Denisse Fitzpatrick

2022-10-02

Probability of events for beta-geometric distribution

If we have a beta-geometric distribution X with pdf $P(X=k\mid \alpha ,\beta )=\frac{Beta(\alpha +1,k+\beta )}{Beta(\alpha ,\beta )}$.

If we have a beta-geometric distribution X with pdf $P(X=k\mid \alpha ,\beta )=\frac{Beta(\alpha +1,k+\beta )}{Beta(\alpha ,\beta )}$.

Reagan Tanner

Beginner2022-10-03Added 8 answers

Step 1

To avoid confusion you should not use $\beta $ for both the parameter and the beta function. I suggest using B for the beta function.

Further $\mathsf{P}(k;\alpha ,\beta )$ 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. $\mathsf{P}(X>x)\text{}=\text{}\mathsf{P}(Xx-1)\text{}\mathsf{P}(X0)\phantom{\rule{0ex}{0ex}}=\text{}(1-t{)}^{x-1}\text{}(1-\mathsf{P}(X=0))$

So the reasoning that follows from there is invalid because it is based on an unjustifiable premise.

To avoid confusion you should not use $\beta $ for both the parameter and the beta function. I suggest using B for the beta function.

Further $\mathsf{P}(k;\alpha ,\beta )$ 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. $\mathsf{P}(X>x)\text{}=\text{}\mathsf{P}(Xx-1)\text{}\mathsf{P}(X0)\phantom{\rule{0ex}{0ex}}=\text{}(1-t{)}^{x-1}\text{}(1-\mathsf{P}(X=0))$

So the reasoning that follows from there is invalid because it is based on an unjustifiable premise.

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