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I have a simple question about conditional probabilities and independence, suppose that $X, Y$ are independent random variables while as well $N$ is a random variable (which both $X$ and $Y$ are independent to). Would it follow that the conditional random variables $X|N$ and $Y|N$ are independent random variables as well? If this is the case, what would be a proof of this? I've had this question for a while.

Thanks for the help.

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If $X$, $Y$ and $N$ are mutually independent, then by definition of independence all the following holds:

$f_{X,Y,N}(x,y,n) = f_X(x)f_Y(y)f_N(n)\tag{1}$ $f_{X,Y}(x,y) = f_X(x)f_Y(y)\tag{2}$ $f_{X,N}(x,n) = f_X(x)f_N(n)\tag{3}$ $f_{Y,N}(y,n) = f_Y(y)f_N(n)\tag{4}$

By the multiplication rule,

$f_{X,Y,N}(x,y,n) = f_N(n)f_{X,Y\mid N}(x,y\mid n) \tag{5}$

Comparing $(1)$ with $(5)$ and using $(2)$, we get

$f_{X,Y\mid N}(x,y\mid n) = f_X(x)f_Y(y) = f_{X,Y}(x,y)\tag{6}$

Using the same reasoning we can find that

$f_{X \mid N}(x \mid n) = f_{X}(x)\tag{7}$ and $f_{Y \mid N}(y \mid n) = f_{Y}(y)\tag{8}$

This allow us to prove that

$$f_{X,Y \mid N}(x,y \mid n) = f_{X \mid N}(x \mid n)f_{Y \mid N}(y \mid n)$$

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  • $\begingroup$ How do you know that $(X,Y)$ and $N$ are independent? That step is not so clear to me. $\endgroup$ – hoyast Dec 7 '15 at 22:48
  • $\begingroup$ Because you said in your question "as well $N$ is a random variable (which both $X$ and $Y$ are independent to)". Wasn't that what you implied? The way I interpreted that sentence is that both $X$ and $Y$ are independent of $N$. Is that right? $\endgroup$ – Carlos Mendoza Dec 7 '15 at 23:07
  • $\begingroup$ I made an edit to my answer, hopefully it's clearer now. $\endgroup$ – Carlos Mendoza Dec 8 '15 at 6:12
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Yes, they are independent. Since $X$ is independent of $N,$ the conditional density of $X | N$ is just the density of $X.$ Similarly for $Y.$

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