If $X,Y$ are mutually dependent binomial random variables, do we know how $Y|X$ and $X|Y$ are distributed?
$X,Y$ are the sum of $n$ iid Bernoulli variables, \begin{align} X&=\sum_{i=1}^{n}X_i, \qquad &X_i \sim Bin(1, P(X_i=1)) \\ Y&=\sum_{i=1}^{n}Y_i, \qquad &Y_i \sim Bin(1, P(Y_i=1)). \end{align}
Wikipedia has a section on conditional binomials, which roughly states:
Take all realizations of $X_i$, which equal $1$ with probability $P(X_i=1)$. Whenever $\{X_i=1\}$, take all realizations of $Y_i$ which equal $1$ with probability $P(Y_i=1|X_i=1)$. Then, $Y$ is binomial with probability $P(Y_i=1)=P(Y_i=1|X_i=1)P(X_i=1)$.
But this seems to hold only if $P(Y_i=1|X_i=0)=0$, since the law of total probability states \begin{equation} P(Y_i=1)=P(Y_i=1|X_i=1)P(X_i=1)+P(Y_i=1|X_i=0)P(X_i=0). \end{equation}
Conversely, the sum of $n$ independent but not identically distributed (i.e. different probabilities) Bernoulli trials has a Poisson-binomial distribution. So my intuition says that if you sum those $Y_i$ where $\{X_i=1\}$ with those where $\{X_i=0\}$, you shouldn't get a binomial distribution since you are summing Bernoulli variables with different probabilities. Thus, if $Y$ is binomially distributed, $Y|X$ cannot be binomially distributed.
