# Bayesian vs. Frequentist inference [duplicate]

I have a $X_1$,$X_2$,...,$X_n$ sequence of binary random interchangeable variables. Given $\theta$*=$\theta$, $X_1$,$X_2$,...,$X_n$ $\thicksim$ iid Ber($\theta$)

With a prior distribution $\pi$($\theta$)=Beta($\theta$).

Thus, I have a posterior distribution $\pi$($\theta$|x)$\propto$Beta($\alpha$+r,$\beta$+n-r)

E($\theta$*|x)=$\frac{\alpha+r}{n+\alpha+\beta}$

Var($\theta$|x)=$\frac{1}{(\alpha+\beta+n)(\beta+n-r+1)(\alpha+r+2)}$

Confidence Interval = $\frac{\alpha+r}{n+\alpha+\beta}$-$Z_{\frac{\alpha}{2}}$$\frac{1}{(\alpha+\beta+n)(\beta+n-r+1)(\alpha+r+2)} < \theta < \frac{\alpha+r}{n+\alpha+\beta}+Z_{\frac{\alpha}{2}}$$\frac{1}{(\alpha+\beta+n)(\beta+n-r+1)(\alpha+r+2)}$

On the other hand, when i do frequentist inference, the maximum likelihood estimator is $\theta_{ML}$=$\frac{r}{n}$ and its variance is Var($\theta_{ML}$)=$\frac{r(1-r)}{n}$

Confidence Interval = $\frac{r}{n}$-$Z_{\frac{\alpha}{2}}$($\frac{r(1-r)}{n})^\frac{1}{2}$ < $\theta$ < $\frac{r}{n}$+$Z_{\frac{\alpha}{2}}$($\frac{r(1-r)}{n})^\frac{1}{2}$

The point estimation is different, the confidence interval also is. The expectations are the same only if $\alpha$=$\beta$=0, but even then, the variances are different.

My question is, when is the frequentist and bayesian inference the same? I read somewhere that it was the same for this example, but I strongly disagree. The only case in which the inference would be the same is if in the bayesian case, n tended to infinity and then the relative weight of the prior distribution would be decline, and it would tend to be the same as the frequentist inference. Am I correct?

Second question: If I have a Beta distribution and I want to construct the confidence intervals, should i use a normal distribution? Is there another option?

Thanks!