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 $\int_{0.1}^1 x \frac{x^{\alpha} (1-x)^{\beta}} {B(\alpha+1, \beta+1)} dx$

Comments:
If the prior distribution has support $[.1,1],$ then the posterior distribution has support contained in or equal to $[.1,1],$ so the posterior distribution could not be any ordinary (two-parameter) beta distribution.
Also, if one tries to use a normal prior for a binomial success
probability then it has to be truncated to have support $[0,1].$
Such a truncated normal prior is not conjugate to a binomial likelihood, so the posterior has to be computed by numerical
methods.
Moreover, there are inferential applications relating to the prevalence of a disease (which must lie in $[0,1])$ where fragmentary information from
medical screening tests in which traditional (non-Bayesian) inference can
lead to point and interval estimates of prevalence with nonsensical values outside of $[0,1].$ However, a Bayesian approach with a beta
prior must give point and interval estimates within $[0,1].$
One example is given here. [In this particular application, a Gibbs sampler is used to find the posterior distribution.]
Note: Undeniably, $\mathsf{Beta}(1,1) \equiv \mathsf{Unif}(0,1)$ is a
"flat" prior distribution for the binomial success probability $\theta.$ However, technically there is controversy whether this distribution is noninformative. The Jeffries prior $\mathsf{Beta}(.5,.5)$ has been
proposed as a more nearly noninformative prior distribution.
