1

As an actuarial student I would like to know why the force of mortality is used to model survival probabilies. It wouldn't be easier to model directly the shape of the survival function?

the force of mortality is defined as: $\mu_x=-\frac{S'(x)}{S(0)}$

while the relationship with the survival probability is, indeed, $S(t)=e^{- \int \limits_{0}^{t} \mu_s \; ds}$

For your understanding of formulas, $T_0$ is the random variable describing the lifetime of a newborn, $S(0)=1$ is the survival function that evaluated in $t=0$.

Enzo
  • 11

1 Answers1

0

One reason this might be interesting from an actuarial point of view, is the obvious parallelism with the computation of bond prices. If we consider a deterministic process for the bond price then we can describe the bond price with the following differential equation

$$dB_t = r_t B_t dt$$

And thus the present value of the bond will be

$$B_0 = B_T \exp\left(-\int_0^T r_t \; dt\right)$$

This has the exact same form as the survival probability. In fact, when you compute actuarial present values, you in effect discount with the interest rate and the mortality. For instance, a pure endowment which pays 1 at maturity will have present value

$$\exp\left(-\int_0^T r_t \; dt\right)\exp\left(-\int_0^T \mu_t \;dt\right)$$

It seems very natural mathematically to merge the two exponentials together and to consider

$$\exp\left(-\int_0^T r_t + \mu_t \; dt\right)$$

So mortality rates work in a sense like interest rates.

Raskolnikov
  • 16,108