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Correlation coefficient

  1. Mar 31, 2013 #1
    Let 'x' and 'y' be two random variables with zero mean.

    We find that 'x' is related to 'y' with a correlation coefficient 'c'.

    Now let us say we are splitting 'x' into correlated term 'xc'and uncorrelated term 'xu'
    Then we have

    [tex]
    x= x_c + x_u
    \\
    \overline{x^2}= \overline{(x_c + x_u)^2} = \overline{x_c^2} + \overline{x_u^2}

    [/tex]
    Then does it mean the following is true? -
    [tex]

    \overline{x_c^2}= \overline{x^2}(|c|^2)
    \\
    \overline{x_u^2}=\overline{x^2}(1-|c|^2)

    [/tex]

    If so how would you prove it?
    I am sure that the above result is true as I saw it in a book.

    Thanks a lot
     
  2. jcsd
  3. Mar 31, 2013 #2

    ssd

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    Please explain your notation of upper bar.
     
  4. Mar 31, 2013 #3

    mathman

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    Gold Member

    x = cy +(x - cy), where xc = cy.

    Implicit is the assumption that |x|2 = |y|2 = 1
     
    Last edited: Mar 31, 2013
  5. Mar 31, 2013 #4
    Upper bar indicates expectation E().
     
  6. Apr 1, 2013 #5


    Are you sure about that implicit assumption that |x|2 = |y|2 because

    In my problem

    [tex]

    \overline{x^2} \neq \overline{y^2}

    [/tex]

    Thanks
     
  7. Apr 1, 2013 #6

    ssd

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    If xc=cy, then correlation coefficient between xc and y is 1 or -1.
     
  8. Apr 1, 2013 #7
    Yes. Good point. Didn't notice.
     
  9. Apr 1, 2013 #8

    ssd

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    I am confused here. How do you take E[xcxu]=0?
     
    Last edited: Apr 1, 2013
  10. Apr 1, 2013 #9
    Because correlated term and uncorrelated term are uncorrelated.
     
  11. Apr 1, 2013 #10

    Stephen Tashi

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    My turn to guess what the question is,

    Let [itex] X [/itex] be a real valued random variable with mean zero and variance [itex] \sigma^2_X [/itex].
    Let [itex] Y [/itex] be a real valued random variable with mean zero and variance [itex] \sigma^2_Y. [/itex].

    Let the [itex] Cov(X,Y) [/itex] denote the covariance of [itex] X [/itex] and [itex] Y [/itex].

    Let [itex] c = \frac {Cov(X,Y)}{\sigma_X \sigma_Y} [/itex].

    Let [itex] a = \frac{ Cov(X,Y)}{\sigma^2_Y} [/itex].

    Define [itex] X_c = a Y [/itex].

    Define [itex] X_u = X - X_c [/itex],

    Define [itex] W = X^2 [/itex]
    Define [itex] W_c = {X_c}^2 [/itex]
    Define [itex] W_u = {X_u}^2 [/itex]

    Is it true that the mean of [itex] W_c [/itex] is equal to the sum of the means of [itex] W_c [/itex] and [itex] W_u [/itex] ?


    How about that?

    Or is the question about sample statistics from random variables rather than about population values?
     
  12. Apr 1, 2013 #11
    But the same question rises...

    when

    Xc= a Y

    Then Xc and Y are completely correlated with a correlation coefficient 1 no matter what the value of a is. (If a is negative then Xc and Y are negatively correlated and if a is positive then they are positively correlated.)
     
  13. Apr 1, 2013 #12

    ssd

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    Can you show this algebraically, please? Long way back I went through descriptive stats. Neither have the references at hand nor my mind readily accepts a linguistic statement like this without proof.

    Simple correlation is nothing but degree of linear dependence. The correlated term may have a partially related (if |r|<1 ) linear part and a non-separable non linearly related part. The uncorrelated term may have a similar type of non linearly related part too. There by, the correlated and uncorrelated part might be linearly related.
     
    Last edited: Apr 1, 2013
  14. Apr 1, 2013 #13
    I don't know how to prove it algebraically basically because I am really confused with how you would represent Xc in terms of Y but I can give you some other kind of proof.

    Let's say Xc is correlated with Y which means that if Y increases by some amount Xc will also increase and how it increases depends on the correlation coefficient. Now we know that Xu is uncorrelated with Y which means that changes in Xu is no way related with changes in Y which in turn means that it is no way related to variations in Xc. So we can say that they both are uncorrelated.
     
  15. Apr 1, 2013 #14

    ssd

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    Unacceptable.

    1/ Xc is correlated with Y which DOES NOT mean that if Y increases by some amount Xc will also always increase. Will always increase if r=1.

    2/ "... how it increases depends on the correlation coefficient..."..... Not at all, if I interpret the word "how" as the rate of increase.

    3/ "Now we know that Xu is uncorrelated with Y which means that changes in Xu is no way related with changes in Y which in turn means that it is no way related to variations in Xc".....
    No, not again..... Xu is uncorrelated with Y but still changes in Xu may be WELL related with changes in Y, but of course non linearly. Or, even strictly linearly in some part of the domain and in some other part of the domain in opposite direction.

    Conclusion: need an algebraic proof before proceeding further.
     
  16. Apr 2, 2013 #15

    Stephen Tashi

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    The term "correlation coefficient" is ambiguous. Can you specify which meaning of "correlation coefficient" you are using? Do you mean Perason's product-moment correlation ? (http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient)

    Or are you talking about the slope of a line in a linear regression?
     
  17. Apr 2, 2013 #16

    ssd

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    Unless otherwise mentioned, we take "correlation coefficient" as simple correlation coefficient (the prod-moment one by Pearson).
     
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