Negative binomial distribution with known parameter k has the following distribution: $$ f(r;k,p)=\binom{k+r-1}{k}(1-p)^{r}p^k~~~~~\text{for}~r=0,1,2,\ldots $$ Then the join probability of $n$ Negative binomial independent variables, has distribution: $$ f(\textbf{r};k,p)=\prod_{i=1}^n\binom{k+ r_i -1}{k}(1-p)^{\sum_{i=1}^{n} r_i}p^{nk} $$ Then it can be rewritten in exponential form as: $$ \begin{align} f(k;r,p) &=\prod_{i=1}^n\binom{k+r_i-1}{k}\exp\left[\ln(p^{kn}(1-p)^{^{\sum_{i=1}^{n} r_i}})\right] \\ &=\prod_{i=1}^n\binom{k+r_i-1}{k}\exp\left[{\ln(1-p)\sum_{i=1}^{n} r_i} + kn\ln(p)\right] \\ \end{align} $$ From above we can see that the sufficient statistic for $p$ is $$T(\textbf{r}) = \sum_{i=1}^{n} r_i $$

My question is what is the distribution of $r_i$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.