Show that this probability distribution is an exponential family

Question

We define a probability distribution on the non-negative integers 0,1,2,...,as having point probabilities $$P_\eta(k)=G(k,\rho)\left(\frac{1}{\rho+\eta}\right)^{\rho}\left(\frac{\eta}{\rho+ \eta}\right)^k$$ for $\eta, \rho>0$. A explicit expression for $G(k,\rho)$ is not needed just that $\sum_{k=0}^{\infty}p_{\eta}(k)=1$. I have to show that this is an exponential family and find cumulant function. I have tried by converters and doing exponential and log tricks to get the expression: $$f(x) = \exp{\left(\frac{\theta x- b(\theta)}{a(\phi)} + c(x,\theta)\right)}$$ (where c is cumulant function), but have not succeeded. Can anyone help me?

Answer 1

Showing the probability distribution is an exponential family: The notation is a bit different from what I'm used to, but this looks similar to the Negative Binomial distribution with parameters $ (\rho,\eta) $; that is, $$ f(k;\rho,\eta)=G(k,\rho)\biggl(\frac{1}{\rho+\eta}\biggr)^{\!\rho}\biggl(\frac{\eta}{\rho+\eta}\biggr)^{\!k},\; k=0,1,\ldots, $$ where $\rho,\eta>0$.

Note that \begin{align*} f(k;\rho,\eta) &=\exp\Bigl\{\ln\bigl(f(k;\rho,\eta)\bigr)\Bigr\}\\ &=\exp\Biggl\{\ln\biggl(G(k,\rho)\biggl(\frac{1}{\rho+\eta}\biggr)^{\!\rho}\biggl(\frac{\eta}{\rho+\eta}\biggr)^{\!k}\biggr)\Biggr\}\\ &=\exp\Biggl\{\ln\bigl(G(k,\rho)\bigr)+\rho\ln\biggl(\frac{1}{\rho+\eta}\biggr)+k\ln\biggl(\frac{\eta}{\rho+\eta}\biggr)\Biggr\}. \end{align*} Now, define \begin{align*} \theta&=\ln\biggl(\frac{\eta}{\rho+\eta}\biggr)\implies \eta=-\frac{e^{\theta}\rho}{e^{\theta}-1}\\ b(\theta)&=-\rho\ln\biggl(\frac{1}{\rho+\eta}\biggr)=-\rho\ln\Biggl(\frac{1}{\rho-\frac{e^{\theta}\rho}{e^{\theta}-1}}\Biggr)=-\rho\ln\biggl(\frac{1-e^\theta}{\rho}\biggr)\\ a(\phi)&=1\\ c(k;\phi)&=\ln\bigr(G(k,\rho)\bigl). \end{align*} Therefore, we have shown that this is a member of the exponential family, that is, $$f(y;\theta,\phi)=\exp\biggl\{\frac{y\theta-b(\theta)}{a(\phi)}+c(y;\phi)\biggr\}.$$