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I am looking at the Wikipedia entry for empirical Bayes, but it's a bit confusing - it seems to me the solution must apply only to the case in which there's only $n=1$ sample $y$ for each $\theta$ and the "sample mean" that's referred to is really just a single value. If this were not the case, I would think that the shrinkage factors would depend on the sample size for each $\theta$.

What if I want to generalize this to where there are multiple samples of $y$ for each Poisson mean $\theta$? Is this the formula for the posterior on $\theta_k$ (the Poisson mean for the $k^{\rm th}$ group): $$ \theta_k = \frac n {(n + \beta)} \times \text{sample_mean}(y_k) + \frac \beta {(n + \beta)} \times (\alpha \times \beta) $$ if so what's the derivation for this?

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  • $\begingroup$ Welcome to the site, @user44285. I took the liberty of editing your post & trying to use the site's $\LaTeX$ markup features to make your equation more readable. Please ensure it still says what you want it to. $\endgroup$ – gung Apr 23 '14 at 2:20
  • $\begingroup$ thanks, i made a correction to my guess. the wikipedia example is strange and i'm starting to wonder if it has a typo $\endgroup$ – user44285 Apr 23 '14 at 3:14
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It appears you are correct that there is an error on wikipedia and their formula only applies when n=1, but your formula is not correct either. If you had multiple observations for each $\theta_k$, then you could derive the posterior for each $\theta_k$ to be

$ p( \theta_k \vert Y_{k1},...,Y_{kn}) \sim gamma(\alpha + \sum_i Y_{ki}, (n + 1/\beta)^{-1}) $

$ E( \theta_k \vert Y_{k1},...,Y_{kn}) = \frac{\alpha + \sum_i Y_{ki}}{n + 1/\beta} = \bar{Y}_k\frac{n\beta}{n\beta+1} + \alpha\beta\frac{1}{n\beta+1} $

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