Hello,
Hmmm actually, you have to prove that , the joint pdf of X and Y+1, can be written in such a form. In what you did, you assumed it was true.
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Let f be the joint pdf of X and . So we'd have , as defined above.
For any measurable function h, we have :
So is the pdf of
And this proves that X and Y+1 are independent.
If you can't see (*), consider you're working on , where and use the general thing :
iff f is the pdf of X.
And as usual, if there's any mistake, tell me... I'm so self-confident you know...
You could use the very definition of independence.
If X and Y are discrete r.v.: for all , we have (the comma means "and" ; I used the independence of X and Y in the second equality). This proves that and are independent.
General case: for all measurable subsets , we have (the comma means "and" ; ; I used the independence of X and Y in the second equality). This proves that and are independent.
No, you can't just mimic the proof.
In fact, there is a very general (and intuitive) result that answers both the initial question and yours. I only give two examples of statements; the generalization is straightforward:
if are independent random variables, and , are (measurable) functions, then and are independent.
if are independent random variables, and , are (measurable) functions, then and are independent.
In the generalized version, there can be any number of random variables, and the functions may depend on variously sized groups of variables provided they are disjoint.
Thank you for your answer Laurent!
This bring us back to Moo's proof I think, in the generalized version I should have that:
Two r.v.s (but this should be true also for n r.v.s, two by two independent) and are independent if and only if for any pair of functions and
provided that the expectations exist.
A special case is of course when two r.v.s and are independent, and you have to check the independence of and which is noname's question...
Am I right?
The problem with Moo's and Captain Black's proofs is that they seemed to assume that the random variables had probability density functions; that's why I posted another elementary and general proof.
The property you're quoting is a (quick) consequence of the definition of independence (the definition is when , indicator function of set ); therefore it can be used to prove noname's question, but not yours. And it is not directly related to my previous post, which gave a way to get new independent r.v. from independent r.v. (by applying functions to disjoint groups of r.v.).
If you would like to answer noname's question using the property you give: let be such that and exist. Remark that where . Therefore, since and are independent, . This proves that and are independent. Compare with my previous proof.
Today I have taken my Statistics exam!
Thank you for your help...
Only one exercise gave me trouble, and it was about the independence of random variables. But it wasn't like the one above.
It was:
X and Y are random variables, check if random variables X+Y , 1 are independent.
The professor tell us that we should have used an appropriate example to check the independence of the variables.
Does anyone know a way to solve this exercise?
Thanks