Author |
Topic: Independent? (Read 1093 times) |
|
archau
Newbie
Posts: 5
|
 |
Independent?
« on: Jan 8th, 2008, 4:55pm » |
 Quote  Modify
|
X and Y are bounded random variables such that for all continuous functions f
E( f(X)f(Y) ) = E(f(X)) E(f(Y))
Are they independent?
|
|
 IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender: 
Posts: 13730
|
|
Re: Independent?
« Reply #1 on: Jan 9th, 2008, 3:21am » |
Quote Modify
|
The best I can say so far is that the extent to which they can be dependent can be shown to be arbitrarily small.
 {p(X&Y) f(X)f(Y) | all X,Y} = {p(X) f(X) | all X} {p(Y) f(Y) | all Y}
{p(X&Y) f(X)f(Y) | all X,Y} = {p(X) f(X) p(Y) f(Y) | all X, Y}
now for all A,B we can choose a continuous function that is 1 in an arbitrarily small neighbourhood of A and B and 0 outside it.
So for all A,B we can make |p(A&B) - p(A) p(B)| arbitrarily small. But, unfortunately, not 0. I'm not sure whether it's sufficient we can do do it in the limit.
Perhaps it would work better to show that the statement can't hold for dependent X,Y.
|
| « Last Edit: Jan 9th, 2008, 3:21am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
william wu
wu::riddles Administrator
Gender:
Posts: 1291
|
|
Re: Independent?
« Reply #2 on: Oct 17th, 2008, 9:12am » |
Quote Modify
|
I think the answer is yes, they are independent.
We aim to show that for two arbitrary measurable sets A and B,
Pr(X  A, Y B) = Pr(X A) Pr(Y B).
By either explicit construction, or perhaps quoting some theorem I don't remember, we can construct a sequence of bounded continuous functions f_n(A) which pointwise converges to the (discontinuous) indicator function 1(X A), so that lim f_n(A) = 1(X A). Similarly, there's a sequence of bounded continuous functions g_n(A) which pointwise converges to the (discontinuous) indicator function 1(Y B), so that lim g_n(B) = 1(Y B).
Then we have the chain of equalities (annotating each line):
Pr(X A, Y B)
(1) = E[ 1(X A) 1(X B) ]
(2) = E[ lim_n f_n(A)  lim_n g_n(B) ]
(3) = E[ lim_n f_n(A) g_n(B) ]
(4) = lim_n E[ f_n(A) g_n(B) ]
(5) = lim_n E[ f_n(A)] E[ g_n(B) ]
(6) = ( lim_n E[ f_n(A)] ) ( lim_n E[ g_n(B)] )
(7) = ( E[ lim_n f_n(A)] ) ( E[ lim_n g_n(B)] )
(8) = ( E[ 1(A)] ) ( E[ 1(B) ] )
(9) = Pr(X A) Pr(Y B).
where each of the equalities are explained as follows:
(1) Fact: Pr(X A) = E[1(X A)]
(2) by definition of f_n and g_n
(3) product of the limits is limit of the products
(4) bounded convergence theorem
(5) invoke the premise
(6) product of the limits is limit of the products
(7) dominated convergence theorem
(8) by definition of f_n and g_n
(9) Fact: Pr(X A) = E[1(X A)].
|
| « Last Edit: Oct 20th, 2008, 10:43am by william wu » |
IP Logged |
[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
|
|
|
archau
Newbie
Posts: 5
|
|
Re: Independent?
« Reply #3 on: Dec 16th, 2008, 10:41pm » |
Quote Modify
|
choose f(x) to be exp(tx) and the result follows from the properties of moment generating functions.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print