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 :

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

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.


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