Let X have the geometric pmf ?(1 ? ?)x?1, x =1, 2, 3, . .

Chapter 6, Problem 3E

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QUESTION:

Let \(X\) have the geometric pmf \(\theta(1-\theta)^{x-1}, x=\) \(1,2,3, \ldots\), where \(\theta\) is beta with parameters \(\alpha\) and \(\beta\). Show that the compound pmf is

\(\frac{\Gamma(\alpha+\beta) \Gamma(\alpha+1) \Gamma(\beta+x-1)}{\Gamma(\alpha) \Gamma(\beta) \Gamma(\alpha+\beta+x)}, \quad x=1,2,3, \ldots\)

With \(\alpha=1\), this is one form of Zipf's law,

\(\frac{\beta}{(\beta+x)(\beta+x-1)}, \quad x=1,2,3, \ldots\) .

Equation Transcription:

 

 ,

 

Text Transcription:

X  

theta(1-theta)^x-1, x= 1,2,3,…,      

Theta

Alpha

Beta

gamma (alpha+beta) gamma (alpha+1) gamma (beta+x-1)/gamma (alpha) gamma (beta)gamma (alpha +beta +x),     x=1,2,3,…

alpha=1

beta/(beta+x)(beta+x-1),      x=1,2,3,…

Questions & Answers

QUESTION:

Let \(X\) have the geometric pmf \(\theta(1-\theta)^{x-1}, x=\) \(1,2,3, \ldots\), where \(\theta\) is beta with parameters \(\alpha\) and \(\beta\). Show that the compound pmf is

\(\frac{\Gamma(\alpha+\beta) \Gamma(\alpha+1) \Gamma(\beta+x-1)}{\Gamma(\alpha) \Gamma(\beta) \Gamma(\alpha+\beta+x)}, \quad x=1,2,3, \ldots\)

With \(\alpha=1\), this is one form of Zipf's law,

\(\frac{\beta}{(\beta+x)(\beta+x-1)}, \quad x=1,2,3, \ldots\) .

Equation Transcription:

 

 ,

 

Text Transcription:

X  

theta(1-theta)^x-1, x= 1,2,3,…,      

Theta

Alpha

Beta

gamma (alpha+beta) gamma (alpha+1) gamma (beta+x-1)/gamma (alpha) gamma (beta)gamma (alpha +beta +x),     x=1,2,3,…

alpha=1

beta/(beta+x)(beta+x-1),      x=1,2,3,…

ANSWER:

Step 1 of 3

Definition geometric probability:

Probability density function of the beta distribution:

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