Solution of equation of binomial random variables Is it possible to find the probability distribution of the random variable $X$ that solves the following equation?
$$
X = Bin(X, p) + Bin(X, 1-p),
$$
where $Bin(X,p)$ is a random variable distributed according to a binomial distribution with parameters $X$ and $p$, $p \in (0, 1)$, the two random variables on the right side are independent.
 A: 
The solutions do not depend on $p$ in $(0,1)$. These are the Poisson distributions and the Dirac mass at $0$.

To show this, first recall that, for every random variable $Y$ binomial $(n,p)$ and every $|s|\leqslant1$, the generating function of $Y$ is $$E(s^Y)=(ps+q)^n.$$ 
Considering $g(s)=E(s^X)$, the identity in distribution in the question translates as 
$$g(s)=g(ps+q)g(qs+p). 
$$
Considering $h(t)=-\log g(1-t)$, one gets
$$h(t)=h(pt)+h(qt).
$$
Iterating this, one gets, for every $n\geqslant1$,
$$
h(t)=\sum_{k=0}^n{n\choose k}h(p^kq^{n-k}t).
$$
Assume that $X$ is integrable with $\lambda=E(X)$, then $g(1-t)=1-\lambda t+o(t)$ when $t\to0$. If $\lambda=0$, $X=0$ with full probability. Otherwise, $h(t)\sim\lambda t$ when $t\to0$. 
Fix some $\lambda'\lt\lambda$, then $h(t)\geqslant\lambda' t$ for every $t$ small enough. When $n\to\infty$, $p^kq^{n-k}\to0$ uniformly in $k$, hence, for every $t$ in $(0,1)$, for every $n$ large enough,
$$
h(t)\geqslant\sum_{k=0}^n{n\choose k}\lambda'p^kq^{n-k}t=\lambda' t.
$$
The same iteration shows that, for every $\lambda''\gt\lambda$, $h(t)\leqslant\lambda''t$, thus, $h(t)=\lambda t$ for every $t$ in $(0,1)$, that is, $g(s)=\mathrm e^{-\lambda(1-s)}$ for every $s$ in $(0,1)$, which means that $X$ is Poisson with parameter $\lambda$.
If $X$ is a non integrable solution, then, for every $\lambda$,  $h(t)\geqslant\lambda t$ for every $t$ small enough, and yet again the same reasoning yields $h(t)\geqslant\lambda t$ for every $t$, hence $h(t)$ is infinite, that is, $g(s)=0$, which is absurd.
