Is there any way to relax the hypothesis of independency of events in a poisson-distributed variable? That is, the occurrence of one event does acutally affect somehow the probability that a second event will occur. May be modifying the pdf? Or is there an other distribution for this? Thanks!
2026-03-30 14:41:15.1774881675
Relaxing the hypothesis of independency in Poisson variables
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Poisson distribution and related exponential distributions are fundamentally based on ideas of independence. That is not to say that conditioning cannot sometimes be useful, but we would have to know more specifically what you have in mind to give a helpful answer. Below are a couple of topics roughly related to your question.
Markov chains (in discrete time) and Markov processes (in continuous time) explicitly treat a what is sometimes loosely called 'one-step dependence', in which the current state (or value) of the process may contain conditional information about the next value in sequence. However, past history before the last known value is not relevant. The first example below is a discrete-time stochastic process and the second is a continuous-time stochastic process.
$1.$ A simple example of a Markov chain is sometimes used to describe sequences of sunny and rainy days. For example, if it is sunny today, then there is an 75% chance of sun tomorrow; if it is rainy today, then there is a 50% chance of rain tomorrow. (But knowing yesterday's weather in addition to today's is not relevant when predicting the weather tomorrow.) Knowing these probabilities it is possible to find the overall percentage of sunny days over a long period of time (8/11). This example has only two 'states' (rainy and sunny), but it is possible to have Markov chains with larger numbers of states. You can find textbooks on 'Markov chains' or search the Internet for information.
$2.$ Another example is a queue (waiting line). Suppose customers arrive at a Poisson rate of 3 per hour, and that there is a single server who can process them at a Poisson rate of 4 per hour (when customers are available). If there are four customers 'in the system' now (one being served and three waiting in line), then one can figure out the probability that no one will be in the system two hours from now. One can also figure out the average number of people in the system at any one time over the long run (3/7).
Simple 'queueing systems' like this one have nice mathematical solutions. But it doesn't take many complications to describe a system that is so messy mathematically that many practitioners would use simulation to see how it evolves. (For example, if the rate of arrivals depends on the number of people already waiting for service or there are several servers working at different rates.) For more on this you can look for books on 'operations research' or 'queueing theory', or search the Internet.