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1) If randome variable $X$ is independent of randome variable $Y_1$ and $X$ is also independent of random variable $Y_2$, is $X$ independent of the joint random variable $(Y_1,Y_2)$?

2) If randome variable $X$ is independent of joint random variable $(Y_1,Y_2)$, is $X$ independent of the randome variable $Y_1$?

If yes, please give a proof. If no, please give a counterexample. Thanks a lot for your answer!

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1 Answer 1

up vote 1 down vote accepted

A fair coin is tossed twice. Let $Y_1=0$ if the first toss is a head, and $1$ if the first toss is a tail. Let $Y_2=0$ if the second toss is a head, and $1$ if the second toss is a tail.

Let $X=0$ if the two tosses are different, and $1$ if they are the same.

It is not hard to verify that $X$ and $Y_1$ are independent, also that $X$ and $Y_2$ are independent. However, given $(Y_1,Y_2)$, we know the value of $X$.

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Thank you. So 1) is not true. What about 2)? –  Zhou Heng Feb 16 '13 at 6:17
    
I thought I would leave it to you, since it is easier. Intuitively, if knowing the joint distribution of $Y_1$ and $Y_2$ tells you nothing about $X$, then knowing $Y_1$ can tell you nothing about $X$, since the joint distribution of $Y_1$ and $Y_2$ tells you the distribution of $Y_1$. –  André Nicolas Feb 16 '13 at 6:30
    
Thank you. I worked it out. We have now $p(x|y_1y_2)=p(x)$. So $p(x|y_1)=\frac{p(xy_1)}{p(y_1)}=\frac{\sum\limits_{y_2}p(xy_1y_2)}{p(y_1)}$$=\f‌​rac{\sum\limits_{y_2}p(x|y_1y_2)p(y_1y_2)}{p(y_1)}=\frac{\sum\limits_{y_2}p(x)p(y‌​_1y_2)}{p(y_1)}=\frac{p(x)\sum\limits_{y_2}p(y_1y_2)}{p(y_1)}=\frac{p(x)p(y_1)}{p‌​(y_1)}=p(x)$. So 2) is true. Thank you again. –  Zhou Heng Feb 16 '13 at 6:50
    
Your proof works fine for discrete distributions. One can write an analogue in the general case, using the cumulative distribution function. –  André Nicolas Feb 16 '13 at 6:58
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