t)\, dt=\displaystyle\int_{0}^{\infty}\,_{t}p_{x}\, dt.$

We have seen another way to describe all this: by using the force of mortality (hazard rate function). By definition, it is $\mu_{t}=-\frac{S_{0}'(t)}{S_{0}(t)},$ the relative rate of decay of the survival function of a newborn. We also use $\mu_{x+t}=-\frac{\frac{d}{dt}\left(\,_{t}p_{x}\right)}{\,_{t}p_{x}}=-\frac{S_{0}'(x+t)}{S_{0}(x+t)}$ as the corresponding force of mortality for $T_{x}$ .

If we know the force of mortality, we can also find the survival function. First, use the Chain Rule to rewrite the equation $\mu_{x+t}=-\frac{\frac{d}{dt}\left(\,_{t}p_{x}\right)}{\,_{t}p_{x}}$ as $\mu_{x+t}=-\frac{d}{dt}\left(\ln(\,_{t}p_{x})\right).$ A bit of algebra and integration then leads to the conclusion that $\,_{t}p_{x}=\exp\left(-\displaystyle\int_{0}^{t}\mu_{x+s}\, ds\right)=e^{-\int_{0}^{t}\mu_{x+s}\, ds}.$

Technically-speaking, all this machinery is best viewed from the perspective of a newborn. If we assume, at “birth”, the newborn will survive to age $x,$ then $T_{x}$ models the remaining lifetime. However, it is common to make the assumption that $T_{x}$ models the remaining lifetime of an entity (person, light bulb, etc..) that is currently age $x$ . This certainly makes the modeling simpler, if nothing else.

The model we look at in the next section has another simplifying feature. It truly does not depend on the assumed attained age $x$ at all!

Constant Force of Mortality (Exponential Decay of Survival)

In survival modeling, it is sometimes best to take the force of mortality as the starting point. Whereas a uniform distribution (De Moivre’s Law) starts with a simple probability density function (a constant function), here we start with a simple force of mortality.

So assume that $\mu_{t}=\lambda$ , a positive constant. This would directly imply that $\mu_{x+t}=\lambda$ as well. So we already see the independence of this model from the value of $x$ .

This can also be seen by computing the survival function $\,_{t}p_{x}.$ First note that $\displaystyle\int_{0}^{t}\mu_{x+s}\, ds=\displaystyle\int_{0}^{t}\lambda\, ds=\lambda t,$ which is already independent of $x$ . Then $\,_{t}p_{x}=e^{-\lambda t},$ which of course is also independent of $x$ .

Since $\,_{t}p_{x}=P[T_{x}>t|T_{0}>x]=P[T_{0}>t]$ , what we have here is an illustration of the fact that this model is “memoryless“. The conditional probability of living another $t$ units of time, based on the assumption that $x$ units of time has already been lived, is independent of $x$ . It also equals the unconditional probability of living at least $t$ units of time.

We see that the conditional PDF in this situation is $f_{x}(t)=-\frac{d}{dt}\left(\,_{t}p_{x}\right)=\lambda e^{-\lambda t}$ and is also independent of $x$ , as it should be. The fact that the PDF is an exponential decay is the reason this distribution is most commonly referred to as an exponential distribution.

The mean (complete expectation of life) in this situation is independent of $x$ and equals the constant function $\stackrel{\circ}e_{x}=\displaystyle\int_{0}^{\infty}e^{-\lambda t}\, dt=\frac{1}{\lambda}.$ Because of this, if we label the mean $\mu$ (don’t confuse this with the force of mortality), the PDF is often represented as $f_{x}(t)=\frac{1}{\mu}e^{-t/\mu}$ instead.

In all of this, there is, of course, a dependence on the parameter $\lambda=\frac{1}{\mu}.$ The animation below shows how the force of mortality (FM), in red, and survival function (SF), in blue, change as $\lambda$ increases from 0.1 to 2 (making the mean decrease from 10 to 0.5).

The graph of $\mu_{x+t}=\lambda$ (red) and $\,_{t}p_{x}=e^{-\lambda t}$ (blue) as $\lambda$ increases from 0.1 to 2. The black dot is at the mean $\stackrel{\circ}e_{x}=\frac{1}{\lambda}$ and decreases from 10 to 0.5. Note that everything is independent of $x,$ illustrating that this distribution is ‘memoryless’.

The graph of $\mu_{x+t}=\lambda$ (red) and $\,_{t}p_{x}=e^{-\lambda t}$ (blue) as $\lambda$ increases from 0.1 to 2. The black dot is at the mean $\stackrel{\circ}e_{x}=\frac{1}{\lambda}$ and decreases from 10 to 0.5. Note that everything is independent of $x,$ illustrating that this distribution is ‘memoryless’.

The Mathematica code for this animation is shown below.

Mathematica code for the animation above

Another Graphical Perspective on the Memoryless Property

Recall that $\,_{t}p_{x}=\frac{S_{0}(x+t)}{S_{0}(x)}$ . It follows from this that, for any any fixed $x>0$ , the graph of $\,_{t}p_{x}$ can be obtained from the graph of $S_{0}(t)$ by the following steps.

1) Shift the graph of $S_{0}(t)$ to the left by $x$ units to obtain the graph of $S_{0}(x+t)$ .

2) Vertically stretch the graph of $S_{0}(x+t)$ by a factor of $\frac{1}{S_{0}(x)}$ to obtain the graph of $\,_{t}p_{x}=\frac{S_{0}(x+t)}{S_{0}(x)}$ .

When a survival random variable $T_{0}$ is memoryless, that means that $\,_{t}p_{x}=S_{0}(t)$ for all $x>0$ . In other words, no matter the value of $x>0,$ the combined graphical two transformations above produce no change in the graph of $S_{0}(t)$ .

This means that the graph of $S_{0}(t)$ is, in a sense, “similar” to itself; at least via these transformations. Any piece of the graph, basically, “looks just like” the whole graph. We can illustrate this with the animation further below.

The piece of the graph to the right and underneath the red lines “looks just like” the whole with respect to these two transformations (the values of $x$ and $\lambda$ here are $x=1$ and $\lambda=1$ ). As the animation parameter $\alpha$ increases, we see the effect of the transformations described in steps 1 and 2 above. The end result is the same as the original whole graph.

Note that the brown horizontal lines all have the same length, representing the horizontal shift from step 1. On the other hand, the vertical orange lines do not all have the same length. Instead, they represent the constant (multiplicative) scaling factor in step 2. Their lengths are all being multiplied by the same constant.

Taking $x=1$ and $\lambda=1,$ this shows the two-step transformation of $S_{0}(t)=e^{-t}$ to $S_{0}(1+t)=e^{-(t+1)}=e^{-1}e^{-t}$ to $\,_{t}p_{x}=\frac{1}{S_{0}(1)}S_{0}(1+t)=\frac{1}{e^{-1}}e^{-1}e^{-t}=e^{-t}=S_{0}(t)$ . In other words, under this two-step transformation, the graph to the right of $x=1$ is “similar to” the whole graph. This is true for this example no matter what $x>0$ is.

Taking $x=1$ and $\lambda=1,$ this shows the two-step transformation of $S_{0}(t)=e^{-t}$ to $S_{0}(1+t)=e^{-(t+1)}=e^{-1}e^{-t}$ to $\,_{t}p_{x}=\frac{1}{S_{0}(1)}S_{0}(1+t)=\frac{1}{e^{-1}}e^{-1}e^{-t}=e^{-t}=S_{0}(t)$ . In other words, under this two-step transformation, the graph to the right of $x=1$ is “similar to” the whole graph. This is true for this example no matter what $x>0$ is.

The Mathematica code for this animation is shown below.

Applications of the Distribution

Intuitively, since the force of mortality is constant, we expect that the “threat of failure” is consistent across the lifetime of the individual (person, light bulb, etc..). We see that this results in a memoryless distribution for which the PDF of the remaining lifetime of an individual $(x)$ at age $x$ is $f_{x}(t)=\lambda e^{-\lambda t}$ , an exponential decay function.

Two implications of this last fact include:

The “majority” of deaths/failures occur at relatively “early” ages.
Every once-in-a-while, an individual will “live” (not fail) for a very long time.

Implication 1 arises from the fact that the “majority” of the area under the graph of the exponential function $f_{x}(t)=\lambda e^{-\lambda t}$ occurs right away, when $t$ is “small”. Implication 2 arises from the fact that $f_{x}(t)>0$ for all $t$ , even though it is “very small” when $t$ is “very large”.

Certainly these properties are not characteristic of human lifetimes. Are they characteristic of lifetimes of light bulbs?

Personally, I have not done any experiments to confirm the answer to this one way or the other. If I had to guess, I’d say that this distribution is not a perfect model of the lifetimes of light bulbs.

So are there any real-life applications of this distribution? Yes. Technically-speaking, it will be a “perfect” model for “wait times” of a so-called Poisson process. This arises when you are counting the number of times some event occurs over a fixed measure of time, under some technical hypotheses that we will not get into.

If $T$ represents the “wait time” between one occurrence of the event and the next in a Poisson process, then $T$ will have an exponential distribution. In other words, $T$ will be a “survival” random variable with a constant force of mortality.

Real-life situations where people have attempted to apply this include: wait times between hurricanes (of any given strength), wait times between arrivals in a line (for example, of people at a ticket counter), and wait times between phone calls.

Do life insurance actuaries ever use this distribution? Yes, though they mostly use it for purposes of interpolation within a given year of the lifetime of an individual person. Over such a short time period, it seems reasonable to assume the “threat of death” remains basically constant.