Show that this probability distribution is an exponential family 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?
 A: 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,
$$f(y;\theta,\phi)=\exp\biggl\{\frac{y\theta-b(\theta)}{a(\phi)}+c(y;\phi)\biggr\}.$$
