Dtjytyk

Given an exponential waiting time with rate $lambda$ we know that the distribution for the waiting time would be
$$ f(T=t) = lambda e^{-lambda t} $$
now, if we assume that the rate is not constant, let´s say $lambda(t)$, I am wondering if the distribution would be just
$$ f(T=t) = lambda(t) e^{-int_0^t lambda(t) dt} $$
or not. And if not, what would be the distribution? I cannot find any references for this distribution, so any help is appreciated.

In actuarial science, the function $lambda(t)$ - more commonly notated as $mu_t$ - is known as the force of mortality or the hazard function. When the force of mortality is not constant, you obtain distributions that are not exponential. For life-insurance context, you may want to look up the Gompertz distribution and Makeham distribution.

Given a force of mortality $mu_t$ for random variable $X$, the relationship between its CDF $F_X$, its PDF $f_X$, and $mu_t$ is
$$mu_t=dfrac{f_X(t)}{1-F_X(t)}=-dfrac{S^{prime}_X(t)}{S_X(t)}=-dfrac{text{d}}{text{d}t}[ln S_X(t)]$$
where $S_X = 1 - F_X$ is commonly called the "survival function." From this, you can integrate to obtain a formula for $S_X$. Given the property that $F_X(0)=0$ (do you see why?), we can obtain $S_X(0) = 1$, hence
$$int_{0}^{t}mu_stext{ d}s=-int_{0}^{t}dfrac{text{d}}{text{d}t}[ln S_X(t)]text{ d}t=-[ln S_X(t)-ln 1]=-ln S_X(t)$$
from which we obtain
$$S_X(t)=expleft[-int_{0}^{t}mu_stext{ d}s right]$$
and thus
$$F_X(t)=1-expleft[-int_{0}^{t}mu_stext{ d}s right]$$
and you can differentiate to obtain
$$f_X(t)=mu_texpleft[-int_{0}^{t}mu_stext{ d}s right]$$
matching your form above.

Except for your error about the meaning of a density, you are correct.

The easy way to figure this out is to consider the discrete version. Suppose the probability of an event happening in $[t,t+dt]$ is $lambda(t) dt$. Then the probability of waiting for at least n intervals with no event is

$$P=prod_{i=1}^n (1-dt lambda((i-1)dt).$$

Now set $dt=t/n$ and send $n to infty$. To compute the limit, compute the exponential of its logarithm. The logarithm is

$$log(P)=sum_{i=1}^n log left ( 1-frac{t}{n} lambda left ( (i-1)frac{t}{n} right ) right ).$$

By linear approximation

$$log(P)=o(1)+sum_{i=1}^n -frac{t}{n} lambda left ( (i-1) frac{t}{n} right ).$$

The sum is a Riemann sum, and the higher order correction vanishes in the $n to infty$ limit, so you get the limit of $log(P)$ as an integral. The quantity you just calculated is then $P(T>t)$.

By the way, the main place you would find this is in references to continuous time, discrete space, time-inhomogeneous Markov chains, where it appears as the holding time distribution.

To sample a value from a non-homogeneous exponential distribution you can follow this steps

S1. Sample $x$ from homogeneous exponential with rate 1

S2. Calculate $Lambda^{-1} (x)$

where $Lambda(t)$ is the intensity function (the integral of the rate).

The random variable $Lambda^{-1} (x)$ has a non-homogeneous distribution with rate $lambda(t)$.

A reference could be the paper "Generating Nonhomogeneous Poisson Processes" by Raghu Pasupathy.