# Bayesian Estimation of CDF

i'm getting pretty confused by the following problem, hope anyone can clarify my mind:

Using a bayesian approach obtain a posteriori and interval estimations for $$\mathbf{F}_{X}(x)$$ using a Uniform(0,1) prior distribution for the parameter $$\theta(x)=F_x(X)$$ for a given sample $$x_1,x_2, \cdots, x_n$$

I have knowledge of Bayesian Statistics, and i know that the CDF is actually a Uniform(0,1) random variable.

But still this kinda confusing for me, given that a point estimate for the $$F_X(x)$$ under the frequentist aproach would be the Empirical Distribution $$F_N(x)=\frac{1}{N}\sum_{i=1}^NI(X_i \leq x)$$, but it´s obvious that this would no be the case for bayesian approach.

And even what does it could mean have the same prior distribution that the one from the model (both prior and $$F_X(x)$$ are $$U$$~$$(0,1)$$

After all is just a product of the Likelihood and the prior (over the integral that makes this integrate one) \begin{align} p(\theta|x_1,...,x_n)=\frac{ L(\theta)\pi(\theta)}{\int_{-\infty}^{\infty}L(\theta)\pi(\theta)d\theta} \end{align}

So are my thoughts correct about $$p(X\mid\theta)$$ being the likelihood of an $$U$$~$$(0,1)$$? And it´s not kinda silly take the same distribution as prior?

https://stats.stackexchange.com/questions/372340/when-is-the-posterior-distribution-equal-to-the-prior

• Also i know there´s a bayesian version of estimating the CDF which has to do with Direchlet Processes, but this looks easier or another way of getting a point estimation. Commented Mar 7, 2021 at 20:08

The random variables $$Z_i:=\mathbb{1}_{\{X_i\,\in\,B\}}$$ for $$i\in\{1,\ldots,n\}$$ are i.i.d $$\text{Bernoulli}$$ with unknown parameter $$\theta=\mathbb{P}(X\in B).$$ Using a $$\text{Uniform}(0,1)$$ non-informative prior distribution for $$\theta$$ the posterior distribution is $$\text{Beta}(1+nT_n(B),1+n(1-T_n(B))).$$ Under a quadratic loss penalization, the bayesian point estimate for $$\theta=\mathbb{P}(X\in B)$$ is the posterior expected value which is the following:
$$\theta^* \,=\, \frac{1+nT_n(B)}{2+n}$$