Question about how to merge poisson distribution

[gokuls]

In general, if A~Po(a) and B~Po(b) are independent random variables, then C = (A+B)~Po(a+b). Can someone please explain the intuition/simple proof of this and a word problem or example would really help to reinforce this concept. Thanks.

 

[chiro]

Hey gokuls and welcome to the forum.

Hint: Find the MGF of a Poisson distribution. What does it mean to calculate the MGF of A + B if A and B are independent variables? What is the form of the product of the two MGF's? What does this say about the distribution?

 

[gokuls]

Hi Chiro,
This concept is used to solve questions as the one following.

The emergency room switchboard has two operators. One operator answers calls for doctors and the other deals with enquiries about patients. The first operator fails to answer 1% of her calls and the second operator fails to answer 3% of his calls. On a typical day, the first and second telephone operators receive 20 and 40 calls respectively during an afternoon session. Using the Poisson distribution find the probability that, between them, the two operators fail to answer two or more calls during an afternoon session.

The two events are independent in the question above and to calculate the probability they both happen, you have to apparently combine the Poisson variables. By the way, what does MGF mean?

Hey gokuls and welcome to the forum.

Hint: Find the MGF of a Poisson distribution. What does it mean to calculate the MGF of A + B if A and B are independent variables? What is the form of the product of the two MGF's? What does this say about the distribution?

 

[chiro]

MGF: Moment Generating Function, a very important thing in probability:

http://en.wikipedia.org/wiki/Moment-generating_function

Take note of the properties regarding multiplication of MGF's.

 

[blue_raver22]

The Poisson distribution is a probability distribution that is often used to model the number of events that occur in a specific time period or space. It is characterized by a single parameter, lambda (λ), which represents the average number of events per time period or space.

To understand how to merge Poisson distributions, it is important to first understand the concept of independence in probability. Two random variables are said to be independent if the occurrence of one does not affect the occurrence of the other. In the context of the Poisson distribution, this means that the number of events in one time period or space does not affect the number of events in another time period or space.

Now, let's consider two independent random variables, A~Po(a) and B~Po(b). This means that A and B have their own respective Poisson distributions, with parameters a and b. The probability of A taking on a specific value is given by P(A=k) = (a^k * e^-a) / k!, and the probability of B taking on a specific value is given by P(B=k) = (b^k * e^-b) / k!.

To find the probability of the merged random variable, C = (A+B), we need to consider all the possible combinations of values that A and B can take on. For example, if A=2 and B=3, then C=5. We can find the probability of this specific combination by multiplying the individual probabilities of A and B, i.e., P(A=2) * P(B=3) = [(a^2 * e^-a) / 2!] * [(b^3 * e^-b) / 3!].

In general, the probability of C taking on a specific value k is given by P(C=k) = Σ P(A=i) * P(B=k-i), where i takes on all possible values from 0 to k.

To simplify this, we can use the binomial theorem to expand (a+b)^k, which gives us (a+b)^k = Σ (k choose i) * a^i * b^(k-i). This is essentially the same as the formula for the probability of C taking on a specific value, with (k choose i) representing the number of ways to choose i events from a total of k events.

Therefore, we can see that the merged random variable C follows a Poisson distribution with parameter (a+b