# A question about independence wrt joint random variable

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 $Y_1$?

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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 $X$.
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