Bayes estimator from Beta prior Geometric data Let $X$ be random variable of geometric distribution $f_{\theta}(x)=(1-
\theta)\cdot \theta^x$ for $x=0,1,\dots .\;$ 
Find Bayes estimator of $
\theta$ calculated on a basis of observation $X=0$ which is $E(\Theta|X=0)$. 
My problem is with a priori distribution which is $\pi(\theta)=3\theta^2,\theta\in(0,1).$ Why is that? 
 A: The prior distribution $\theta \sim \mathsf{Beta}(3,1)$ with density
function $\pi(\theta) = 3\theta^2 = 3\theta^{3-1}(1-\theta)^{1-1},$ for 
$0 < \theta < 1,$  has
mean $E(\theta) = 3/(3+1) = 3/4$ and places probability 0.875 in $(.5, 1).$
You can learn more about beta distributions from your text or from the
Wikipedia article. The probability 0.875 can be obtained by integration
of the density function or from software. The computation in R statistical
software is shown below.
1-pbeta(.5, 3, 1)
## 0.875


Thus you begin this inferential procedure with the prior information or
  personal belief that $\theta,$ which must lie in $(0,1)$ is noticeably greater
  than $1/2.$ All Bayesian inferences begin with a prior distribution.

The geometric likelihood function corresponding to data $x = 0$ is 
$\pi(x|\theta) = (1 - \theta)\theta^x = \pi(0|\theta) = (1-\theta).$
Following my Comment, by Bayes' Theorem, the posterior distribution is
$$\pi(\theta|x) \propto \pi(\theta) = \theta^2 \times (1 - \theta)
\propto \theta^2(1-\theta) = \theta^{3-1}(1-\theta)^{2-1},$$
where the symbol $\propto$ (read 'proportional to') recognizes that we
are using the kernels of the prior and posterior density functions
and omitting the constants, which are not necessary. 
We recognize the
posterior kernel as that of $\mathsf{Beta}(3,2),$ which has mean $3/(3+2) = 0.6.$
Thus $E(\Theta|0) = 0.6.$ The central idea of Bayesian estimation is that
the information in the prior distribution and the data are combined to 
yield a posterior distribution. In this case the data $X=0$ has reduced the
estimated mean
from 0.75 to 0.6.
This easy identification of the posterior distribution is possible because
the prior distribution and the likelihood function are conjugate (that is
'mathematically compatible') so that the kernel of the posterior density
is easily recognized.
Note: A 95% Bayesian posterior interval estimate of $\Theta$ is $(.194, .932).$
qbeta(c(.025,.975), 3, 2)
## 0.1941204 0.9324140

