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Random Talk on Random Thoughts

Compared Two Poisson Variables

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Background

Last Friday, I had to submit a homework which required me to evaluate $\Pr(A > B)$ and $\Pr(A = B)$, where $A$ and $B$ were two independent Poisson random variables with parameters $\alpha$ and $\beta$ respectively.

Problem

I then started evaluating the sum.

\[ \Pr(A > B) = \sum_{i = 1}^\infty \sum_{j = 0}^{i - 1} \frac{e^{-\alpha} \alpha^i}{i!} \cdot \frac{e^{-\beta} \beta^j}{j!} \]

Then I was stuck. I couldn’t compute this sum also.

\[ \Pr(A = B) = \sum_{i = 0}^\infty \frac{e^{-(\alpha + \beta)} \alpha^i \beta^i}{(i!)^2} \]

Fact

I googled for a solution for hours, and after I saw equation (3.1) in a paper, I gave up finding exact solutions.1 As a supporter of free software, I avoided using M$ Ex*, and wrote a program in C++ to approximate the above probabitities by directly adding them term by term.

Source code

Sample output

Assume that Poisson r.v. A and B are indepedent
Parameter for A: 1.6
Parameter for B: 1.4
Number of terms to be added (100 <= N <= 1000): 8
P(A > B) = 0.423023, P(A < B) = 0.335224, P(A = B) = 0.241691

Lessons learnt

  1. A one-line method for writing the content of a function which returns the factorial of a number.

    URL: http://progopedia.com/example/factorial/

  2. Evaluation of a function inside GDB

    URL: http://stackoverflow.com/q/1354731/


  1. Keller, J. B. (1994). A characterization of the Poisson distribution and the probability of winning a game. The American Statistician, 48(4), 294–298.

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