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Thread: The sign of COV( f(x,y) , g(x,y) )

  1. #1
    Stole
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    Smile The sign of COV( f(x,y) , g(x,y) )




    I am a PhD student in Economics, I am reading a paper on statistics, and the author states the following proposition without proof. My question is is it correct? Do we need to impose more conditions to get the result? Thanks!!!

    Suppose x and y are random variables. If
    (1) E[f(x,y)]=0
    (2) f(x,y) and g(x,y) are increasing in x.
    (3) f(x,y) and g(x,y) are decreasing in y.
    Then
    COV( f(x,y), g(x,y) )>0.
    Last edited by Stole; 11-01-2008 at 02:11 PM.

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    Quote Originally Posted by Stole View Post
    Do we need to impose more conditions to get the result?
    Well, no, we don't.

    The author (based on what you have provided) is just simply implying that both f(x,y) and g(x,y) are monotone functions in both x (increasing) and y (decreasing).

    And, therefore, the covariance between these two functions will be positive.

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    Quote Originally Posted by Stole View Post
    I am a PhD student in Economics, I am reading a paper on statistics, and the author states the following proposition without proof. My question is is it correct? Do we need to impose more conditions to get the result? Thanks!!!

    Suppose x and y are random variables. If
    (1) E[f(x,y)]=0
    (2) f(x,y) and g(x,y) are increasing in x.
    (3) f(x,y) and g(x,y) are decreasing in y.
    Then
    COV( f(x,y), g(x,y) )>0.
    I think you need a small correction here.
    It should be COV( f(x,y), g(x,y) )>=0.

    Consider the following situation.
    Let
    f(x,y) = x-y , g(x,y) = x-2y and E(x-y) =0

    it holds 2&3 conditions

    But f & g are independent( we can make this combination). So this lead Cov(f,g) = 0
    Last edited by vinux; 11-02-2008 at 12:29 PM.
    In the long run, we're all dead.

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    O.o

    How are f ang g independent? g = f - y.

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    Ok. Think about this..

    (X,Y) are follows bivariate normal rv. mean =(0,0)
    Var(X) = 6 Cov(X,Y) = 4 Var(Y) = 3

    So what will be the distribution for f and g?
    In the long run, we're all dead.

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    Re:

    hrm, I'll consider that, but what will you do with conditional probabilities like P(f < 0 | g > 0) = P(x - y < 0 | x - 2y > 0)? This is a different event than f < 0.

  7. #7
    Stole
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    Thank you for your reply.
    Is there a way to prove it? or is there a textbook for reference to this proposition?
    I know if f(x) and g(x) are monotonic in x in the same direction, then COV(f(x),g(x)) are positive. But I have doubt on this when it comes to two variables, x and y.

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    Quote Originally Posted by Dragan View Post
    Well, no, we don't.

    The author (based on what you have provided) is just simply implying that both f(x,y) and g(x,y) are monotone functions in both x (increasing) and y (decreasing).

    And, therefore, the covariance between these two functions will be positive.

    I think you need a condition like x and y are independent.
    In the long run, we're all dead.

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    Quote Originally Posted by zmogggggg View Post
    hrm, I'll consider that, but what will you do with conditional probabilities like P(f < 0 | g > 0) = P(x - y < 0 | x - 2y > 0)? This is a different event than f < 0.
    Did you calculated this? what is the answer
    In the long run, we're all dead.

  10. #10
    Stole
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    Quote Originally Posted by vinux View Post
    I think you need a small correction here.
    It should be COV( f(x,y), g(x,y) )>=0.

    Consider the following situation.
    Let
    f(x,y) = x-y , g(x,y) = x-2y and E(x-y) =0

    it holds 2&3 conditions

    But f & g are independent( we can make this combination). So this lead Cov(f,g) = 0
    But the covariance can never be negative, right?

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    It can.. Cov(f,g) is between f & g. Not between X & Y

    The same example.
    (X,Y) are follows bivariate normal rv. mean =(0,0)
    Var(X) = 6 Cov(X,Y) = sqrt(17) Var(Y) = 3

    Here you will get Cov(f,g) <0
    In the long run, we're all dead.

  12. #12
    Stole
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    Quote Originally Posted by vinux View Post
    It can.. Cov(f,g) is between f & g. Not between X & Y

    The same example.
    (X,Y) are follows bivariate normal rv. mean =(0,0)
    Var(X) = 6 Cov(X,Y) = sqrt(17) Var(Y) = 3

    Here you will get Cov(f,g) <0
    That is right!
    Then we need more assumptions. Except independence between x and y, is there any other assumption can guarantee this?

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    May be this .
    Cov(f, g ) >= 0
    ( Otherwise you have conditions like strictly increasing /decreasing )
    In the long run, we're all dead.

  14. #14
    Stole
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    Quote Originally Posted by vinux View Post
    It can.. Cov(f,g) is between f & g. Not between X & Y

    The same example.
    (X,Y) are follows bivariate normal rv. mean =(0,0)
    Var(X) = 6 Cov(X,Y) = sqrt(17) Var(Y) = 3

    Here you will get Cov(f,g) <0
    Quote Originally Posted by vinux View Post
    May be this .
    Cov(f, g ) >= 0
    ( Otherwise you have conditions like strictly increasing /decreasing )
    Unfortunately, this assumption is exactly the result.

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    Sorry for the confusion. I was talking about the Then part

    Then
    COV( f(x,y), g(x,y) )>=0.
    It is not the condition
    In the long run, we're all dead.

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