# Intensity

Intensities are extremely useful in the insurance industry. In this post I will introduce intensities in the setting of pricing out a simple life insurance.

## Life insurance

Take the most simple life insurance: We have to pay $1$ to the customer if he/she dies before the age of $n$ years old.

When tasked with coming up with a price for this insurance, we want to know how likely the customer is to die before reaching the age of $n$ years. When we know that, and a few other things like how much interest we can accumulate in the time span and our expenses etc. we are ready to price the insurance.

We hopefully have access to some statistics that will tell us at what age people normally die, allowing us to figure out a density function for the distribution of life lengths of new-borns. Lets call it $f(t)$. Let $X$ be the time of death.

$$Pr[X\leq n] = \int_0^n f(t) dt$$

This is all well and good, but often the customer will not be $0$ years old when taking up the insurance, and as such should expect a lower price since he/she can promise not to die before xers current age, call it $a$. So really we then want to know:

$$Pr[X\leq n | X>a] =\frac {Pr[a < X\leq n]}{Pr[X>a]} = {\int_a^n f(t) dt}/{\int_0^a 1 - f(t) dt}$$

This is not too bad, but it will be a lot cleaner when using intensities.

## Cumulative functions

To avoid messing with too many integrals lets define $F(t)=\Pr[X\leq t]$ and $\bar F(t)=\Pr[X>t]=1-F(t)$. Actuarians call these the mortality and survival functions. We can restate the probability above as

$$\Pr[X\leq n | X>a]=\int_a^n \bar F(t|X>a) f(t|X>t)dt=\frac{\bar{F}(a+t)}{\bar F(a)}\frac{f(t)}{\bar(F)(t)}$$

## Intensities

Lets start with defining the intensity $\mu(t)$:

$$\mu(t)=\frac{f(t)}{\bar F(t)}$$

This definition implies that $\mu(t)=\frac{-\frac d{dt}\bar F(t)}{\bar F(t)}=\frac d{dt} -\ln(\bar F(t))$ and so:

$$\bar F (t) = e^{-\int_0^t \mu(\tau) d\tau}$$

So our expression reduces to

$$\Pr[X\leq n | X>a]=\int_a^n e^{-\int_a^{a+t}\mu(\tau) d\tau}\mu(t) dt$$

If we redefine $n$ so instead of saying we want an insurance until age $n$, we say that we want $n$ years of insurance it becomes even simpler:

$$\Pr[X\leq a+n | X>a]=\int_0^n e^{-\int_0^t\mu(a+\tau) d\tau}\mu(t) dt$$

## Why does it work?

What we need to figure out when selling insurances to a person of $a>0$ years of age really is a new density function, for the distribution of remaining life length of a $a$-year old.

$f(t|a) = \frac {f(a+t)}{\bar F(a)}$.

The probabiliy of living $t$ years after reaching age $a$ should be the same as a newborn living $a+t$ years, conditioned on him/her surviving $a$ years. However conditioning $\mu(t)$ is a lot simpler, consider first what happens to the survival function:

$$\bar F(t| a) = \frac {\bar F(a+t)} {\bar F(a)}$$

Plug these in to our definition of $\mu$ and we see that

$$\mu(t|a)=\frac {f(a+t)} {\bar F(a+t)}=\mu(a+t)$$

Which is why intensities are so handy.