Definition 4.1. Let be nonnegative, independent and identically distributed random variables. The random variable counting the number of events until time t, that is

is called renewal process, and its mean is referred to as renewal function.

Theorem 4.2. If are exponentially distributed with parameter then

Proof. It can be seen from the construction that is Erlang distributed with parameters thus its distribution function is

In the proof we use that if an event A involves even B, denoted as in probability theory, then . Clearly, in our case event A can be defined as and event B is .

It can easily be seen that exactly events occur if . Hence

As we can see the number of events that happened until time is Poisson distributed with parameter and it is called a Poisson process with rate .

It is not difficult to verify that

1. ,

2. ,

3. .

Definition 4.3. Rarity condition

Let denote the number of events (number of renewals) that occurred in the interval . Due to the construction of the Poisson process and the memoryless property of the exponential distribution one can easily see that the distribution of depends only on irrespective to the position of the interval (time-homogeneous). In addition, the number of renewals happened during non-intersected intervals are independent random variables (independent increments).

The Poisson process has been introduced as a counting process with probability distribution for the number of arrival during a given interval of length , namely we have

Let us investigate the joint distribution of the arrival epochs during a given time interval of length t when it is known in advance that exactly arrivals have occurred during that interval. Let us divide the interval into nonoverlapping intervals in the following way. Intervals of length always precede the interval of length , , and the interval is of length and in addition

Let denote the event that exactly one arrival occurs in each of the intervals , , and that no arrival occurs in any of the intervals , . We would like to calculate the probability of event given that exactly arrivals have occurred in the interval .

By the definition of the conditional probability thus we have

When the number of arrivals of a Poisson process during nonoverlapping intervals are considered, they can be viewed as independent random variables with Poisson distribution. Thus the probability of the joint events may be calculated as the product of the individual probabilities. ( Poisson process has independent increments ) Therefore

and

By using these probabilities we immediately get

On the other hand, let us consider another process that selects points in the interval independently where each point has uniform distribution over this interval. It can easily be verified that

where the term comes about because the permutations of the points are not distinguished. Since these two conditional distributions are the same we can conclude that if in an interval of length here are exactly arrivals from a Poisson process, then the joint distribution of the moments when these arrivals have occurred is the same as the distribution of points independent and uniformly distributed over the same interval.

Example 4.1. What is the rate ( intensity ) of the renewals

Solution:

Definition 4.4. (Stochastic convergence, convergence in probability) A sequence of random variables is said to converge in probability to a random variable if , for any ,

Theorem 4.5. converges in probability to .

Proof. By applying the Chebychev inequality and observing that

we get

which implies that

Definition 4.6. The rate of renewals is defined as

Theorem 4.7. (The Elementary Renewal Theory)

Derivation of system of differential equations for the Poisson process

Let denote the probability that until time t, events occurred. Due to the properties of the Poisson process the following system of equations can be written

The first equation can be rewritten as

which implies

Similarly to this the other equations imply the following system of differential equations with initial condition

Notice that we have solved this system at Examples 3.7 and 3.14 and the solution is the Poisson distribution, that is