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Suppose that, for year $t$, the data $y$ is Poisson with mean $a + bt$. Assume also a uniform prior on $(a,b)$. If we have $n$ years of data then I think the posterior for $(a,b)$ will be

\begin{align*} p(a,b | y) &\propto p(y|a,b)p(a,b) \\ &\propto p(y | a, a) \\ &=\prod_{i=1}^n {\rm Poisson}(y_i | a+ bt_i) \\ &\propto \prod_{i=1}^n (a+ bt_i)^{y_i} e^{-(a+ bt_i)} \\ &= e^{-(na + b\sum_{i=1}^n t_i)}\prod_{i=1}^n (a + bt_i)^{y_i} \\ \end{align*}

What I'm struggling with is how to find the sufficient statistics. At first I thought maybe continue like this

\begin{align*} e^{-(na + b\sum_{i=1}^n t_i)}\prod_{i=1}^n (a + bt_i)^{y_i} &=e^{-(na + b\sum_{i=1}^n t_i)}e^{\sum_{i=1}^n y_i \log(a+bt_i)} \end{align*}

which seems promising but I'm confused about how to deal with the argument of the log since it also depends on $i$.

Any suggestions?

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    $\begingroup$ There is no one-dimensional sufficient statistics for this model. $\endgroup$ – Zen Feb 14 '14 at 22:39
  • $\begingroup$ @Zen Thanks Zen. Would we be able to find multidimensional sufficient statistics? $\endgroup$ – tony2785 Feb 15 '14 at 23:09
  • $\begingroup$ There is no factorisation in the likelihood, hence no fixed dimension sufficient statistic. $\endgroup$ – Xi'an Dec 9 '14 at 13:30

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