Getting from Binomial to Poisson

In many binomial problems, the number of Bernoulli trials n is large, relatively speaking, and the probability of success p is small such that n p is of moderate magnitude. For example, consider problems that deal with rare events where the probability of occurrence is small (as a concrete example, counting the number of people with July 1 as birthday out of a random sample of 1000 people). It is often convenient to approximate such binomial problems using the Poisson distribution. The justification for using the Poisson approximation is that the Poisson distribution is a limiting case of the binomial distribution. Now that cheap computing power is widely available, it is quite easy to use computer or other computing devices to obtain exact binomial probabiities for experiments up to 1000 trials or more. Though the Poisson approximation may no longer be necessary for such problems, knowing how to get from binomial to Poisson is important for understanding the Poisson distribution itself.

Consider a counting process that describes the occurrences of a certain type of events of interest in a unit time interval subject to three simplifying assumptions (discussed below). We are interested in counting the number of occurrences of the event of interest in a unit time interval. As a concrete example, consider the number of cars arriving at an observation point in a certain highway in a period of time, say one hour. We wish to model the probability distribution of how many cars that will arrive at the observation point in this particular highway in one hour. Let X be the random variable described by this probability distribution. We wish to konw the probability that there are k cars arriving in one hour. We start with using a binomial distribution as an approximation to the probability P(X=k). We will see that upon letting n \rightarrow \infty, the P(X=k) is a Poisson probability.

Suppose that we know E(X)=\alpha, perhaps an average obtained after observing cars at the observation points for many hours. The simplifying assumptions alluded to earlier are the following:

  1. The numbers of cars arriving in nonoverlapping time intervals are independent.
  2. The probability of one car arriving in a very short time interval of length h is \alpha h.
  3. The probability of having more than one car arriving in a very short time interval is esstentially zero.

Assumption 1 means that a large number of cars arriving in one period does not imply fewer cars will arrival in the next period and vice versa. In other words, the number of cars that arrive in any one given moment does affect the number of cars that will arrive subsequently. Knowing how many cars arriving in one minute will not help predict the number of cars arriving at the next minute. Assumption 2 means that the rate of cars arriving is dependent only on the length of the time interval and not on when the time interval occurs (e.g. not on whether it is at the beginning of the hour or toward the end of the hour). The assumptions 2 and 3 allow us to think of a very short period of time as a Bernoulli trial. Thinking of the arrival of a car as a success, each short time interval will result in only one success or one failure.

To start, we can break up the hour into 60 minutes (into 60 Bernoulli trials). We then consider the binomial distribution with n=60 and p=\frac{\alpha}{60}. So the following is an approximation to our desired probability distribution.

\displaystyle (1) \ \ \ \ \ P(X=k) \approx \binom{60}{k} \biggl(\frac{\alpha}{60}\biggr)^k \biggr(1-\frac{\alpha}{60}\biggr)^{60-k} \ \ \ \ \ k=0,1,2,\cdots, 60

Conceivably, there can be more than 1 car arriving in a minute and observing cars in a one-minute interval may not be a Bernoulli trial. For a one-minute interval to qualify as a Bernoulli trial, there is either no car arriving or 1 car arriving in that one minute. So we can break up an hour into 3,600 seconds (into 3,600 Bernoulli trials). We now consider the binomial distribution with n=3600 and p=\frac{\alpha}{3600}.

\displaystyle (2) \ \ \ \ \ P(X=k) \approx \binom{3600}{k} \biggl(\frac{\alpha}{3600}\biggr)^k \biggr(1-\frac{\alpha}{3600}\biggr)^{3600-k} \ \ \ \ \ k=0,1,2,\cdots, 3600

It is also conceivable that more than 1 car can arrive in one second and observing cars in one-second interval may still not qualify as a Bernoulli trial. So we need to get more granular. We can divide up the hour into n equal subintervals, each of length \frac{1}{n}. The assumptions 2 and 3 ensure that each subinterval is a Bernoulli trial (either it is a success or a failure; one car arriving or no car arriving). Assumption 1 tells us that all the n subintervals are independent. So breaking up the hour into n moments and counting the number of moments that are successes will result in a binomial distribution with parameters n and p=\frac{\alpha}{n}. So we are ready to proceed with the following approximation to our probability distribution P(X=k).

\displaystyle (3) \ \ \ \ \ P(X=k) \approx \binom{n}{k} \biggl(\frac{\alpha}{n}\biggr)^k \biggr(1-\frac{\alpha}{n}\biggr)^{n-k} \ \ \ \ \ k=0,1,2,\cdots, n

As we get more granular, n \rightarrow \infty. We show that the limit of the binomial probability in (3) is the Poisson distribution with parameter \alpha. We show the following.

\displaystyle (4) \ \ \ \ \ P(X=k) = \lim \limits_{n \rightarrow \infty} \binom{n}{k} \biggl(\frac{\alpha}{n}\biggr)^k \biggr(1-\frac{\alpha}{n}\biggr)^{n-k}=\frac{e^{-\alpha} \alpha^k}{k!} \ \ \ \ \ \ k=0,1,2,\cdots

In the derivation of (4), we need the following two mathematical tools. The statement (5) is one of the definitions of the mathematical constant e. In the statement (6), the integer n in the numerator is greater than the integer k in the denominator. It says that whenever we work with such a ratio of two factorials, the result is the product of n with the smaller integers down to (n-(k-1)). There are exactly k terms in the product.

\displaystyle (5) \ \ \ \ \ \lim \limits_{n \rightarrow \infty} \biggl(1+\frac{x}{n}\biggr)^n=e^x

\displaystyle (6) \ \ \ \ \ \frac{n!}{(n-k)!}=n(n-1)(n-2) \cdots (n-k+1) \ \ \ \ \ \ \ \  k<n

The following is the derivation of (4).

\displaystyle \begin{aligned}(7) \ \ \ \ \  P(X=k)&=\lim \limits_{n \rightarrow \infty} \binom{n}{k} \biggl(\frac{\alpha}{n}\biggr)^k \biggr(1-\frac{\alpha}{n}\biggr)^{n-k} \\&=\lim \limits_{n \rightarrow \infty} \ \frac{n!}{k! (n-k)!} \biggl(\frac{\alpha}{n}\biggr)^k \biggr(1-\frac{\alpha}{n}\biggr)^{n-k} \\&=\lim \limits_{n \rightarrow \infty} \ \frac{n(n-1)(n-2) \cdots (n-k+1)}{n^k} \biggl(\frac{\alpha^k}{k!}\biggr) \biggr(1-\frac{\alpha}{n}\biggr)^{n} \biggr(1-\frac{\alpha}{n}\biggr)^{-k} \\&=\biggl(\frac{\alpha^k}{k!}\biggr) \lim \limits_{n \rightarrow \infty} \ \frac{n(n-1)(n-2) \cdots (n-k+1)}{n^k} \\&\times \ \ \ \lim \limits_{n \rightarrow \infty} \biggr(1-\frac{\alpha}{n}\biggr)^{n} \ \lim \limits_{n \rightarrow \infty} \biggr(1-\frac{\alpha}{n}\biggr)^{-k} \\&=\frac{e^{-\alpha} \alpha^k}{k!} \end{aligned}

In (7), we have \displaystyle \lim \limits_{n \rightarrow \infty} \ \frac{n(n-1)(n-2) \cdots (n-k+1)}{n^k}=1. The reason being that the numerator is a polynomial where the leading term is n^k. Upon dividing by n^k and taking the limit, we get 1. Based on (5), we have \displaystyle \lim \limits_{n \rightarrow \infty} \biggr(1-\frac{\alpha}{n}\biggr)^{n}=e^{-\alpha}. For the last limit in the derivation we have \displaystyle \lim \limits_{n \rightarrow \infty} \biggr(1-\frac{\alpha}{n}\biggr)^{-k}=1.

We conclude with some comments. As the above derivation shows, the Poisson distribution is at heart a binomial distribution. When we divide the unit time interval into more and more subintervals (as the subintervals get more and more granular), the resulting binomial distribution behaves more and more like the Poisson distribution.

The three assumtions used in the derivation are called the Poisson postulates, which are the underlying assumptions that govern a Poisson process. Such a random process describes the occurrences of some type of events that are of interest (e.g. the arrivals of cars in our example) in a fixed period of time. The positive constant \alpha indicated in Assumption 2 is the parameter of the Poisson process, which can be interpreted as the rate of occurrences of the event of interest (or rate of changes, or rate of arrivals) in a unit time interval, meaning that the positive constant \alpha is the mean number of occurrences in the unit time interval. The derivation in (7) shows that whenever a certain type of events occurs according to a Poisson process with parameter \alpha, the counting variable of the number of occurrences in the unit time interval is distributed according to the Poisson distribution as indicated in (4).

If we observe the occurrences of events over intervals of length other than unit length, say, in an interval of length t, the counting process is governed by the same three postulates, with the modification to Assumption 2 that the rate of changes of the process is now \alpha t. The mean number of occurrences in the time interval of length t is now \alpha t. The Assumption 2 now states that for any very short time interval of length h (and that is also a subinterval of the interval of length t under observation), the probability of having one occurrence of event in this short interval is (\alpha t)h. Applyng the same derivation, it can be shown that the number of occurrences (X_t) in a time interval of length t has the Poisson distribution with the following probability mass function.

\displaystyle (8) \ \ \ \ \ P(X_t=k)=\frac{e^{-\alpha t} \ (\alpha t)^k}{k!} \ \ \ \ \ \ \ \  k=0,1,2,\cdots

Relating Binomial and Negative Binomial

The negative binomial distribution has a natural intepretation as a waiting time until the arrival of the rth success (when the parameter r is a positive integer). The waiting time refers to the number of independent Bernoulli trials needed to reach the rth success. This interpretation of the negative binomial distribution gives us a good way of relating it to the binomial distribution. For example, if the rth success takes place after k failed Bernoulli trials (for a total of k+r trials), then there can be at most r-1 successes in the first k+r trials. This tells us that the survival function of the negative binomial distribution is the cumulative distribution function (cdf) of a binomial distribution. In this post, we gives the details behind this observation. A previous post on the negative binomial distribution is found here.

A random experiment resulting in two distinct outcomes (success or failure) is called a Bernoulli trial (e.g. head or tail in a coin toss, whether or not the birthday of a customer is the first of January, whether an insurance claim is above or below a given threshold etc). Suppose a series of independent Bernoulli trials are performed until reaching the rth success where the probability of success in each trial is p. Let X_r be the number of failures before the occurrence of the rth success. The following is the probablity mass function of X_r.

\displaystyle (1) \ \ \ \ P(X_r)=\binom{k+r-1}{k} p^r (1-p)^k \ \ \ \ \ \ k=0,1,2,3,\cdots

Be definition, the survival function and cdf of X_r are:

\displaystyle (2) \ \ \ \ P(X_r > k)=\sum \limits_{j=k+1}^\infty \binom{j+r-1}{j} p^r (1-p)^j \ \ \ \ \ \ k=0,1,2,3,\cdots

\displaystyle (3) \ \ \ \ P(X_r \le k)=\sum \limits_{j=0}^k \binom{j+r-1}{j} p^r (1-p)^j \ \ \ \ \ \ k=0,1,2,3,\cdots

For each positive integer k, let Y_{r+k} be the number of successes in performing a sequence of r+k independent Bernoulli trials where p is the probability of success. In other words, Y_{r+k} has a binomial distribution with parameters r+k and p.

If the random experiment requires more than k failures to reach the rth success, there are at most r-1 successes in the first k+r trails. Thus the survival function of X_r is the same as the cdf of a binomial distribution. Equivalently, the cdf of X_r is the same as the survival function of a binomial distribution. We have the following:

\displaystyle \begin{aligned}(4) \ \ \ \ P(X_r > k)&=P(Y_{k+r} \le r-1) \\&=\sum \limits_{j=0}^{r-1} \binom{k+r}{j} p^j (1-p)^{k+r-j} \ \ \ \ \ \ k=0,1,2,3,\cdots \end{aligned}

\displaystyle \begin{aligned}(5) \ \ \ \ P(X_r \le k)&=P(Y_{k+r} > r-1) \ \ \ \ \ \ k=0,1,2,3,\cdots \end{aligned}

Remark
The relation (4) is analogous to the relationship between the Gamma distribution and the Poisson distribution. Recall that a Gamma distribution with shape parameter \alpha and scale parameter n, where n is a positive integer, can be interpreted as the waiting time until the nth change in a Poisson process. Thus, if the nth change takes place after time t, there can be at most n-1 arrivals in the time interval [0,t]. Thus the survival function of this Gamma distribution is the same as the cdf of a Poisson distribution. The relation (4) is analogous to the following relation.

\displaystyle (5) \ \ \ \ \int_t^\infty \frac{\alpha^n}{(n-1)!} \ x^{n-1} \ e^{-\alpha x} \ dx=\sum \limits_{j=0}^{n-1} \frac{e^{-\alpha t} \ (\alpha t)^j}{j!}

A previous post on the negative binomial distribution is found here.

The Poisson Distribution

Let \alpha be a positive constant. Consider the following probability distribution:

\displaystyle (1) \ \ \ \ \ P(X=j)=\frac{e^{-\alpha} \alpha^j}{j!} \ \ \ \ \ j=0,1,2,\cdots

The above distribution is said to be a Poisson distribution with parameter \alpha. The Poisson distribution is usually used to model the random number of events occurring in a fixed time interval. As will be shown below, E(X)=\alpha. Thus the parameter \alpha is the rate of occurrence of the random events; it indicates on average how many events occur per unit of time. Examples of random events that may be modeled by the Poisson distribution include the number of alpha particles emitted by a radioactive substance counted in a prescribed area during a fixed period of time, the number of auto accidents in a fixed period of time or the number of losses arising from a group of insureds during a policy period.

Each of the above examples can be thought of as a process that generates a number of arrivals or changes in a fixed period of time. If such a counting process leads to a Poisson distribution, then the process is said to be a Poisson process.

We now discuss some basic properties of the Poisson distribution. Using the Taylor series expansion of e^{\alpha}, the following shows that (1) is indeed a probability distribution.

\displaystyle . \ \ \ \ \ \ \ \sum \limits_{j=0}^\infty \frac{e^{-\alpha} \alpha^j}{j!}=e^{-\alpha} \sum \limits_{j=0}^\infty \frac{\alpha^j}{j!}=e^{-\alpha} e^{\alpha}=1

The generating function of the Poisson distribution is g(z)=e^{\alpha (z-1)} (see The generating function). The mean and variance can be calculated using the generating function.

\displaystyle \begin{aligned}(2) \ \ \ \ \ &E(X)=g'(1)=\alpha \\&\text{ } \\&E[X(X-1)]=g^{(2)}(1)=\alpha^2 \\&\text{ } \\&Var(X)=E[X(X-1)]+E(X)-E(X)^2=\alpha^2+\alpha-\alpha^2=\alpha \end{aligned}

The Poisson distribution can also be interpreted as an approximation to the binomial distribution. It is well known that the Poisson distribution is the limiting case of binomial distributions (see [1] or this post).

\displaystyle (3) \ \ \ \ \ \lim \limits_{n \rightarrow \infty} \binom{n}{j} \biggl(\frac{\alpha}{n}\biggr)^j \biggl(1-\frac{\alpha}{n}\biggr)^{n-j}=\frac{e^{-\alpha} \alpha^j}{j!}

One application of (3) is that we can use Poisson probabilities to approximate Binomial probabilities. The approximation is reasonably good when the number of trials n in a binomial distribution is large and the probability of success p is small. The binomial mean is n p and the variance is n p (1-p). When p is small, 1-p is close to 1 and the binomial variance is approximately np \approx n p (1-p). Whenever the mean of a discrete distribution is approximately equaled to the mean, the Poisson approximation is quite good. As a rule of thumb, we can use Poisson to approximate binomial if n \le 100 and p \le 0.01.

As an example, we use the Poisson distribution to estimate the probability that at most 1 person out of 1000 will have a birthday on the New Year Day. Let n=1000 and p=365^{-1}. So we use the Poisson distribution with \alpha=1000 \times 365^{-1}. The following is an estimate using the Poisson distribution.

\displaystyle . \ \ \ \ \ \ \ P(X \le 1)=e^{-\alpha}+\alpha e^{-\alpha}=(1+\alpha) e^{-\alpha}=0.2415

Another useful property is that the independent sum of Poisson distributions also has a Poisson distribution. Specifically, if each X_i has a Poisson distribution with parameter \alpha_i, then the independent sum X=X_1+\cdots+X_n has a Poisson distribution with parameter \alpha=\alpha_1+\cdots+\alpha_n. One way to see this is that the product of Poisson generating functions has the same general form as g(z)=e^{\alpha (z-1)} (see The generating function). One interpretation of this property is that when merging several arrival processes, each of which follow a Poisson distribution, the result is still a Poisson distribution.

For example, suppose that in an airline ticket counter, the arrival of first class customers follows a Poisson process with a mean arrival rate of 8 per 15 minutes and the arrival of customers flying coach follows a Poisson distribution with a mean rate of 12 per 15 minutes. Then the arrival of customers of either types has a Poisson distribution with a mean rate of 20 per 15 minutes or 80 per hour.

A Poisson distribution with a large mean can be thought of as an independent sum of Poisson distributions. For example, a Poisson distribution with a mean of 50 is the independent sum of 50 Poisson distributions each with mean 1. Because of the central limit theorem, when the mean is large, we can approximate the Poisson using the normal distribution.

In addition to merging several Poisson distributions into one combined Poisson distribution, we can also split a Poisson into several Poisson distributions. For example, suppose that a stream of customers arrives according to a Poisson distribution with parameter \alpha and each customer can be classified into one of two types (e.g. no purchase vs. purchase) with probabilities p_1 and p_2, respectively. Then the number of “no purchase” customers and the number of “purchase” customers are independent Poisson random variables with parameters \alpha p_1 and \alpha p_2, respectively. For more details on the splitting of Poisson, see Splitting a Poisson Distribution.

Reference

  1. Feller W. An Introduction to Probability Theory and Its Applications, Third Edition, John Wiley & Sons, New York, 1968

The hazard rate function

In this post, we introduce the hazard rate function using the notions of non-homogeneous Poisson process.

In a Poisson process, changes occur at a constant rate \lambda per unit time. Suppose that we interpret the changes in a Poisson process from a mortality point of view, i.e. a change in the Poisson process mean a termination of a system, be it biological or manufactured, and this Poisson process counts the number of of terminations as they occur. Then the rate of change \lambda is interpreted as a hazard rate (or failure rate or force of mortality). With a constant force of mortality, the time until the next change is exponentially distributed. In this post, we discuss the hazard rate function in a more general setting. The process that counts of the number of terminations will no longer have a constant hazard rate, and instead will have a hazard rate function \lambda(t), a function of time t. Such a counting process is called a non-homogeneous Poisson process. We discuss the survival probability models (the time to the next termination) associated with a non-homogeneous Poisson process. We then discuss several important examples of survival probability models, including the Weibull distribution, the Gompertz distribution and the model based on the Makeham’s law. We aso comment briefly the connection between the hazard rate function and the tail weight of a distribution.

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The Poisson Process
We start with the three postulates of a Poisson process. Consider an experiment in which the occurrences of a certain type of events are counted during a given time interval. We call the occurrence of the type of events in question a change. We assume the following three conditions:

  1. The numbers of changes occurring in nonoverlapping intervals are independent.
  2. The probability of two or more changes taking place in a sufficiently small interval is essentially zero.
  3. The probability of exactly one change in the short interval (t,t+\delta) is approximately \lambda \delta where \delta is sufficiently small and \lambda is a positive constant.

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When we interpret the Poisson process in a mortality point of view, the constant \lambda is a hazard rate (or force of mortality), which can be interpreted as the rate of failure at the next instant given that the life has survived to time t. With a constant force of mortality, the survival model (the time until the next termination) has an exponential distribution with mean \frac{1}{\lambda}. We wish to relax the constant force of mortality assumption by making \lambda a function of t instead. The remainder of this post is based on the non-homogeneous Poisson process defined below.

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The Non-Homogeneous Poisson Process
We modifiy condition 3 above by making \lambda(t) a function of t. We have the following modified counting process.

  1. The numbers of changes occurring in nonoverlapping intervals are independent.
  2. The probability of two or more changes taking place in a sufficiently small interval is essentially zero.
  3. The probability of exactly one change in the short interval (t,t+\delta) is approximately \lambda(t) \delta where \delta is sufficiently small and \lambda(t) is a nonnegative function of t.

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We focus on the survival model aspect of such counting processes. Such process can be interpreted as models for the number of changes occurred in a time interval where a change means “termination” or ‘failure” of a system under consideration. The rate of change function \lambda(t) indicated in condition 3 is called the hazard rate function. It is also called the failure rate function in reliability engineering and the force of mortality in life contingency theory.

Based on condition 3 in the non-homogeneous Poisson process, the hazard rate function \lambda(t) can be interpreted as the rate of failure at the next instant given that the life has survived to time t.

Two random variables naturally arise from a non-homogeneous Poisson process are described here. One is the discrete variable N_t, defined as the number of changes in the time interval (0,t). The other is the continuous random variable T, defined as the time until the occurrence of the first change. The probability distribution of T is called a survival model. The following is the link between N_t and T.

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\displaystyle \begin{aligned}(1) \ \ \ \ \ \ \ \ \ &P[T > t]=P[N_t=0] \end{aligned}

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Note that P[T > t] is the probability that the next change occurs after time t. This means that there is no change within the interval (0,t). We have the following theorems.

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Theorem 1.
Let \displaystyle \Lambda(t)=\int_{0}^{t} \lambda(y) dy. Then e^{-\Lambda(t)} is the probability that there is no change in the interval (0,t). That is, \displaystyle P[N_t=0]=e^{-\Lambda(t)}.

Proof. We are interested in finding the probability of zero changes in the interval (0,y+\delta). By condition 1, the numbers of changes in the nonoverlapping intervals (0,y) and (y,y+\delta) are independent. Thus we have:

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\displaystyle (2) \ \ \ \ \ \ \ \ P[N_{y+\delta}=0] \approx P[N_y=0] \times [1-\lambda(y) \delta]

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Note that by condition 3, the probability of exactly one change in the small interval (y,y+\delta) is \lambda(y) \delta. Thus [1-\lambda(y) \delta] is the probability of no change in the interval (y,y+\delta). Continuing with equation (2), we have the following derivation:

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\frac{P[N_{y+\delta}=0] - P[N_y=0]}{\delta} \approx -\lambda(y) P[N_y=0] \\&\text{ } \\&\frac{d}{dy} P[N_y=0]=-\lambda(y) P[N_y=0] \\&\text{ } \\&\frac{\frac{d}{dy} P[N_y=0]}{P[N_y=0]}=-\lambda(y) \\&\text{ } \\&\int_0^{t} \frac{\frac{d}{dy} P[N_y=0]}{P[N_y=0]} dy=-\int_0^{t} \lambda(y)dy \end{aligned}

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Evaluating the integral on the left hand side with the boundary condition of P[N_0=0]=1 produces the following results:

\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &ln P[N_t=0]=-\int_0^{t} \lambda(y)dy \\&\text{ } \\&P[N_t=0]=\displaystyle e^{\displaystyle -\int_0^{t} \lambda(y)dy} \end{aligned}

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Theorem 2
As discussed above, let T be the length of the interval that is required to observe the first change. Then the following are the distribution function, survival function and pdf of T:

\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &F_T(t)=\displaystyle 1-e^{-\int_0^t \lambda(y) dy} \\&\text{ } \\&S_T(t)=\displaystyle e^{-\int_0^t \lambda(y) dy} \\&\text{ } \\&f_T(t)=\displaystyle \lambda(t) e^{-\int_0^t \lambda(y) dy} \end{aligned}

Proof. In Theorem 1, we derive the probability P[N_y=0] for the discrete variable N_y derived from the non-homogeneous Poisson process. We now consider the continuous random variable T, the time until the first change, which is related to N_t by (1). Thus S_T(t)=P[T > t]=P[N_t=0]=e^{-\int_0^t \lambda(y) dy}. The distribution function and density function can be derived accordingly.

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Theorem 3
The hazard rate function \lambda(t) is equivalent to each of the following:

\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\lambda(t)=\frac{f_T(t)}{1-F_T(t)} \\&\text{ } \\&\lambda(t)=\frac{-S_T^{'}(t)}{S_T(t)} \end{aligned}

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Remark
Theorem 1 and Theorem 2 show that in a non-homogeneous Poisson process as described above, the hazard rate function \lambda(t) completely specifies the probability distribution of the survival model T (the time until the first change) . Once the rate of change function \lambda(t) is known in the non-homogeneous Poisson process, we can use it to generate the survival function S_T(t). All of the examples of survival models given below are derived by assuming the functional form of the hazard rate function. The result in Theorem 2 holds even outside the context of a non-homogeneous Poisson process, that is, given the hazard rate function \lambda(t), we can derive the three distributional items S_T(t), F_T(t), f_T(t).

The ratio in Theorem 3 indicates that the probability distribution determines the hazard rate function. In fact, the ratio in Theorem 3 is the usual definition of the hazard rate function. That is, the hazard rate function can be defined as the ratio of the density and the survival function (one minus the cdf). With this definition, we can also recover the survival function. Whenever \displaystyle \lambda(x)=\frac{f_X(x)}{1-F_X(x)}, we can derive:

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &S_X(x)=\displaystyle e^{-\int_0^t \lambda(y) dy} \end{aligned}

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As indicated above, the hazard rate function can be interpreted as the failure rate at time t given that the life in question has survived to time t. It is the rate of failure at the next instant given that the life or system being studied has survived up to time t.

It is interesting to note that the function \Lambda(t)=\int_0^t \lambda(y) dy defined in Theorem 1 is called the cumulative hazard rate function. Thus the cumulative hazard rate function is an alternative way of representing the hazard rate function (see the discussion on Weibull distribution below).

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Examples of Survival Models

–Exponential Distribution–
In many applications, especially those for biological organisms and mechanical systems that wear out over time, the hazard rate \lambda(t) is an increasing function of t. In other words, the older the life in question (the larger the t), the higher chance of failure at the next instant. For humans, the probability of a 85 years old dying in the next year is clearly higher than for a 20 years old. In a Poisson process, the rate of change \lambda(t)=\lambda indicated in condition 3 is a constant. As a result, the time T until the first change derived in Theorem 2 has an exponential distribution with parameter \lambda. In terms of mortality study or reliability study of machines that wear out over time, this is not a realistic model. However, if the mortality or failure is caused by random external events, this could be an appropriate model.

–Weibull Distribution–
This distribution is an excellent model choice for describing the life of manufactured objects. It is defined by the following cumulative hazard rate function:

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\Lambda(t)=\biggl(\frac{t}{\beta}\biggr)^{\alpha} \end{aligned} where \alpha > 0 and \beta>0

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As a result, the hazard rate function, the density function and the survival function for the lifetime distribution are:

\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\lambda(t)=\frac{\alpha}{\beta} \biggl(\frac{t}{\beta}\biggr)^{\alpha-1} \\&\text{ } \\&f_T(t)=\frac{\alpha}{\beta} \biggl(\frac{t}{\beta}\biggr)^{\alpha-1} \displaystyle e^{\displaystyle -\biggl[\frac{t}{\beta}\biggr]^{\alpha}} \\&\text{ } \\&S_T(t)=\displaystyle e^{\displaystyle -\biggl[\frac{t}{\beta}\biggr]^{\alpha}} \end{aligned}

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The parameter \alpha is the shape parameter and \beta is the scale parameter. When \alpha=1, the hazard rate becomes a constant and the Weibull distribution becomes an exponential distribution.

When the parameter \alpha<1, the failure rate decreases over time. One interpretation is that most of the defective items fail early on in the life cycle. Once they they are removed from the population, failure rate decreases over time.

When the parameter 1<\alpha, the failure rate increases with time. This is a good candidate for a model to describe the lifetime of machines or systems that wear out over time.

–The Gompertz Distribution–
The Gompertz law states that the force of mortality or failure rate increases exponentially over time. It describe human mortality quite accurately. The following is the hazard rate function:

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\lambda(t)=\alpha e^{\beta t} \end{aligned} where \alpha>0 and \beta>0.

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The following are the cumulative hazard rate function as well as the survival function, distribution function and the pdf of the lifetime distribution T.

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\Lambda(t)=\int_0^t \alpha e^{\beta y} dy=\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta} \\&\text{ } \\&S_T(t)=\displaystyle e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}\biggr)} \\&\text{ } \\&F_T(t)=\displaystyle 1-e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}\biggr)} \\&\text{ } \\&f_T(t)=\displaystyle \alpha \ e^{\beta t} \ e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}\biggr)} \end{aligned}

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–Makeham’s Law–
The Makeham’s Law states that the force of mortality is the Gompertz failure rate plus an age-indpendent component that accounts for external causes of mortality. The following is the hazard rate function:

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\lambda(t)=\alpha e^{\beta t}+\mu \end{aligned} where \alpha>0, \beta>0 and \mu>0.

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The following are the cumulative hazard rate function as well as the survival function, distribution function and the pdf of the lifetime distribution T.

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\Lambda(t)=\int_0^t (\alpha e^{\beta y}+\mu) dy=\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t \\&\text{ } \\&S_T(t)=\displaystyle e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t\biggr)} \\&\text{ } \\&F_T(t)=\displaystyle 1-e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t\biggr)} \\&\text{ } \\&f_T(t)=\biggl( \alpha e^{\beta t}+\mu t \biggr) \ e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t\biggr)} \end{aligned}

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The Tail Weight of a Distribution
The hazard rate function can provide information about the tail of a distribution. If the hazard rate function is decreasing, it is an indication that the distribution has a heavy tail, i.e., the distribution significantly puts more probability on larger values. Conversely, if the hazard rate function is increasing, it is an indication of a lighter tail. In an insurance context, heavy tailed distributions (e.g. the Pareto distribution) are suitable candidates for modeling large insurance losses (see this previous post or [1]).

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Reference

  1. Klugman S.A., Panjer H. H., Wilmot G. E. Loss Models, From Data to Decisions, Second Edition., Wiley-Interscience, a John Wiley & Sons, Inc., New York, 2004