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Jan 3, 2019 at 18:58 comment added Xi'an The mean of the $\text{Beta}(k\theta,k(1-\theta))$ distribution is $\theta$. And $k$ sets the imprecision about this value.
Jan 3, 2019 at 18:23 comment added Student Agreed. But do you seen any reason to have a prior distribution $\text{Beta}(k\theta,k(1-\theta))$ with its specific parameters?
Jan 3, 2019 at 17:39 comment added Xi'an (a) use sufficiency as $\sum_i X_i\sim \text{Neg}(n,\theta)$ and (b) this is not a fully Bayesian approach to the problem
Jan 3, 2019 at 17:28 comment added Student If $X$ is a random vector of samples such that $X = (X_1,X_2,\cdots,X_n)$ then each $X_i \sim \text{Geo}(\theta).$ I don't see why $X_i\sim \text{Neg}(n,\theta)?$ Furthermore it is not certain the proportion in $2018$ is the same as that in $2017.$ We are modelling the unknown parameter $p$ (in $2018$) with beta distribution whose parameters use the proportion obtained in $2017.$
Jan 2, 2019 at 8:06 history answered Xi'an CC BY-SA 4.0