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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.

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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.

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  • $\begingroup$ It seems to me that the solution (i.e. the parameter $\lambda$ of the poisson distribution) does not depend on $p$, right? $\endgroup$ – QuantumLogarithm Mar 11 '14 at 8:31
  • $\begingroup$ Correct, for every value of $p$, the set of solutions is the same. $\endgroup$ – Did Mar 11 '14 at 8:33
  • $\begingroup$ In case the equation is $X_1=Bin(X_2,p) + Bin(X_3,1−p)$, where $X_1$, $X_2$ and $X_3$ are identically distributed, but $X_2$ and $X_3$ are not independent, nothing can be done to determine the distribution of those variables, right? $\endgroup$ – QuantumLogarithm Mar 11 '14 at 10:29
  • $\begingroup$ Unless one specifies the dependence of $(X_2,X_3)$ (and even then this is not assured). $\endgroup$ – Did Mar 11 '14 at 10:32
  • $\begingroup$ If $X$ is integrable then did you use a Taylor expansion for $g(1-t)$? Don't you have then $o(t^2)$? $\endgroup$ – QuantumLogarithm Mar 14 '14 at 12:45

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