What happens if I change the range of a flat prior for Bayesian inference? I am working through an example on doing Bayesian inference on binomial distribution using a flat prior, and trying to understand the impact of choosing a prior. I know that if we work with a flat prior $\theta \in [0, 1]$, it is equivalent to $\theta \sim B(1,1)$, and we can use the conjugate pair of binomial distribution to conclude the posterior as $\theta \sim B(\alpha+1, \beta+1)$. Therefore our point estimate of $\theta$ equals to the mean of the distribution, $= \frac {\alpha+1} {\alpha + \beta + 2}$. However, I want to know what would happens if we choose the prior with a different uniform distribution, say $\theta \in [0.1,1]$. I don't know if the concluded posterior would be the same ($\theta \sim B(\alpha+1, \beta+1)$), but the point estimate would change; or if the concluded posterior would be different. I have thought either:

*

*do $E[\frac{X+a}{b}] = \frac {E[X] + a}{b}$, with the assumption of the new $\theta = \frac{x-0.1}{1-0.1}$; or,

*changing the calculation of the mean by changing the integration range

but both method seems wrong. So I'm stuck at this problem and would like some advice on this.
Thanks a lot for your time!
* Note: I know in general we should not use flat prior, and using normal prior gives a much better result, but this is kind of the point of the example I'm working on, on investigating the effect of selecting the prior.
Edit1: by changing the calculation of the mean, I mean normally, we use $\int_0^1 x \frac{x^{\alpha} (1-x)^{\beta}} {B(\alpha+1, \beta+1)} dx = \frac {\alpha+1} {\alpha + \beta + 2}$. To calculate the mean where $\theta \in [0,1]$, we do $\int_{0.1}^1 x \frac{x^{\alpha} (1-x)^{\beta}} {B(\alpha+1, \beta+1)} dx$

The standard proportionality result for the posterior still holds, but the posterior is now concentrated on the same restricted set as the prior.  To see this, consider the general case where you restrict your prior to the set $\mathscr{D}$.  If you use a prior proportionate to $\pi(\theta) \cdot \mathbb{I}(\theta \in \mathscr{D})$ then you get the posterior:
$$\pi(\theta|\mathbf{x}) \propto L_\mathbf{x}(\theta) \cdot \pi(\theta) \cdot \mathbb{I}(\theta \in \mathscr{D}),$$
which is proportionate to the unrestricted posterior, but restricted over the set $\mathscr{D}$.
