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Consider a discrete distribution $P(\mathbf{X},Y)$ with $X = \{ X_1, \dots, X_N \}$. I use the shorthand notation $p(\mathbf{x}, y)$ for $P(\mathbf{X}{=}\mathbf{x}, Y{=}y)$. Consider $P_{ind}(\mathbf{X},Y)$ defined as $p_{ind}(\mathbf{x}, y) = p_{ind}(\mathbf{x} \vert y) p(y)$ where $$ p_{ind}(\mathbf{x} \vert y) := \prod_{n=1}^N p(x_n \vert y) $$ Is there a way to generate a random variable $\mathbf{X}'$ through some (deterministic or non-deterministic) transformation of $\mathbf{X}$ such that $P(\mathbf{X}', Y) = P_{ind}(\mathbf{X},Y)$?

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  • $\begingroup$ Are you satisfied with the answer below? $\endgroup$ Aug 4 at 14:32
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You want to simulate the joint discrete distribution of $(X_1,\dots,X_N,Y)$, knowing the following: (i) $X_1,\dots,X_N$ are conditionally independent and identically distributed given $Y$, (ii) the probability mass function (pmf) $p_Y$ of $Y$, and (iii) the conditional pmf's $p_{X|Y}$ of each $X_n$ given $Y$. This simulation is done in a straightforward manner.

Indeed, let $y_1,y_2,\dots$ be the distinct possible values of $Y$ and let $x_1,x_2,\dots$ be the distinct possible values of each $X_n$. For $m=0,1,\dots$, $k=0,1,\dots$, and $j=1,2,\dots$, let $$t_m:=\sum_{j=1}^m p_Y(y_j),\quad s_{k|j}:=\sum_{i=1}^k p_{X|Y}(x_i|y_j), $$ so that $t_0=0$ and $s_{0|j}=0$ for all $j$.

Let $U_1,\dots,U_N,V$ be independent random variables (r.v.'s) each uniformly distributed on $[0,1)$. Define the r.v.'s $X'_1,\dots,X'_N,Y'$ as follows: for any $m=1,2,\dots$, $k=1,2,\dots$, and $n=1,\dots,N$, $$Y':=y_m\quad\text{if}\quad V\in[t_{m-1},t_m),$$ $$X'_n:=x_k\quad\text{if}\quad Y'=y_m\ \text{and}\ U_n\in[s_{k-1|m},s_{k|m}).$$

Then the joint distribution of $(X'_1,\dots,X'_N,Y')$ will be the same as that of $(X_1,\dots,X_N,Y)$.

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