1. Random
2. 4. Special Distributions
3. The Gompertz Distribution

## The Gompertz Distribution

The Gompertz distributon, named for Benjamin Gompertz, is a continuous probability distribution on $$[0, \infty)$$ that has exponentially increasing failure rate. Unfortunately, the death rate of adult humans increases exponentially, so the Gompertz distribution is widely used in actuarial science.

### The Basic Gompertz Distribution

#### Distribution Functions

We will start by giving the reliability function, since most applications of the Gompertz distribution deal with mortality.

Random variable $$X$$ has the basic Gompertz distribution with shape parameter $$a \in (0, \infty)$$ if $$X$$ has a continuous distribution on $$[0, \infty)$$ with reliability function $$G^c$$ given by $G^c(x) = \P(X \gt x) = \exp\left[-a\left(e^x - 1\right)\right], \quad x \in [0, \infty)$

Proof that $$G^c$$ is a reliability function:

Note that $$G^c$$ is continuous and decreasing on $$[0, \infty)$$ with $$G^c(0) = 1$$ and $$G^c(x) \to 0$$ as $$x \to \infty$$.

In the special case $$a = 1$$, $$X$$ has the standard Gompertz distribution.

$$X$$ has distribution function $$G$$ given by $G(x) = \P(X \le x) = 1 - \exp\left[-a\left(e^x - 1\right)\right], \quad x \in [0, \infty)$

Proof:

This follows trivially from the reliability function above, since $$G = 1 - G^c$$.

$$X$$ has quantile function $$G^{-1}$$ given by $G^{-1}(p) = \ln\left[1 - \frac{1}{a} \ln(1 - p)\right], \quad p \in [0, 1)$

1. The first quartile is $$q_1 = \ln\left\{1 + [\ln(4) - \ln(3)] \big/ a\right\}$$.
2. The median is $$q_2 = \ln\left[1 + \ln(2) \big/ a\right]$$.
3. The third quartile is $$q_3 = \ln\left[1 + \ln(4) \big/ a\right]$$.
Proof:

The formula for $$G^{-1}$$ follows from the distribution function above by solving $$p = G(x)$$ for $$x$$ in terms of $$p$$.

Open the special distribution calculator and select the Gompertz distribution. Vary the shape parameter and note the shape of the distribution function. For selected values of the shape parameter, computer a few values of the distribution function and the quantile function.

$$X$$ has probability density function $$g$$ given by $g(x) = a e^x \exp\left[-a\left(e^x - 1\right)\right], \quad x \in [0, \infty)$

1. If $$a \lt 1$$ then $$g$$ is increasing and then decreasing with mode $$x = -\ln(a)$$.
2. If $$a \ge 1$$ then $$g$$ is decreasing with mode $$x = 0$$.
3. If $$a \lt (3 - \sqrt{5})\big/2 \approx 0.382$$ then $$g$$ is concave up and then down then up again, with inflection points at $$x = \ln\left[(3 \pm \sqrt{5})\big/2 a\right]$$.
4. If $$(3 - \sqrt{5})\big/2 \le a \lt (3 + \sqrt{5})\big/2 \approx 2.618$$ then $$g$$ is concave down and then up, with inflection point at $$x = \ln\left[(3 + \sqrt{5})\big/2 a\right]$$.
5. If $$a \ge (3 + \sqrt{5})\big/2$$ then $$g$$ is concave up.
Proof:

The formula for $$g$$ follows from the distribution function since $$g = G^\prime$$ Parts (a)–(d) follow from \begin{align} g^\prime(x) & = a e^x (1 - a e^x) \exp\left[-a\left(e^x - 1\right)\right] \\ g^{\prime\prime}(x) & = a e^x (1 - 3 a e^x + a^2 e^{2x}) \exp\left[-a\left(e^x - 1\right)\right] \end{align}

Open the special distribution simulator and select the Gompertz distribution. Vary the shape parameter and note the shape of the probability density function. For selected values of the shape parameter, run the simulation 1000 times and compare the empirical density function to the probability density function.

Finally, as promised, $$X$$ has exponentially increasing failure rate.

$$X$$ has failure rate function $$r$$ given by $$r(x) = a e^x$$ for $$x \in [0, \infty)$$

Proof:

Recall that the is $$r(x) = g(x) \big/ G^c(x)$$ so the result follows from the distribution function above and the probability density function above.

#### Moments

The moments of the basic Gompertz distribution cannot be given in simple closed form, but the mean and moment generating function can at least be expressed in terms of a special function known as the exponential integral. There are many variations on the exponential integral, but for our purposes, the following version is best:

The exponential integral with parameter $$a \in (0, \infty)$$ is the function $$E_a: \R \to (0, \infty)$$ defined by $E_a(t) = \int_1^\infty u^t e^{-a u} du, \quad t \in \R$

As before, we assume that $$X$$ has the basic Gompertz distribution with shape parameter $$a \in (0, \infty)$$.

$$X$$ has moment generating function $$m$$ given by $m(t) = \E\left(e^{t X}\right) = a e^a E_a(t), \quad t \in \R$

Proof:

Using the substitution $$u = e^x$$ we have $m(t) = \int_0^\infty e^{t x} a e^x e^a \exp\left(-a e^x \right) dx = a e^a \int_1^\infty u^t e^{-a u} du = a e^a E_a(t)$

It follows that $$X$$ has moments of all orders. Here is the mean:

$$X$$ has mean $$\E(X) = e^a E_a(-1)$$.

Proof:

First we use the substitution $$y = e^x$$ to get $\E(X) = \int_0^\infty x a e^x e^a \exp\left(-a e^x\right) dx = a e^a \int_1^\infty \ln(y) e^{-a y} dy$ Next, integration by parts with $$u = \ln(y)$$, $$dv = e^{-a y} dy$$ gives $\E(X) = e^a \int_1^\infty \frac{1}{y} e^{-a y} dy = e^a E_a(-1)$

Open the special distribution simulator and select the Gompertz distribution. Vary the shape parameter and note the size and location of the mean $$\pm$$ standard deviation bar. For selected values of the parameter, run the simulation 1000 times and compare the empirical mean and standard deviation to the distribution mean and standard deviation.

#### Related Distributions

The basic Gompertz distribution has the usual connections to the standard uniform distribution by means of the distribution and quantile functions.

Suppose that $$a \in (0, \infty)$$.

1. If $$U$$ has the standard uniform distribution then $$X = \ln\left[1 - \frac{1}{a} \ln(U) \right]$$ has the basic Gompertz distribution with shape parameter $$a$$.
2. If $$X$$ has the basic Gompertz distribution with shape parameter $$a$$ then $$U = \exp\left[-a\left(e^X - 1\right)\right]$$ has the standard uniform distribution.
Proof:
1. Recall that if $$U$$ has the standard uniform distribution, then $$1 - U$$ also has the standard uniform distribution, and hence $$X = G^{-1}(1 - U)$$ has the basic Gompertz distribution with shape parameter $$a$$.
2. If $$X$$ has the basic Gompertz distribution with shape parameter $$a$$ then $$G(X)$$ has the standard uniform distribution, and hence so does $$U = 1 - G(X)$$.

Since the quantile function of the basic Gompertz distribution has a simple closed form, the distribution can be simulated using the random quantile method.

Open the random quantile experiment and select the Gompertz distribution. Vary the shape parameter and note the shape of the distribution and probability density functions. For selected values of the parameter, run the simulation 1000 times and compare the empirical density function, mean, and standard deviation to their distributional counterparts.

The basic Gompertz distribution also has simple connections to the exponential distribution.

Suppose that $$a \in (0, \infty)$$.

1. If $$X$$ has the basic Gompertz distribution with shape parameter $$a$$, then $$Y = e^X - 1$$ has the exponential distribution with rate parameter $$a$$.
2. If $$Y$$ has the exponential distribution with rate parameter $$a$$, then $$X = \ln(Y + 1)$$ has the Gompertz distribution with shape parameter $$a$$.
Proof:

These results follow from the standard change of variables formula. The transformations, which are inverses of each other, are $$y = e^x - 1$$ and $$x = \ln(y + 1)$$ for $$x, \, y \in [0, \infty)$$. Let $$g$$ and $$h$$ denote PDFs of $$X$$ and $$Y$$ respectively.

1. We start with $$g(x) = a e^x \exp\left[-a\left(e^x - 1\right)\right]$$ for $$x \in [0, \infty)$$ and then $h(y) = g(x) \frac{dx}{dy} = a \exp[\ln(y + 1)] \exp\{-a [\exp(\ln(y + 1)) - 1]\} \frac{1}{y + 1} = a e^{-a y}, \quad y \in [0, \infty)$ which is the PDF of the exponential distribution with rate parameter $$a$$.
2. We start with $$h(y) = a e^{-a y}$$ for $$y \in [0, \infty)$$ and then $g(x) = h(y) \frac{dy}{dx} = a \exp\left[-a (e^x - 1)\right] e^x, \quad x \in [0, \infty)$ which is the PDF of the Gompertz distribution with shape parameter $$a$$.

In particular, if $$Y$$ has the standard exponential distribution (rate parameter 1), then $$X = \ln(Y + 1)$$ has the standard Gompertz distribution (shape parameter 1). Since the exponential distribution is a scale family (the scale parameter is the reciprocal of the rate parameter), we can construct an arbitrary basic Gompertz variable from a standard exponential variable. Specifically, if $$Y$$ has the standard exponential variable and $$a \in (0, \infty)$$, then $X = \ln\left(\frac{1}{a}Y + 1 \right)$ has the Gompertz distribution with shape parameter $$a$$.

The extreme value distribution (Gumbel distribution) is also related to the Gompertz distribution.

If $$X$$ has the standard extreme value distribution for minimums, then the conditional distribution of $$X$$ given $$X \ge 0$$ is the standard Gompertz distribution.

Proof:

By definition, $$X$$ has PDF $$f$$ given by $$f(x) = e^x \exp\left(-e^x\right)$$ for $$x \in \R$$. The conditional PDF of $$X$$ given $$X \ge 0$$ is $g(x) = \frac{f(x)}{\P(X \ge 0)} = \frac{e^x \exp\left(-e^x\right)}{e^{-1}} = e^x \exp\left[-\left(e^x - 1\right)\right], \quad x \in [0, \infty)$ which is the PDF of the standard Gompertz distribution.

### The General Gompertz Distribution

The basic Gompertz distribution is generalized, like so many distributions on $$[0, \infty)$$, by adding a scale parameter. Recall that scale transformations often correspond to a change of units (minutes to hours, for example) and thus are fundamental.

If $$Z$$ has the basic Gompertz distribution with shape parameter $$a \in (0, \infty)$$ and $$b \in (0, \infty)$$ then $$X = b Z$$ has the Gompertz distribution with shape parameter $$a$$ and scale parameter $$b$$.

#### Distribution Functions

$$X$$ has reliability function $$F^c$$ given by $F^c(x) = \P(X \gt x) = \exp\left[-a \left(e^{x / b} - 1\right)\right], \quad x \in [0, \infty)$

Proof:

Recall that $$F^c(x) = G^c(x / b)$$ where $$G^c$$ is the reliability function of the basic Gompertz distribution with shape parameter $$a$$, given above.

$$X$$ has distribution function $$F$$ given by $F(x) = \P(X \le x) = 1 - \exp\left[-a \left(e^{x / b} - 1\right)\right], \quad x \in [0, \infty)$

Proof:

As before, $$F = 1 - F^c$$. Also, $$F(x) = G(x / b)$$ where $$G$$ is the distribution function of the basic Gompertz distribution with shape parameter $$a$$, given above.

$$X$$ has quantile function $$F^{-1}$$ given by $F^{-1}(p) = b \ln\left[1 - \frac{1}{a} \ln(1 - p)\right], \quad p \in [0, 1)$

1. The first quartile is $$q_1 = b \ln\{1 + [\ln(4) - \ln(3)] \big/ a\}$$.
2. The median is $$q_2 = b \ln\left[1 + \ln(2) \big/ a\right]$$.
3. The third quartile is $$q_3 = b \ln\left[1 + \ln(4) \big/ a\right]$$.
Proof:

Recall that $$F^{-1}(p) = b G^{-1}(p)$$ where $$G^{-1}$$ is the quantile function of the basic Gompertz distribution with shape parameter $$a$$, given above.

Open the special distribution calculator and select the Gompertz distribution. Vary the shape and scale parameters and note the shape and location of the distribution function. For selected values of the parameters, computer a few values of the distribution function and the quantile function.

$$X$$ has probability density function $$f$$ given by $f(x) = \frac{a}{b} e^{x/b} \exp\left[-a\left(e^{x/b} - 1\right)\right], \quad x \in [0, \infty)$

1. If $$a \lt 1$$ then $$f$$ is increasing and then decreasing with mode $$x = -b \ln(a)$$.
2. If $$a \ge 1$$ then $$f$$ is decreasing with mode $$x = 0$$.
3. If $$a \lt (3 - \sqrt{5}) \big/ 2 \approx 0.382$$ then $$f$$ is concave up and then down then up again, with inflection points at $$x = b \ln\left[(3 \pm \sqrt{5}) \big/ 2 a\right]$$.
4. If $$(3 - \sqrt{5}) \big/ 2 \le a \lt (3 + \sqrt{5}) \big/ 2 \approx 2.618$$ then $$f$$ has is concave down and then up, with inflection point at $$x = b \ln\left[(3 + \sqrt{5}) \big/ 2 a\right]$$.
5. If $$a \ge (3 + \sqrt{5}) \big/ 2$$ then $$f$$ is concave up.
Proof:

Recall that $$f(x) = \frac{1}{b} g\left(\frac{x}{b}\right)$$ where $$g$$ is the PDF of the basic Gompertz distribution with shape parameter $$a$$, given above.

Open the special distribution simulator and select the Gompertz distribution. Vary the shape and scale parameters and note the shape and location of the probability density function. For selected values of the parameters, run the simulation 1000 times and compare the empirical density function to the probability density function.

Once again, $$X$$ has exponentially increasing failure rate.

$$X$$ has failure rate function $$R$$ given by $R(x) = \frac{a}{b} e^{x / b}, \quad x \in [0, \infty)$

Proof:

Recall that $$R(x) = f(x) \big/ F^c(x)$$. Also, $$R(x) = \frac{1}{b} r\left(\frac{x}{b}\right)$$ where $$r$$ is the failure rate function for the basic Gompertz distribution with shape parameter $$a$$, given above.

#### Moments

As with the basic distribution, the moment generating function and mean of the general Gompertz distribution can be expressed in terms of the exponential integral. Suppose again that $$X$$ has the Gompertz distribution with shape parameter $$a \in (0, \infty)$$ and scale parameter $$b \in (0, \infty)$$.

$$X$$ has moment generating function $$M$$ given by $M(t) = \E\left(e^{t X}\right) = a e^a E_a(b t), \quad t \in \R$

Proof:

Recall that $$M(t) = m(b t)$$ where $$m$$ is the MGF of the basic Gompertz distribution with shape parameter $$a$$, given above.

$$X$$ has mean $$\E(X) = b e^a E_a(-1)$$.

Proof:

This follows from the mean of the basic Gompertz distribution with shape parameter $$a$$ given above, and the standard property $$\E(X) = b \E(Z)$$.

Open the special distribution simulator and select the Gompertz distribution. Vary the shape and scale parameters and note the size and location of the mean $$\pm$$ standard deviation bar. For selected values of the parameters, run the simulation 1000 times and compare the empirical mean and standard deviation to the distribution mean and standard deviation.

#### Related Distributions

Since the Gompertz distribution is a scale family for each value of the shape parameter, it is trivially closed under scale transformations.

If $$X$$ has the Gompertz distribution with shape parameter $$a$$ and scale parameter $$b$$, and if $$c \in (0, \infty)$$, then $$Y = c X$$ has the Gompertz distribution with shape parameter $$a$$ and scale parameter $$b c$$.

As with the basic distribution, the Gompertz distribution has the usual connections with the standard uniform distribution by means of the distribution and quantile functions.

Suppose that $$a, \, b \in (0, \infty)$$.

1. If $$U$$ has the standard uniform distribution then $$X = b \ln\left[1 - \frac{1}{a} \ln(U) \right]$$ has the Gompertz distribution with shape parameter $$a$$ and scale parameter $$b$$.
2. If $$X$$ has the Gompertz distribution with shape parameter $$a$$ and scale parameter $$b$$, then $$U = \exp\left[-a\left(e^{X / b} - 1\right)\right]$$ has the standard uniform distribution.
Proof:

This follows from the corresponding result above for the basic distribution and the definition of the general Gompertz variable as $$X = b Z$$ where $$Z$$ has the basic Gompertz distribution with shape parameter $$a$$.

Again, since the quantile function of the Gompertz distribution has a simple closed form, the distribution can be simulated using the random quantile method.

Open the random quantile experiment and select the Gompertz distribution. Vary the shape and scale parameters and note the shape and location of the distribution and probability density functions. For selected values of the parameters, run the simulation 1000 times and note the agreement between the empirical density function and the probability density function.

The following result is a slight generalization of the connection between the basic Gompertz distribution and the extreme value distribution.

If $$X$$ has the extreme value distribution for minimums with scale parameter $$b \gt 0$$, then the conditional distribution of $$X$$ given $$X \ge 0$$ is the Gompertz distribution with shape parameter 1 and scale parameter $$b$$.

Proof:

We can take $$X = b V$$ where $$V$$ has the standard extreme value distribution for minimums. Note that $$X \ge 0$$ if and only if $$V \ge 0$$. Hence the conditional distribution of $$X$$ given $$X \ge 0$$ is the same as the conditional distribution of $$b V$$ given $$V \ge 0$$. But by the result above the conditional distribution of $$V$$ given $$V \ge 0$$ has the standard Gompertz distribution.

Finally, with a slight generalization of the result above, we can construct a general Gompertz variable from a standard exponential variable.

If $$Y$$ has the standard exponential distribution and if $$a, \, b \in (0, \infty)$$ then $X = b \ln\left(\frac{1}{a} Y + 1\right)$ has the Gompertz distribution with shape parameter $$a$$ and scale parameter $$b$$.