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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.

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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.

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  • $\begingroup$ This is all exactly right. The question is what choice of priors on p, lambda_a, and lambda_b yield equivalent posterior distributions for my quantities R and R1. I am guessing that identical gamma priors on the lambdas and a Jeffrey's prior--i.e. beta(0.5,0.5)--on p will do it, but I am having trouble verifying this. $\endgroup$ – Jacob Socolar May 12 '16 at 22:12
  • $\begingroup$ +1, by the way, for an exceptionally good explanation of the equivalence. $\endgroup$ – Jacob Socolar May 12 '16 at 22:27

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