Question about how to merge poisson distribution

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Discussion Overview

The discussion revolves around the merging of Poisson distributions, specifically how to combine independent Poisson random variables. Participants seek clarification on the underlying principles, proofs, and practical applications of this concept, including a word problem to illustrate the idea.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant states that if A~Po(a) and B~Po(b) are independent, then C = (A+B)~Po(a+b) and requests an intuitive explanation and example.
  • Another participant suggests finding the Moment Generating Function (MGF) of a Poisson distribution to understand the merging process and poses questions about the implications of calculating the MGF for independent variables.
  • A third participant provides a practical example involving two operators in an emergency room, detailing their call answering rates and asking how to calculate the probability of failing to answer calls using the combined Poisson variables.
  • A later reply defines MGF as Moment Generating Function and emphasizes its importance in probability, directing participants to additional resources regarding its properties.

Areas of Agreement / Disagreement

Participants express different aspects of the merging process and its applications, but there is no consensus on a single method or example. The discussion remains exploratory with multiple viewpoints presented.

Contextual Notes

Participants have not fully resolved the mathematical steps involved in merging the distributions or the implications of using MGF in this context. The example provided raises additional questions about the assumptions made in the scenario.

Who May Find This Useful

This discussion may be useful for students and practitioners in probability and statistics, particularly those interested in Poisson distributions and their applications in real-world scenarios.

gokuls
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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.
 
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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?
 
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?
chiro said:
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?
 

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