# Poisson process

could someone explain why p=lambda*delta
(edited 10 months ago)
Original post by Student 999
could someone explain why p=lambda*delta

As far as I got, p stands for the probability how many arrivals occur to a certain time. Iambda in this case means the expected value and variance at the same time, so how many arrivals to unit time are expected, although that value may differ to the top or to the bottom. Delta in turn stands for the time slot, so the time span in which a number of (expected) arrivals may happen.
(edited 10 months ago)
Original post by Student 999
could someone explain why p=lambda*delta

If you have (on average) lambda events per unit time, then the expected number of events in an interval of length delta is going to be lambda * delta. For a Bernoulli trial, you either have 0 or 1 (successful) events so the expectation equals the probability of one successful event.
Many thanks, is the idea essentially just that since the expectation of a Bernoulli trial is equivalent by definition to the expected number of successes in a trial which can be derived to p<=1. The expected number of successes in a time interval delta is just lambda*delta hence p=lambda*delta.

Original post by DFranklin
If you have (on average) lambda events per unit time, then the expected number of events in an interval of length delta is going to be lambda * delta. For a Bernoulli trial, you either have 0 or 1 (successful) events so the expectation equals the probability of one successful event.
I don't understand what you're trying to say, to be honest (in particular " can be derived to p<=1" makes no sense to me).
Original post by DFranklin
I don't understand what you're trying to say, to be honest (in particular " can be derived to p<=1" makes no sense to me).

Consider the first time interval (0,delta], we want to find the probability of the Bernoulli trial that represents this interval to be succesful? Note the expectation of a Bernoulli rv is p ,(which has to be less than or equal to 1 since p is a probability).

Now to get p=lambda*delta, since the expectation of a Bernoulli rv is analogous to the weighted average of successful events of the Bernoulli trial? This is in other words lambda*delta since lambda is the rate, delta is the time interval thus lambda*delta is the expected number of successful events for a time interval of length delta.
(edited 10 months ago)