# Distribution of empirical frequency

Suppose that, for a given $$n \in \mathbb{N}$$, I draw points $$x_1,...,x_n$$ uniformly in $$[0,1]$$ and independently from each other.

What would be the distribution of the empirical frequency of points falling before $$a \in [0,1]$$, i.e the distribution of the statistic :

$$f_{n,a} = \frac{\#\{i \in 1,...,n \text{ such that }x_i

Obviously, it's mean will be $$a$$, but i cant work out the proper distribution of this stat for all $$a,n$$.

Someone ?

Let $$X_1,\dots,X_n \sim \mathcal{U}_{[0,1]}$$.
For $$a \in [0,1]$$, $$\mathbb P(X_i \leq a)=a$$.

Thus, $$Y_n = \sum_{i=1}^n I_{X_i \leq a} \sim \text{Bin}(n,a).$$

The support of the empirical frequency, $$\frac{Y_n}{n}$$, is $$B_n:=\{ \frac{i}{n}, 0 \leq i \leq n \}$$ and for $$s \in B_n$$:

\begin{align*} \mathbb{P}( n^{-1} Y_n = s ) &= \mathbb{P}(Y_n = ns ) \\ &= \binom{n}{ns} a^{ns}(1-a)^{n(1-s)} \end{align*}

This result generalizes for any cumulative distribution function $$F$$ and for $$t \in \mathbb R$$ if $$X_i \sim F$$ then, $$Y_n = \sum_{i=1}^n I_{X_i \leq t } \sim \text{Bin}(n, F(t))$$

If I understood the question correctly, it's a Beta distribution, where parameters are:

$$\alpha$$ is the count of items < $$a$$,

$$\beta$$ is the count of items >= $$a$$.

You could think of the procedure as a Bernoulli process, where a success is drawing a sample above $$a$$ and failure - below $$a$$, which obviously has $$p(success) = a$$, for $$a \in [0,1]$$.

I've flipped the logic of success and failure above to avoid writing $$(1-a)$$ for clarity's sake, but the cases are symmetrical.