# 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.

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.

• 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. – Jacob Socolar May 12 '16 at 22:12
• +1, by the way, for an exceptionally good explanation of the equivalence. – Jacob Socolar May 12 '16 at 22:27