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Consider the real-valued random variables $X,Y,Z,W,A$.

Suppose that $Z,W,A$ are mutually independent and that $Y:=g(Z,W)$ for some Borel measurable function $g:\mathbb{R}^2\rightarrow \mathbb{R}$.

Assume that $X$ has the same distribution as $Y$.

Does this imply that

(1) The distribution of $X$ conditional on $W$ is equivalent to the distribution of $g(Z,W)$ conditional on $W$? Could you give some explanations on why yes or why no and whether the independence of $Z$ from $W$ is used?

(2) The distribution of $X$ conditional on $(W,A)$ is equivalent to the distribution of $g(Z,W)$ conditional on $(W,A)$? Could you give some explanations on why yes or why no and whether the independence of $A$ from $(Z,W)$ is used?

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Assertions (1) and (2) need not be true given your assumptions. The reason is that knowing only that $X$ has the same distribution as $Y$ says nothing about the joint distribution of $X$ with $(Z,W,A)$.

An example where (1) and (2) are false: Say $Z$, $W$, $A$ are mutually independent coin tosses, each taking value $0$ or $1$ with equal probability. Let $Y:= Z+W$, and let $X$ have the same distribution as $Y$ (so $X$ has Binomial($n=2,p=1/2$) distribution), but independent of $(Z,W,A)$. Then $$P(X=0\mid W=1) = P(X=0)=\textstyle\frac14$$ while $$P(Y=0\mid W=1)=P(Z+W=0\mid W=1)=0.$$ Similarly $P(X=0\mid W=1, A=1)=\frac14$ while $P(Y=0\mid W=1, A=1)=0$. The independence of $Z$ from $W$, or the independence of $A$ from $(Z,W)$ doesn't come into play.

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  • $\begingroup$ Thanks a lot. May I ask you a further clarification? Since $X$ is distributed as $Y$, sampling from the distribution of $X$ is equivalent to sampling from the distribution of $Y$. Assuming that $g$ is known and that the distribution $f_{Z,W}$ of $(Z,W)$ is known, we can draw a realisation $(z,w)$ of $(Z,W)$ from $f_{Z,W}$ and then set $x=g(z,w)$. Is this procedure correct? Doesn't this mean assuming that $X|W=w$ is distributed as $g(Z,W)|W=w$? $\endgroup$ – STF May 3 '17 at 17:43
  • $\begingroup$ @STF Until you specify the joint distribution of $X, W, Z, A$ you don't really know how $X\mid W$ behaves. If you set $X:=Y$ (which is one way to specify the joint distribution) then for sure $X\mid W=w$ is distributed as $g(Z,W)\mid W=w$. $\endgroup$ – grand_chat May 3 '17 at 17:51
  • $\begingroup$ Are you saying that the sampling procedure above assumes $X\equiv Y$? $\endgroup$ – STF May 3 '17 at 18:00
  • $\begingroup$ @STF Yes, because your procedure is explicitly generating $y$ from $z$ and $w$, and then setting $x=y$. Knowing that $X$ is distributed like $Y$ doesn't mean that $x$ has to be generated from the same $z$ and $w$ that generated $y$. See the binomial example, where we generate $x$ from a pair of $z$ and $w$ that are independent from $y$. $\endgroup$ – grand_chat May 3 '17 at 18:05
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    $\begingroup$ @STF Yes, since $X$ has the same distribution as $g(Z,W)$. Exactly what 'close' means is up to debate but this result is just saying that the histogram for a random sample will approximate the population density as the sample size increases. $\endgroup$ – grand_chat May 3 '17 at 19:28

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