# Show that the posterior mean can be written as a weighted average of the prior mean and the MLE

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 $$S = \sum_{i=1}^n x_i$$.

• What did you find as the prior and posterior distributions, the prior and posterior means and the MLE? Commented Aug 19, 2022 at 9:09
• 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 $S = \sum_{i=1}^n x_i$.
– user365863
Commented Aug 19, 2022 at 9:38
• That posterior density (or something proportional to it) looks strange. What do you get for the the prior and posterior means and the MLE? Commented Aug 19, 2022 at 9:59