Force of mortality

In actuarial science, force of mortality represents the instantaneous rate of mortality at a certain age measured on an annualized basis. It is identical in concept to failure rate, also called hazard function, in reliability theory.

Motivation and definition
In a life table, we consider the probability of a person dying from age x to x + 1, called qx. In the continuous case, we could also consider the conditional probability of a person who has attained age (x) dying between ages x and x + Δx, which is


 * $$P_{x}(\Delta x)=P(xx)=\frac{F_X(x+\Delta\;x)-F_X(x)}{(1-F_X(x))}$$

where FX(x) is the cumulative distribution function of the continuous age-at-death random variable, X. As Δx tends to zero, so does this probability in the continuous case. The approximate force of mortality is this probability divided by Δx. If we let Δx tend to zero, we get the function for force of mortality, denoted by $$\mu(x)$$:


 * $$\mu\,(x)= \lim_{\Delta x \rightarrow 0} \frac{F_X(x+\Delta\;x)-F_X(x)}{\Delta x (1-F_X(x))} = \frac{F'_X(x)}{1-F_X(x)}$$

Since fX(x)=F 'X(x) is the probability density function of X, and S(x) = 1 - FX(x) is the survival function, the force of mortality can also be expressed variously as:


 * $$\mu\,(x)=\frac{f_X(x)}{1-F_X(x)}=-\frac{S'(x)}{S(x)}=-{\frac{d}{dx}}\ln[S(x)].$$

To understand conceptually how the force of mortality operates within a population, consider that the ages, x, where the probability density function fX(x) is zero, there is no chance of dying. Thus the force of mortality at these ages is zero. The force of mortality μ(x) uniquely defines a probability density function fX(x).

The force of mortality $$\mu(x)$$ can be interpreted as the conditional density of failure at age x, while f(x) is the unconditional density of failure at age x. The unconditional density of failure at age x is the product of the probability of survival to age x, and the conditional density of failure at age x, given survival to age x.

This is expressed in symbols as


 * $$\,\mu(x)S(x) = f_X(x)$$

or equivalently


 * $$\mu(x) = \frac{f_X(x)}{S(x)}.$$

In many instances, it is also desirable to determine the survival probability function when the force of mortality is known. To do this, integrate the force of mortality over the interval x to x + t


 * $$ \int_{x}^{x+t} \mu(y) \, dy = \int_{x}^{x+t} -\frac{d}{dy} \ln[S(y)]\, dy $$.

By the fundamental theorem of calculus, this is simply


 * $$ -\int_{x}^{x+t} \mu(y) \, dy = \ln[S(x + t)] - \ln[S(x)]. $$

Let us denote


 * $$ S_x(t) = \frac{S(x+t)}{S(x)}, $$

then taking the exponent to the base e, the survival probability of an individual of age x in terms of the force of mortality is


 * $$ S_x(t) = \exp \left(-\int_x^{x+t}\mu(y)\, dy\, \right). $$

Examples

 * The simplest example is when the force of mortality is constant:


 * $$\mu(y) = \lambda,$$


 * then the survival function is


 * $$S_x(t) = e^{-\int_x^{x+t} \lambda dy} = e^{-\lambda t},$$


 * is the exponential distribution.


 * When the force of mortality is


 * $$\mu(y) = \frac{y^{\alpha-1} e^{-y}}{\Gamma(\alpha) - \gamma(\alpha, y)}, $$


 * where γ(α,y) is the lower incomplete gamma function, the probability density function that of Gamma distribution


 * $$f(x) = \frac{x^{\alpha - 1} e^{-x}}{\Gamma(\alpha)}.$$


 * When the force of mortality is


 * $$ \mu(y) = \alpha \lambda^\alpha y^{\alpha-1},$$


 * where α ≥ 0, we have


 * $$\int_x^{x+t}\mu(y) dy = \alpha \lambda^\alpha \int_x^{x+t} y^{\alpha-1} dy = \lambda^\alpha( (x+t)^\alpha - x^\alpha).$$


 * Thus, the survival function is


 * $$S_x(t) = e^{-\int_x^{x+t}\mu(y) dy} = A(x) e^{ - (\lambda (x+t))^\alpha }, $$


 * where $$A(x) = e^{(\lambda x)^{\alpha}}.$$ This is the survival function for Weibull distribution. For α = 1, it is same as the exponential distribution.


 * Another famous example is when the survival model follows Gompertz–Makeham law of mortality. In this case, the force of mortality is


 * $$\mu(y) = A + Bc^{y} \quad\text{for } y \geqslant 0.$$


 * Using the last formula, we have


 * $$\int_x^{x+t} (A + Bc^{y}) dy = At + B(c^{x+t}-c^{x})/\ln[c].$$


 * Then


 * $$S_x(t) = e^{-(At + B(c^{x+t}-c^{x})/\ln[c])} = e^{-At} g^{c^{x} (c^{t}-1)} $$


 * where $$g=e^{-B/\ln[c]}.$$