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If I have a Poisson regression such that $\lambda = \alpha + \beta t$, $\alpha + \beta t \geq 0$ $\forall t, \alpha, \beta$ and $Y_t \sim \textrm{Poisson}(\lambda_t)$ for which I have 10 observations from $t=0$ to $t=9$ then the likelihood is:

$$\prod_{t=0}^9 \left\{\frac{(\alpha+\beta t)^{y_t}}{y_t!}e^{-(\alpha+\beta t)} I_{\alpha+\beta t} \geq 0\right\}$$

If I set the prior to be proportional to 1, then the likelihood is proportional to the posterior, so mathematically the expressions for deriving the MLE of $(\alpha, \beta)$ and the posterior distribution are the same.

However, are the sufficient statistics the same? I know that sufficient statistics for $(\alpha, \beta)$ with regards to obtaining the MLE in the linear model are:

$$\sum_{t=0}^9 y_t\ \text{ and }\ \sum_{t=0}^9 y_tt$$ for $\alpha$ and $\beta$, respectively (which I also have trouble seeing). But does this also apply for the posterior distribution?

Xi'an
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user3821273
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    Yes, the concept of sufficient statistics can be defined both in classical and bayesian setting, and gives equivalent results. (except, maybe, in some exotic infinite-dimensional settings). – kjetil b halvorsen Mar 09 '15 at 12:45
  • I edited your question. However the statistic $(\sum_t y_t,\sum y_t t)$ is not sufficient for the Poisson regression, where there is no non-trivial sufficient statistic... – Xi'an Mar 09 '15 at 17:44

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