# How should I deduce the conjugate prior and corresponding posterior for a geometric distribution

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?

• What do you get if you expand and simplify $\exp \left\{ b \log \left(\frac{\theta}{1 - \theta} \right) + a \log (1-\theta) \right\}$? Aug 19, 2022 at 12:55
• $b \log(\theta) + \frac{a}{b} \log(1-\theta)$
– user365863
Aug 19, 2022 at 12:59
• 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.; $B(\alpha, \beta)$
– user365863
Aug 19, 2022 at 13:02
• 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 Aug 19, 2022 at 13:13
• Aug 19, 2022 at 13:47