# Probability of events for beta-geometric distribution

If we have a beta-geometric distribution $X$ with pdf

$$P(X= k\mid\alpha, \beta) = \frac{Beta(\alpha+1,k + \beta)}{Beta(\alpha, \beta)}$$

Answer in a different forum without explanation: alt_forum

Edit: Actually, I realised the solution is to look at the definition of $P(\cdot)$ more closely. Using as reference ( Muprhy, "Machine Learning, A Probabilistic Perspective", p. 78, equation 3.31) and replacing the binomial distribution with a geometric distribution $Y$ starting at $0$ i.e. $Y = Geom(k| \theta)$ we get:

$$P(X> k\mid\alpha, \beta) = \int_{0}^{1}P(Y>k)Beta(\theta|\alpha, \beta)d\theta = \int_{0}^{1}(1-\theta)^kBeta(\theta|\alpha, \beta)d\theta = \int_{0}^{1}Beta(\theta|\alpha, \beta + k)d\theta = \frac{Beta(\alpha, \beta + k)}{Beta(\alpha, \beta)}$$

The last equality just follows from integrating the beta distribution. Also note that $Beta(\cdot|\cdot)$ is the probability density, while $Beta(\cdot)$ is the beta function.

• You seem to be using the lower-case $\beta$ to refer to either of two different things. Also, where you wrote $X = P(k\mid\alpha, \beta) = \cdots,$ I wonder if you know that "$=$" means "equals", i.e. $X$ is the same thing as what comes to the right of the "equals" sign. That is clearly not right. – Michael Hardy Aug 7 '17 at 23:33
• To bs honest with you, I did, I just thought mathematicians don't care about overloading of names (I have seen notation in papers a lot more confusing then what I wrote). I know that the Random variable is not the same as its pdf, and I know the beta function is not the same as the parameter. – baibo Aug 8 '17 at 6:26
• I figured out the answer myself and will post it in the evening. – baibo Aug 8 '17 at 12:32

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.
So, the claims you make at the very beginning of your chain are complete absurdities. $$\mathsf P(X>x)~{=~\mathsf P(X>x-1)~\mathsf P(X>0) \\ = ~(1-t)^{x-1}~(1-\mathsf P(X=0))}$$