Let's give a generic discrete probability space $(\Omega,P)$ where $|\Omega| \le \aleph_0$.
A random variable $X\colon \Omega \to \{0,\dots,n\}$ is a Binomial one if, by definition, $p_X(k):=P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$ where $k \in \{0,\dots,n\}$ and $p \in [0,1]$ (Is this a correct definition? Is there a more general one?).
I know that if we have $X_1,\dots,X_n\colon\Omega \to \{0,1\}$ independent Bernoulli random variables, then $X:=X_1+\dots+X_n\colon \Omega \to \{0,\dots,n\}$ is a Binomial random variable.
I'm wondering if the converse holds. Namely, if $X\colon \Omega \to \{0,\dots,n\}$ is a Binomial random variable, then can it be written as the sum of independent Bernoulli random variables?
Thank you for your time!
The answer is no.
Consider the probability space $(\Omega, \mathcal{F}, \mathbb{P})$, where $\Omega = \{0,\ldots,n\}$, $\mathcal{F} = \mathcal{P}(\Omega)$ and $\mathbb{P}: \mathcal{F} \to [0,1]$ defined as $\mathbb{P}(A) = \sum\limits_{i \in A} \binom ni p^i(1-p)^{n-i}$ for every $A \in \mathcal{F}$.
Consider now $X: \Omega \to \{0,\ldots, n\}$ as $X(i)=i$ (the identity function).
The distribution of $X$ is clearly binomial.
Suppose now that $X = X_1 + \ldots + X_n$ with $X_i$ iid with law Bernoulli of parameter $p$, all defined on $(\Omega, \mathcal{F}, \mathbb{P})$.
This would mean that there is $A \in \mathcal{F}$ such that $\mathbb{P}(A) = p$.
And this must be true for every $p \in (0,1)$.
Consider the map $F: (0,1) \to \mathcal{P}(\{0,\ldots,n\})$ that for a given $p$ returns such a set $A$. Since the domain is infinite and the codomain is finite there must be a given $A \subseteq \{0,\ldots,n\}$ such that for infinite many $p$ we have $$\sum\limits_{i \in A} \binom ni p^i(1-p)^{n-i} = p$$ Therefore we have $\sum\limits_{i \in A} \binom ni x^i(1-x)^{n-i} = x$ for every real $x$ (polynomial identity).
And now, for example with $n=2$, we obtain that the LHS can be equal to $0,1,x^2,(1-x)^2,2x(1-x),1-x^2,1-(1-x)^2,1-2x(1-x)$, none of which is equal to $x$.