Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 1 down vote favorite

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!

share|improve this question

1 Answer

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 $X$.

share|improve this answer
Thank you. So 1) is not true. What about 2)? – Zhou Heng Feb 16 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 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 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 at 6:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.