# A problem related to Poisson process

1. Mar 19, 2009

### quacam09

Hi all, I have a probability problem. Can you help me? Thank you!

Here is the problem:

Consider the queueing system, there are n customers 1, 2, ...N.
Customer 1 arrives in accordance with a Poisson process with rate Lamda, customer 2 arrives in accordance with a Poisson process with rate Lamda,..., customer N arrives in accordance with a Poisson process with rate Lamda .

What is the distribution in which at least one customer arrive at time t?

2. Mar 19, 2009

### Focus

You can condition on the number of customers that have arrived and the exponential holding times of the Poisson process, then use the law of total probability.

3. Mar 20, 2009

### quacam09

Thank you for your response! Can you explain your idea in detail?

Is the following solution correct?

Consider the queueing system, there are n customers 1, 2, ...N.
Customer 1 arrives in accordance with a Poisson process with rate Lamda, customer 2 arrives in accordance with a Poisson process with rate Lamda,..., customer N arrives in accordance with a Poisson process with rate Lamda .

N processes are mutually independent and homogeneous Poisson processes with rate Lamda

=> at least one of custumers arrive the system, which occurs at the rate Lamda*N

4. Mar 20, 2009

### Focus

Yes but that is given that there is no one in the queue. Given that k people are in the queue, the arrival rate is (N-k)*lambda (k not greater than N). Now you have to use $\mathbb{P}(A)=\sum_{k \in \mathbb{N}}\mathbb{P}(A|B=k)\mathbb{P}(B=k)$.

Sorry do you actually mean the probability of at least one customer arrive by time t? At time t, you can't have two customers arriving.

5. Mar 20, 2009

### quacam09

Sorry, "At least one customer" mean we can have one, two, ...or N customer arrive by time t. N processes are mutually independent and homogeneous Poisson processes with rate Lamda. So at time t, we can have two customers arriving.

6. Mar 20, 2009

### Focus

Ok well then work out the probability of no customers arriving by time t. If N people are arriving with independent Poisson, then you have N*lambda Poisson, so the probability of no people arriving by time t is exp distributed with parameter N*lambda. This is due to the holding times of Poisson processes (that is the times between jumps) are exponential.

Sorry if my posts aren't making sense, I have been somewhat tired recently.

7. Mar 21, 2009

### quacam09

OK. Thank you for your help!