If we have a beta-geometric distribution X with pdf P(X=k|alpha, beta)=(beta (alpha+1,k+beta))/(beta(alpha,beta))

Denisse Fitzpatrick 2022-10-02 Answered
Probability of events for beta-geometric distribution
If we have a beta-geometric distribution X with pdf P ( X = k α , β ) = B e t a ( α + 1 , k + β ) B e t a ( α , β ) .
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Answers (1)

Reagan Tanner
Answered 2022-10-03 Author has 8 answers
Step 1
To avoid confusion you should not use β for both the parameter and the beta function. I suggest using B for the beta function.
Further P ( k ; α , β ) is a probability density function. X is a continuous random variable realised within the support of [0;1]. It certainly is not equal to its own pdf.
Step 2
So, the claims you make at the very beginning of your chain are complete absurdities. P ( X > x )   =   P ( X > x 1 )   P ( X > 0 ) =   ( 1 t ) x 1   ( 1 P ( X = 0 ) )
So the reasoning that follows from there is invalid because it is based on an unjustifiable premise.
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