The given pmf is for a geometric distribution and is $f(x_i|\theta) = (1-\theta)^{x_i - 1}\theta; ~x_i = 1, 2 ,\cdots, $ and the 1-parameter exponential family I have obtained is; $$f(x|\theta) = \exp \left\{ \log \left(\frac{\theta}{1 - \theta} \right) + x_i \log (1-\theta)\right\}$$I have tried to obtain the conjugate prior and have arrived at this step; $$f(x_i|\theta) \propto \ \left\{ b \log \left(\frac{\theta}{1 - \theta} \right) + a \log (1-\theta) \right\}$$I know the conjugate prior has to be a Beta distribution. How do I deduce it from this last step? Also, how do I obtain the corresponding posterior distribution?

  • $\begingroup$ What do you get if you expand and simplify $\exp \left\{ b \log \left(\frac{\theta}{1 - \theta} \right) + a \log (1-\theta) \right\}$? $\endgroup$
    – Henry
    Aug 19, 2022 at 12:55
  • $\begingroup$ $b \log(\theta) + \frac{a}{b} \log(1-\theta)$ $\endgroup$
    – user365863
    Aug 19, 2022 at 12:59
  • $\begingroup$ And that gives me a Beta with $\theta^{a-1} (1-\theta)^{b-1}$ but when I turn it into $B(\alpha, \beta)$, I get $\alpha = a - 1$ and $\beta = b$ but I am not sure of the denominator i.e.; $B(\alpha, \beta)$ $\endgroup$
    – user365863
    Aug 19, 2022 at 13:02
  • 1
    $\begingroup$ Your details are not quite correct: for example $\exp \left\{ b \log \left(\frac{\theta}{1 - \theta} \right) + a \log (1-\theta) \right\} = \theta ^b (1-\theta)^{a-b}$. But yes, this is proportional to the density of a Beta distribution $\endgroup$
    – Henry
    Aug 19, 2022 at 13:13
  • $\begingroup$ See also en.wikipedia.org/wiki/Conjugate_prior $\endgroup$ Aug 19, 2022 at 13:47


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