Take the 2-minute tour ×
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.

There are three statements:

  1. $(X_1,X_2)$ is a uniform random vector over some subset $S \subset \mathbb{R}^2$,
  2. $X_1$ and $X_2$ are two uniform random variables over some $S_1 \subset \mathbb{R}$ and $S_2 \subset \mathbb{R}$ respectively,
  3. $X_1$ and $X_2$ are two independent random variables.

where $S_1$ and $S_2$ are projections of $S$ onto the two axes for $X_1$ and $X_2$ respectively.

  1. I was wondering if any two of the three statements imply the other one? I know that 3 and 2 imply 1, and suspect other implications are also true.
  2. What is the necessary (and sufficient) condition for 2 and 1 to imply each other? Is 3? Or some requirements on $S$?
  3. Also I wonder if the above true statements can be generalized to cases when there are finite random variables $(X_1, X_2, \dots, X_n)$, or even infinitely many (countably or uncountably) random variables? Is 3 required to be modified to be mutually independent, or pairwise independence will work?

Thanks!

share|improve this question
    
If 3. is true, then I think 1. and 2. are equivalent. Otherwise neither 1. nor 2. implies the other, nor do they (even taken together) imply 3. –  mjqxxxx Oct 25 '12 at 14:33
    
@mjqxxxx: How do you show that 3 and 1 imply 2? Why 2 and 1 do not imply 3? –  Tim Oct 25 '12 at 14:36
    
@Tim if they are independent random variables, then the joint density is the product of marginals. In other words, if a joint density is given with the independence conditions they should be generated by two uniform marignals. –  Seyhmus Güngören Oct 25 '12 at 14:47

1 Answer 1

3 + either 1 or 2 implies the other. If you also have $S=S_1$x$S_2$, 1 and 2 imply 3.

Independence of random variables is equivalent to the statement that the joint density function $f_{X,Y}(x,y)$ is equal to product of the marginal densities $f_X(x)\cdot f_Y(y)$. If you have 1) and 2) and $S=S_1$x$S_2$, then you can just work out both expressions and you'll see they're the same.

You already noted that 2 and 3 imply 1 so let's assume we have 1 and 3 now. Write $f_{X,Y}(x,y)=f_X(x)\cdot f_Y(y)$; considering that equation for fixed $y\in S_2$ we see that $f_X(x)$ must be uniform over $S_1$ and similarly $f_Y(y)$ must be uniform over $S_2$.

share|improve this answer
1  
Sorry but 1. and 2. DO NOT imply 3. –  Did Oct 25 '12 at 15:34
    
They do. 1. and 2. specify both the joint distributions and the marginal distributions. That's all the information you need. –  anonymous Oct 25 '12 at 15:37
    
Sure about that? Consider $X_1=X_2$ uniform on $(0,1)$ (more sophisticated examples exist, in particular such that the joint distribution of $(X_1,X_2)$ has a density). –  Did Oct 25 '12 at 15:39
    
1 and 2 dont imply 3. You can have dependent case where 1 and 2 holds. –  Seyhmus Güngören Oct 25 '12 at 15:39
    
Whoops. Misread the part about $S_1$ and $S_2$ being projections. Fixed. –  anonymous Oct 25 '12 at 15:43

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.