Probability of events for beta-geometric distributionIf 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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Probability of events for beta-geometric distributionIf 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      )      .

Reagan Tanner · Accepted Answer

Step 1To 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 2So, the claims you make at the very beginning of your chain are complete absurdities.       P    (  X  &amp;gt;  x  )         =               P        (    X    &amp;gt;    x    −    1    )               P        (    X    &amp;gt;    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.

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

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