Proof of independence of random variables

**noname** · September 17th 2009, 12:15 AM

Hi, i have a question

Two random variables X and Y are independent. How can i prove that X,Y+1 are independent?

Thanks

**CaptainBlack** · September 17th 2009, 11:00 AM

Originally Posted by noname

Hi, i have a question

Two random variables X and Y are independent. How can i prove that X,Y+1 are independent?

Thanks

You can either show that:

$f(x,y+1)=g(x)h(y+1)$

or that:

$f(y+1|x)=h(y+1)$

Where $f(x,y+1)$ is the pdf of the joint distribution, $g(x)$ is the pdf of $X$ , $h(y+1)$ the pdf of $Y+1$ and $f(y+1|x)$ is the condition distribution of $Y+1$

CB

**noname** · September 17th 2009, 12:40 PM

Hi, thank you for your answer.

Is this correct?

**Moo** · September 18th 2009, 07:46 AM

Hello,

Hmmm actually, you have to prove that $f(x,y+1)$ , the joint pdf of X and Y+1, can be written in such a form. In what you did, you assumed it was true.

-----
Let f be the joint pdf of X and . So we'd have $f(x,y)=f_1(x)f_2(y)$ , as defined above.
For any measurable function h, we have :

$\begin{aligned}\mathbb{E}(h(X,Y+1))&=\int_{\mathbb {R}^2} h(x,y+1)f(x,y) ~dxdy \quad (*) \\<br /> &=\int_{\mathbb{R}^2} h(x,y)f(x,y-1) ~dxdy \\<br /> &=\int_{\mathbb{R}^2} h(x,y)f_1(x)f_2(y-1) ~dxdy<br /> \end{aligned}$

So $f_1(x)f_2(y-1)$ is the pdf of $(X,Y+1)$
And this proves that X and Y+1 are independent.

If you can't see (*), consider you're working on $\mathbb{E}((h\circ p)(X,Y))$ , where $p(a,b)=(a,b+1)$ and use the general thing :
$\mathbb{E}(h(X))=\int_{\mathbb{R}} h(x)f(x) ~dx$ iff f is the pdf of X.

And as usual, if there's any mistake, tell me... I'm so self-confident you know...

**Laurent** · September 18th 2009, 09:20 AM

Originally Posted by noname

Hi, i have a question

Two random variables X and Y are independent. How can i prove that X,Y+1 are independent?

Thanks

You could use the very definition of independence.

If X and Y are discrete r.v.: for all $x,y$ , we have $P(X=x,Y+1=y)=P(X=x,Y=y-1)$ $=P(X=x)P(Y=y-1)=P(X=x)P(Y+1=y)$ (the comma means "and" ; I used the independence of X and Y in the second equality). This proves that $X$ and $Y+1$ are independent.

General case: for all measurable subsets $A,B$ , we have $P(X\in A,Y+1\in B)=P(X\in A,Y\in B-1)$ $=P(X\in A)P(Y\in B-1)=P(X\in A)P(Y+1\in B)$ (the comma means "and" ; $B-1=\{y-1|y\in B\}$ ; I used the independence of X and Y in the second equality). This proves that $X$ and $Y+1$ are independent.

**Ruby** · September 18th 2009, 12:19 PM

Originally Posted by Laurent

You could use the very definition of independence.

[...snip]

General case: for all measurable subsets $A,B$ , we have $P(X\in A,Y+1\in B)=P(X\in A,Y\in B-1)$ $=P(X\in A)P(Y\in B-1)=P(X\in A)P(Y+1\in B)$ (the comma means "and" ; $B-1=\{y-1|y\in B\}$ ; I used the independence of X and Y in the second equality). This proves that $X$ and $Y+1$ are independent.

Hi,

Can I use the same to proof the following?

$X_1, X_2, Y_1, Y_2$ are indipendent random variables, expectations and variances are $E(X_i)=E(X), E(Y_j)=E(Y), var(X_i)=var(X), var(Y_i)=var(Y)$ where $i=1,2$ and $j=1,2$

Is $Z_1=X_1*Y_1$ indipendent from $Z_2=X_2*Y_2$ ?

Thanks!

**Laurent** · September 18th 2009, 01:40 PM

Originally Posted by Ruby

Can I use the same to proof the following?

$X_1, X_2, Y_1, Y_2$ are indipendent random variables, expectations and variances are $E(X_i)=E(X), E(Y_j)=E(Y), var(X_i)=var(X), var(Y_i)=var(Y)$ where $i=1,2$ and $j=1,2$

Is $Z_1=X_1*Y_1$ indipendent from $Z_2=X_2*Y_2$ ?

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 $X_1,X_2,X_3,X_4$ are independent random variables, and $f_1:\mathbb{R}^2\to\mathbb{R}$ , $f_2:\mathbb{R}^2\to\mathbb{R}$ are (measurable) functions, then $f_1(X_1,X_2)$ and $f_2(X_3,X_4)$ are independent.

if $X_1,X_2,X_3,X_4$ are independent random variables, and $f_1:\mathbb{R}^3\to\mathbb{R}$ , $f_2:\mathbb{R}\to\mathbb{R}$ are (measurable) functions, then $f_1(X_1,X_2,X_3)$ and $f_2(X_4)$ 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.

**Ruby** · September 19th 2009, 02:06 AM

Originally Posted by Laurent

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 $X_1,X_2,X_3,X_4$ are independent random variables, and $f_1:\mathbb{R}^2\to\mathbb{R}$ , $f_2:\mathbb{R}^2\to\mathbb{R}$ are (measurable) functions, then $f_1(X_1,X_2)$ and $f_2(X_3,X_4)$ are independent.

[...snip]

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) $X_1$ and $X_2$ are independent if and only if for any pair of functions $f_1(x_1)$ and $f_2(x_2)$

$E[f_1(x_1)f_2(x_2)] = E[f_1(x_1)]E[f_2(x_2)]$

provided that the expectations exist.

A special case is of course when two r.v.s $X$ and $Y$ are independent, and you have to check the independence of $f_1(x) = x$ and $f_2(y) = y+1$ which is noname's question...

Am I right?

**Laurent** · September 19th 2009, 04:57 AM

Originally Posted by Ruby

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) $X_1$ and $X_2$ are independent if and only if for any pair of functions $f_1(x_1)$ and $f_2(x_2)$

$E[f_1(x_1)f_2(x_2)] = E[f_1(x_1)]E[f_2(x_2)]$

provided that the expectations exist.

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 $f_i=1_{A_i}$ , indicator function of set $A_i$ ); 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 $f_1,f_2$ be such that $E[f_1(X)]$ and $E[f_2(Y+1)]$ exist. Remark that $f_2(Y+1)=\widetilde{f_2}(Y)$ where $\widetilde{f_2}(y)=f_2(y+1)$ . Therefore, since $X$ and $Y$ are independent, $E[f_1(X)f_2(Y+1)]=E[f_1(X)\widetilde{f_2}(Y)]$ $=E[f_1(X)]E[\widetilde{f_2}(Y)]=E[f_1(X)]E[f_2(Y+1)]$ . This proves that $X$ and $Y+1$ are independent. Compare with my previous proof.

**Ruby** · September 19th 2009, 06:07 AM

Thanks for your time Laurent, it's all becoming clearer to me now!

**noname** · September 28th 2009, 12:31 PM

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

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