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How can I write the unnormalized posterior

$ f(p_1, p_2 | Y) = (z_1-1)*log(p_1) + (n_1-z_1-1)*log(1-p_1) + (z_2-1)*log(p_2) + (n_2-z_2-1)*log(1-p_2) $

in terms of the log odds-ratio $\alpha$ and the log odds-product $\eta$?

$\alpha = log \left(\frac{p_2/(1-p_2)}{p_1/(1-p_1}\right) $ and $\eta = log \left(\frac{p_2}{1-p_2} \times \frac{p_1}{1-p_1}\right)$?

where z_1, n_1, z_2, and n_2 are constants.

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1 Answer 1

up vote 2 down vote accepted

Note that we have $$ \alpha = \log p_2 - \log(1-p_2) - \log p_1 + \log(1-p_1),$$ $$ \eta=\log p_2 -\log (1-p_2)+\log p_1 -\log(1-p_1).$$ Then $$\eta-\alpha = 2(\log p_1 - \log (1-p_1),$$ $$\eta +\alpha=2(\log p_2 -\log(1-p_2).$$

EDIT: previously I wrote that your unnormalized posterior couldn't be written in terms of $\alpha$ and $\eta$. However I just noticed that $$\frac{\eta - \alpha}{2}=\log \frac{p_1}{1-p_1}$$ so that $$p_1 = \frac{e^{(\eta-\alpha)/2}}{e^{(\eta-\alpha)/2}+1},$$ and similarly $$\frac{\eta + \alpha}{2}=\log \frac{p_2}{1-p_2}$$ so that $$p_2 = \frac{e^{(\eta+\alpha)/2}}{e^{(\eta+\alpha)/2}+1}.$$ Now since your unnormalized posterior is a function of $p_1$ and $p_2$, and of the constants $z_1,n_1,z_2,n_2$, you can using the above expressions write the posterior in terms of $\alpha$ and $\eta$.

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Note to the OP: please reread this, after the EDIT, since I figured out you can rewrite as you wanted in terms of $\alpha,\eta$. –  coffeemath Dec 9 '12 at 2:55

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