Probability generating function (binomial distribution)

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SUMMARY

The probability generating function (PGF) for a binomial distribution is defined as G(s) = E(s^Y) = [(1-p) + ps]^n, where Y is a counting random variable representing the number of successes in n trials, with success probability p. The derivation involves recognizing that G(s) can be expressed as a sum of the binomial probabilities, leading to the formulation G(s) = ∑ (n choose y)(sp)^y q^(n-y). By substituting q = 1 - p, the PGF simplifies to the established form. This discussion clarifies the transition from the summation to the final PGF expression.

PREREQUISITES
  • Understanding of binomial distribution and its parameters (n, p)
  • Familiarity with probability generating functions (PGFs)
  • Knowledge of combinatorial notation (n choose y)
  • Basic calculus concepts related to expected values
NEXT STEPS
  • Study the derivation of probability generating functions for different distributions
  • Learn about the applications of PGFs in combinatorial problems
  • Explore the relationship between PGFs and moment generating functions
  • Investigate the use of PGFs in solving recurrence relations
USEFUL FOR

Students and professionals in statistics, mathematicians, and data scientists interested in probability theory and its applications, particularly in understanding binomial distributions and generating functions.

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Homework Statement


The probability generating funtion G is definied for random varibles whos range are [tex]\subset[/tex] {0,1,2,3,...}. If Y is such a random variable we will call it a counting random varible. Its probabiltiy generating function is [tex]G(s) = E(s^{y})[/tex] for those s's such that [tex]E(|s|^{y})[/tex]) < [tex]\infty[/tex].

Homework Equations



binomial distribution = [tex]\left(\stackrel{n}{y}\right)[/tex][tex]p^{y}[/tex][tex]q^{n-y}[/tex] , y = 0,1,2,3,...n and 0 [tex]\leq[/tex] p [tex]\leq[/tex] 1

The Attempt at a Solution



What i have so far is...

[tex]G(s) = E(s^{y})[/tex] = [tex]\sum[/tex] [tex]s^{y}[/tex][tex]\left(\stackrel{n}{y}\right)[/tex][tex]p^{y}[/tex][tex]q^{n-y}[/tex]

[tex]G(s) = E(s^{y})[/tex] = [tex]\sum[/tex] [tex]\left(\stackrel{n}{y}\right)[/tex][tex](sp)^{y}[/tex][tex]q^{n-y}[/tex]

not sure where to go from that. i managed to do it for the geometric random variable distribution b/c there was no "n choose y". Thanks to wiki, I know what the answer should be. The answer is G(s) = [tex][(1-p) + ps]^{n}[/tex]. I can't see how they went from what i have above to that.
 
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well i just realized that [tex]G(s) = E(s^{y})[/tex] = [tex]\sum[/tex] [tex]\left(\stackrel{n}{y}\right)[/tex][tex](sp)^{y}[/tex][tex]q^{n-y}[/tex]

is the same thing as [tex](q + sp)^{n}[/tex] .

Also by definition [tex]p + q = 1 \Rightarrow q = 1-p[/tex] which means...

[tex]G(s) = E(s^{y}) = [(1-p) + ps]^{n}[/tex]
 
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