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From the geometric distribution pmf, $f(x_i |\theta) = (1-\theta)^{x_i -1} \theta; x_i = 1, 2, \cdots$, I have obtained the 1-parameter exponential family as $$exp \left\{\log \frac{\theta}{1-\theta} + x_i \log (1- \theta) \right\} $$How can I show that the posterior mean can be written a a weighted average of the prior mean and the MLE. I have found this but can't relate it to my particular question. Thanks in advance.

EDIT: I got the prior as $\theta^{a-1} (1-\theta)^{b -1}$ and the posterior as $\theta^{a + n -1} (1-\theta)^{b + S + n -1}$, with $S = \sum_{i=1}^n x_i$.

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    $\begingroup$ What did you find as the prior and posterior distributions, the prior and posterior means and the MLE? $\endgroup$
    – Henry
    Commented Aug 19, 2022 at 9:09
  • $\begingroup$ I got the prior as $Beta(\alpha, \beta)$ i.e. $\theta^{a-1} (1-\theta)^{b -1}$ and the posterior as a beta still with $\theta^{a + n -1} (1-\theta)^{b + S + n -1}$, with $S = \sum_{i=1}^n x_i$. $\endgroup$
    – user365863
    Commented Aug 19, 2022 at 9:38
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    $\begingroup$ That posterior density (or something proportional to it) looks strange. What do you get for the the prior and posterior means and the MLE? $\endgroup$
    – Henry
    Commented Aug 19, 2022 at 9:59

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