Question

Suppose the number of customers arriving at a store obeys a Poisson distribution with an average of λ customers per unit time. That is, if Y is the number of customers arriving in an interval of length t, then Y∼Poisson(λt). Suppose that the store opens at time t=0. Let X be the arrival time of the first customer. Show that X∼Exponential(λ).

Answer 1

This is a simple problem related to having an understanding of poisson's distirbution and exponential distribution.

We actually need to first of all understand the significiance and relationship between the two.

The Poisson distribution models the number of arrivals in a certain fixed time. It is a discrete distribution, taking on values 0,1,2,…0,1,2,….

Whereas,

The exponential distribution models the waiting time between consecutive arrivals. It is a continuous distribution.

Now Think !

Is there any connection between the two ??

There is a connection, since they are used in modelling two different features of the same phenomenon.

But they are defined over two different aspects of the same process and are hence quite different distributions.

Another way to understand hte connection between the two is that if the waiting times between any two consecutive arrivals are independent exponentially distributed with parameter λ, then the number of events in unit time has Poisson distribution with parameter λ.

This explains our inference that if

Y is the number of customers arriving in an interval of length t, then Y∼Poisson(λt). (general poisson distirbution dipiction)

For the same distirbution if we consider time as the parameter of reference then then the arrival time of the first customer. will be shown by an exponential distirbution

X∼Exponential(λ).