Solution Found!
Let X have the geometric pmf ?(1 ? ?)x?1, x =1, 2, 3, . .
Chapter 6, Problem 3E(choose chapter or problem)
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: