log ratio of Poisson means versus logit of binomial probability Assume that two observed counts C1 and C2 are independent realizations of two Poisson processes:
C1 ~ Poisson(lambda1)
C2 ~ Poisson(lambda2)

I am interested in inference on the log-ratio R
R = log(lambda1/lambda2)

What relationship does R bear to R1, where
C1 ~ binomial(p, C1+C2)
R1 = logit(p)

?
To sharpen the question, I have two sub-questions:
1) If I were to put priors on lambda1, lambda2, and p in precisely such a way that the priors on R and R1 are identical, is it guaranteed that my posteriors for R and R1 (given observed C1 and C2) would be identical?
2) Is there a natural choice of prior distributions that produces the behavior pre-supposed in question 1; i.e. identical priors for R and R1?  I'm thinking (hoping) that it might be the case for some set of standard priors, like identical gamma priors for the lambdas, and beta(0.5,0.5) for p.
 A: Maybe you already know that, but I think that could help a little bit.
Let $A$ and $B$ are two independent random variables of Poisson distributions:
$$
A \sim Poisson(\lambda_A) ~~~~ B \sim Poisson(\lambda_B)
$$
Let $a$ and $b$ denote their actual (empirically measured) values of $A$ and $B$, respectively. Let $n = a + b$. Then, the conditional distribution of $A$ under condition $A+B = n$ is Binomial:
$$
(A | A+B = n) \sim Binomial(p, n), ~~ \text{where} ~~
p = \frac{\lambda_A}{\lambda_A + \lambda_B}
$$
$$
\mathbb{P}(A = x | A+B=n) = {n \choose x} \cdot p^x \cdot (1-p)^{n-x}
$$
This result is originally from an article by Przyborowski & Wileński, 1940, Homogeneity of results in testing samples from Poisson series. I've also found it later in book by Lehmann & Romano, Testing Statistical Hypotheses.
Going back to your question, note that $1 - p = \frac{\lambda_B}{\lambda_A + \lambda_B}$, so we get:
$$
logit(p) = \log(\frac{p}{1-p}) = \log(\frac{\lambda_A}{\lambda_B})
$$
I hope it'll help.

If you need a quick proof ot the property above:
According to joint probability distributions (here we use independence):
$$
\mathbb{P}(A=a, B=b) = \frac{(\lambda_A)^a \cdot e^{-\lambda_A}}{a!}
\cdot \frac{(\lambda_B)^b \cdot e^{-\lambda_B}}{b!}
$$
Assuming that $n=a+b$, $\mu=\lambda_A+\lambda_B$ and
$p = \frac{\lambda_A}{\lambda_A + \lambda_B}$, we can rewrite it as:
$$
\mathbb{P}(A=a, B=b) = (\frac{\mu^n \cdot e^{-\mu}}{n!}) \cdot
(\frac{n!}{a! \cdot (n-a)!} \cdot p^a \cdot (1-p)^{n-a})
$$
The first parenthesis is responsible for reaching the total value $n$, whereas the second splits this value between two random variables.
If you put a condition on the sum of variables $A+B=n$, it simplifies to:
$$
\mathbb{P}(A=a| A+B=n) =
{n \choose a} \cdot p^a \cdot (1-p)^{n-a}
$$ QED

Your sugesstion about two gamma and beta look promising. This sum of lambdas looks
familiar to a convolution at specific point $1$.
If PDF of Gammma($\alpha$, $\beta$) is:
$$
f_1(x) = 
\frac{\beta^{\alpha} \cdot x^{\alpha-1} \cdot e^{-\beta x}}{\Gamma(\alpha)} = 
\frac{\beta}{\Gamma(\alpha-1)} \cdot \frac{(\beta x)^{\alpha-1} \cdot
e^{-\beta x}}{\alpha-1}
$$
I use $\Gamma(\alpha) = (\alpha-1) \cdot \Gamma(\alpha-1)$ and we have Poisson multiplied by a constant in which we have $\Gamma(\alpha-1)$, which is good, because PDF of
$$
Beta(x, \alpha-1, \beta-1) =
\frac{\Gamma(\alpha + \beta -2)}{\Gamma(\alpha-1) \cdot \Gamma(\beta-1)}
\cdot x^{\alpha-1}(1-x)^{\beta -1}
$$
Now PDF of Gammma($\beta$, $\alpha$) not for $x$, but $1-x$ is:
$$
f_2(1-x) = 
\frac{\alpha^{\beta} \cdot (1-x)^{\beta-1} \cdot e^{-\alpha (1-x)}}{\Gamma(\beta)} = 
\frac{\alpha}{\Gamma(\beta-1)} \cdot \frac{(\alpha (1-x))^{\beta-1} \cdot
e^{-\alpha (1-x)}}{\beta-1}
$$
For two independent random variables, the PDF of their sum is simply a convolution.
$$
(f_1 * f_2)(1) = \int\limits_{-\infty}^{+\infty} f_1(x) \cdot f_2(1-x) dx
$$
I haven't checked whether it would lead to beta, but your way of thinking looks promising.
