# Conditional probability containing two random variables

-Define $X$ and $X_1$ to be two dependent non-negative random variables.
Specifically, $X_1$ depends on two (continuous) random variables, termed $X_1^{(1)}$ and $X_1^{(2)}$, and $X$ depends on these two random variables in addition to a third one, termed $X^{(3)}$ (which is also continous r.v.).
-In a similar fashion, we define $Y$ and $Y_1$ to be two dependent non-negative random variables: $Y_1$ depends on $Y_1^{(1)}$ and $Y_1^{(2)}$, and $Y$ depends on $Y_1^{(1)}$, $Y_1^{(2)}$ and $X^{(3)}$.
-Note that $X$ and $Y$ are dependent on each other; they both depends on the same random variable $X^{(3)}$. We also note that $X_1$ is independent of $Y$, and $Y_1$ is independent of $X$.
-Let $a$, $b$, $c$, and $d$ be some non-negative constant values that the random variables can take.

Based on the above, is the following relation correct ? if so, how can we prove it ?
$$P(X_1=a,Y_1=b \mid X=c,Y=d)= P(X_1=a \mid X=c) P(Y_1=b \mid Y=d)$$

Attempt:
I tried to use Bayes’ theorem:
$$P(X_1=a,Y_1=b \mid X=c,Y=d)= \frac{P(X_1=a, X=c, Y_1=b, Y=d)}{P(X=c,Y=d)}$$
But how to proceed from here? How to decompose $P(X_1=a, X=c, Y_1=b, Y=d)$ ?

I also tried to use the following relation (but I cannot prove it analytically):
$P(X_1=a,Y_1=b \mid X=c,Y=d)= P(X_1=a \mid X=c, Y=d) P(Y_1=b \mid X=c,Y=d)$. But even if this relation is correct, can we claim that $P(X_1=a \mid X=c, Y=d)=P(X_1=a \mid X=c)$ ?

PS: Clearly, $X$, $X_1$, $Y$ and $Y_1$ are discrete random variables. $X_1^{(1)}$, $X_1^{(2)}$, $X^{(3)}$, $Y_1^{(1)}$, $Y_1^{(2)}$ are continuous random variables.

#### Solutions Collecting From Web of "Conditional probability containing two random variables"

Well if you have trouble proving something, you may want to look for a counterexample.

Maybe something like this helps. Say $X_1$, $Y_1$ and $X^{(3)}$ are independent and $X=X_1+[X^{(3)}]$ and $Y=Y_1+[X^{(3)}]$, where $[\_]$ is the greatest integer function.

Then $P(X_1=a,Y_1=b \mid X=c,Y=d)=0$ for $a\neq b+c-d$.

However, you can probably (depending on the distribution) find a combination of numbers $a\neq b +c-d$ for which $P(X_1=a \mid X=c) P(Y_1=b \mid Y=d) \neq 0.$