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

Denisse Fitzpatrick

Answered question


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 ) .

Answer & Explanation

Reagan Tanner

Reagan Tanner

Beginner2022-10-03Added 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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