Why does the MAP differ from the MLE for the uniform prior in Laplace's Rule? Laplace's Rule of Succession produces an estimate for the probability $p$ of a Bernoulli distribution. It starts with a $Beta(1,1)$ prior (equivalent to a uniform distribution prior on $(0,1)$), and then obtain the Maximum A Posterior (MAP) estimator $\frac{k+1}{n+2}$, where $k$ is the number of success in $n$ trials.
Why does this MAP estimator differ from the MLE estimator of $\frac{k}{n}$ despite having a uniform prior? Especially when it is obvious from the definition of the MAP that if the prior distribution $g(p)$ is a constant function, then MLE = MAP.
Attempts:

*

*We could say that the $Beta(1,1)$ prior assigns $0$ to the endpoints $p = 0,1$ and hence is not truly uniform/constant. But aren't the endpoints irrelevant for the uniform distribution? The uniform distribution on $(0,1)$ and on $[0,1]$ only differs on a set of measure zero and so should be equivalent?


*The derivative of the likelihood function in the derivation of the MLE contains $p$ and $1-p$ in the denominator of fractions. Which means the endpoints have to be excluded from the MLE calculations. Why then do we need to include these endpoints for MLE = MAP to hold?


*In the definition of the MAP, how do I know what the domain of the prior $g(p)$ should be?

There is no such thing as an "uninformative" prior, that brings no information to a model.
For beta-binomial model, with uniform prior $\mathcal{B}(1, 1)$ the mode of the posterior (MAP) is $\frac{x+1-1}{x+1+x-n+1-2} = \frac{x}{x+n-x} = \frac{x}{n}$, so it's the same as MLE.
If you want the mean of the posterior to be equal to MLE, there's another prior, though an improper one, that leads to the same solution as MLE: it's the Haldane's prior $\mathcal{B}(0, 0)$.
