Posted on by Bill

The Force of Mortality (Hazard Rate Function)

Studying for Exam LTAM, Part 1.2

Photo by Markus Spiske on Unsplash

We all face potential risks to our lives every day. Daily risks include traffic accidents, lightening strikes, and poisoning. These risks can be quantified mathematically in numerous ways.

Fortunately, succumbing to a life-endangering risk on any given day has a low probability of occurrence. However, if you have people who are dependent on you and do lose your life, financial hardships for them can follow. Life insurance is meant to help to lessen the financial risks to them associated with your passing. Though it cannot take away the emotions that flow from their loss, it can help them to get back on their feet.

Actuaries often work for life insurance companies and they attempt to models these risks with mathematical functions. One such function is called the “force of mortality“, or “hazard (rate) function“. This function is related to the standard probability functions (PDFs, CDFs, and SFs) that I discussed in the post “Families of Continuous Survival Random Variables, Studying for Exam LTAM, Part 1.1“.

Mathematical Definition of the Force of Mortality

Let T_{0} be a continuous survival random variable for a life “at birth”. For any given x>0, assuming survival to age x, the corresponding “remaining life” continuous random variable T_{x} is defined by T_{x}=T_{0}-x. We have thus generated an infinite family of related continuous random variables \{T_{x}\}_{x\geq 0}.

Following standard actuarial notation, for any fixed x, we let \,_{t}p_{x}=P[T_{x}>t|T_{0}>x] and \,_{t}q_{x}=1-\,_{t}p_{x}=P[T_{x}\leq t|T_{0}>x] be the corresponding survival function (SF) and cumulative distribution function (CDF), respectively, for the life denoted by (x). The corresponding probability density function (PDF) is f_{x}(t)=\frac{d}{dt}\left(\,_{t}q_{x}\right)=-\frac{d}{dt}\left(\,_{t}p_{x}\right).

It is also customary to denote \,_{t}p_{x} by S_{x}(t) and to note that, by the formula for conditional probability, \,_{t}p_{x}=\frac{S_{0}(x+t)}{S_{0}(x)}.

The force of mortality (FM) \mu_{t} at an arbitrary value of t>0 is defined to be the relative rate of decay of the survival function S_{0}(t) at that point. In other words, \mu_{t}=-\frac{S_{0}'(t)}{S_{0}(t)}.

For a life of current age x (again, denoted by (x)) the force of mortality is assumed to be well-modeled by the function \mu_{x+t}=-\frac{S_{0}'(x+t)}{S_{0}(x+t)}=-\frac{\frac{d}{dt}\left(\,_{t}p_{x}\right)}{\,_{t}p_{x}}. Visually, this just takes the graph of \mu_{t} and shifts it to the left by x units.

By the Chain Rule, we can also write \mu_{x+t}=-\frac{d}{dt}\left(\ln(\,_{t}p_{x})\right) so that integration and exponentiation will lead to \,_{t}p_{x}=\exp\left(-\displaystyle\int_{0}^{t}\mu_{x+s}\, ds\right).

Interpretation of the Force of Mortality

These mathematical definitions are all well-and-good, but what do they mean? How should they be interpreted? Are they applicable to real-life?

Recall first the fact that a PDF gives probabilities by integration: P[a\leq T_{x}\leq b]=\displaystyle\int_{a}^{b}f_{x}(t)\, dt. If \Delta t=b-a is sufficiently small, then we obtain the approximation P[a\leq T_{x}\leq b]\approx f_{x}(t)\cdot \Delta t.

For example, suppose someone who has just turned 60 years old is wondering about her probability of dying within one month after their 70th birthday. If f_{60}(10)=0.01, then this probability can be approximated as 0.01\cdot \frac{1}{12}\approx 0.000833 (about a 1 in 1200 chance). This is a “pure” (unconditional) probability that is computed without knowing whether she will even live to be 70 or not.

The force of mortality, on the other hand, gives a conditional probability rate of death. If she assumes she lives to her 70th birthday, what then is her probability of dying within one month after that? Presumably, this probability could very well be larger because it has ignored the chances of dying between the ages of 60 and 70. If, for example, \mu_{60+10}=\mu_{70}=0.02, then the probability can be approximated as 0.02\cdot \frac{1}{12}\approx 0.00167 (about a 1 in 600 chance).

We can verify that the force of mortality is the correct function to use here by the calculations below. For the approximation in the previous paragraph to be “good”, we want the instantaneous death rate to be a limit of conditional death rates as \Delta t\rightarrow 0.

By the definition of conditional probability, P[t<T_{x}\leq t+\Delta t|T_{x}>t]=\frac{P[t<T_{x}\leq t+\Delta t]}{P[T_{x}>t]}. But P[T_{x}>t]=\,_{t}p_{x} and P[t<T_{x}\leq t+\Delta t]=\,_{t+\Delta t}q_{x}-\,_{t}q_{x}=\,_{t}p_{x}-\,_{t+\Delta t}p_{x}.

Therefore, \displaystyle\lim_{\Delta t\rightarrow 0}\frac{P[t<T_{x}\leq t+\Delta t|T_{x}>t]}{\Delta t}=-\frac{1}{\,_{t}p_{x}}\displaystyle\lim_{\Delta t\rightarrow 0}\frac{\,_{t+\Delta t}p_{x}-\,_{t}p_{x}}{\Delta t}.

But by the definition of the derivative, this is the same as -\frac{\frac{d}{dt}\left(\,_{t}p_{x}\right)}{\,_{t}p_{x}}=\mu_{x+t}.

The Force of Mortality for a Uniform Distribution (De Moivre’s Law)

We end this post with an example. We consider a continuous uniform distribution (De Moivre’s Law), as done in the post “Families of Continuous Survival Random Variables, Studying for Exam LTAM, Part 1.1“.

For a uniform distribution, there is a limiting age \omega>0 (a “drop-dead age” that a person cannot live past). For any fixed 0\leq x<\omega, the PDF of T_{x}=T_{0}-x is f_{x}(t)=\begin{cases} \frac{1}{\omega-x} & \mbox{if } 0\leq t\leq \omega-x \\ 0 & \mbox{if } t>\omega\end{cases}. The CDF is then \,_{t}q_{x}=\begin{cases} \frac{t}{\omega-x} & \mbox{if } 0\leq t\leq \omega-x \\ 1 & \mbox{if } t>\omega\end{cases} and the SF is \,_{t}p_{x}=\begin{cases} 1-\frac{t}{\omega-x} & \mbox{if } 0\leq t\leq \omega-x \\ 0 & \mbox{if } t>\omega\end{cases}.

From this we see that, for 0<t<\omega-x, we have \frac{d}{dt}\left(\,_{t}p_{x}\right)=-\frac{1}{\omega-x} so that the FM is \mu_{x+t}=\frac{1/(\omega-x)}{(\omega-x-t)/(\omega-x)}=\frac{1}{\omega-x-t}. For x=0 this is \mu_{t}=\frac{1}{\omega-t}.

The graph of \mu_{x+t}=\frac{1}{\omega-x-t}, as a function of t over 0<t<\omega-x has a vertical asymptote as t\rightarrow \omega-x from the left. This is reflecting the fact that the maximum remaining lifetime \omega-x truly is a “drop-dead time” in this model.

Taking \omega=100, the animation below shows the graph of \mu_{x+t} shifting to the left as x increases from x=0 to x=90.

The graph of \mu_{x+t}=\frac{1}{\omega-x-t} over the interval 0<t<\omega-x when \omega=100. In the animation, x is increasing from x=0 to x=90. This is the family \{\mu_{x+t}\}_{x\geq 0} of force of mortality functions for a continuous survival function with a uniform distribution. The vertical asymptote occurs at the “drop dead time” 100-x (the maximum remaining lifetime of (x)).

Here is the Mathematica code that made the animation above. Note that there is a computation which is done using Solve before Manipulate is used.

The Mathematica code for the animation above. Note that a computation involving Solve was done before using Manipulate.

We end by re-emphasizing the meaning of the force of mortality for this example. Suppose that \omega=100 and that we are considering an individual (60) (an individual of age x=60). Using the formula, the value of \mu_{60+10}=\mu_{70} is \frac{1}{100-70}=\frac{1}{30}. Therefore, if, at her 60th birthday, we make the assumption that this person lives to see her 70th birthday, she will have an approximate probability of \frac{1}{30}\cdot \frac{1}{12}\approx .00278 of dying within one month after her 70th birthday.

This can be contrasted with the unconditional probability. If, at her 60th birthday, we do not assume that she makes it to her 70th birthday, her approximate probability of dying within one month after her 70th birthday is f_{60}(10)\cdot \frac{1}{12}=\frac{1}{40}\cdot \frac{1}{12}\approx .00208.

Next: The Complete Expectation of Life, Studying for Exam LTAM, Part 1.3