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   Author  Topic: Independent?  (Read 1083 times)
archau
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Independent?  
« on: Jan 8th, 2008, 4:55pm »
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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?
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towr
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Re: Independent?  
« Reply #1 on: Jan 9th, 2008, 3:21am »
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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

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william wu
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Re: Independent?  
« Reply #2 on: Oct 17th, 2008, 9:12am »
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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


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archau
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Re: Independent?  
« Reply #3 on: Dec 16th, 2008, 10:41pm »
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choose f(x) to be exp(tx) and the result follows from the properties of moment generating functions.
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