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Help with variance sum + correlation coefficient formula

  1. Sep 29, 2007 #1

    Simfish

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    [SOLVED] Help with variance sum + correlation coefficient formula

    This is a worked example

    The objective is to prove

    [tex]-1 \leq \rho(X,Y) \leq 1[/tex]

    Then the book uses this formula...

    (2) [tex]0 \leq Var(\left \frac{X}{\sigma_x} + \frac{Y}{\sigma_y} \right)[/tex]

    (3) [tex]= \frac{Var(X)}{{\sigma_x}^2} + \frac{Var(Y)}{{\sigma_y}^2} + \frac{2Cov(X,Y)}{\sigma_x \sigma_y}[/tex]

    The question is, how does 2 lead to 3? Namely, how does [tex]Var(\frac{X}{\sigma_x} ) => \frac{Var(X)}{{\sigma_x}^2}[/tex]?

    Also, how does one get the idea to use formula (2) to prove (1)? It doesn't seem like a natural step
     
    Last edited: Sep 29, 2007
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  3. Sep 29, 2007 #2

    Simfish

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    sorry, I edit my posts a lot - so somehow, edited posts on PF don't edit the tex code any longer once you edit the posts enough...

    Namely, how does [tex]Var(\frac{X}{\sigma_x})[/tex] => [tex]\frac{Var(X)}{{\sigma_x}^2}[/tex]?
     
  4. Sep 29, 2007 #3

    EnumaElish

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    What is Var(aX), where a is constant?
     
  5. Sep 29, 2007 #4

    Simfish

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    aVar(X)

    holy crap
    i never knew my attention lapses were that bad
     
  6. Sep 29, 2007 #5

    D H

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    That's not right. The variance of a one-dimensional random variable X is defined as [itex]\text{Var}(X) = \text{E}[(X-\text{E}(X))^2][/itex]. What does this mean in terms of scaling X by some quantity a?
     
  7. Sep 29, 2007 #6

    Simfish

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    Oh, I see.

    [tex]a^2 Var(X)[/tex]
     
  8. Sep 29, 2007 #7

    D H

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    Now onto the next question: Given two random variables [itex]X[/itex] and [itex]Y[/itex], what is [itex]\text{Var}(X+Y)[/itex]? Apply the definition of [itex]\text{Var}(X)[/itex] to the new random variable [itex]X+Y[/itex].
     
  9. Sep 29, 2007 #8

    Simfish

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    [tex]= {Var(X)} + {Var(Y)} + {2Cov(X,Y)}[/tex]

    but that's from memorization - I'll try to derive it now
     
    Last edited: Sep 29, 2007
  10. Sep 29, 2007 #9

    Simfish

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    [tex]Var(X+Y)[/tex]

    [tex]= E[(X+Y)^2] - E[X+Y]^2[/tex]

    [tex]= E[X^2 + 2XY + Y^2] - E[X+Y]^2[/tex]
    [tex]= E[X^2] + 2E[XY] + E[Y^2] - (E[X] + E[Y])^2[/tex]
    [tex]= E[X^2] + 2E[XY] + E[Y^2] - E[X]^2 - E[Y]^2 - 2E[X]E[Y][/tex]
    [tex]= (E[X^2]- E[X]^2) + (E[Y^2] - E[Y]^2) + (2E[XY]- 2E[X]E[Y])[/tex]
    [tex]= Var(X) + Var(Y) + 2Cov(X,Y)[/tex] IF dependent
    IF independent, 2E[XY] = 2E[X]E[Y]

    ==
    Okay, can someone please address my second question?
    Also, how does one get the idea to use formula (2) to prove (1)? It doesn't seem like a natural step
     
  11. Sep 29, 2007 #10

    D H

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    What is [itex]\text{Var}\left(X/{{\sigma_x}}\right)[/itex]?
     
  12. Sep 29, 2007 #11

    Simfish

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    [tex]\frac{Var(X)}{{\sigma_x}^2} = 1 [/tex] since [tex] Var(X) = \sigma_x, Var(Y) = \sigma_y[/tex]. I have the entire proof in the book - but the first step seems unnatural (how does one get the inspiration to use [tex]0 \leq Var(\left \frac{X}{\sigma_x} + \frac{Y}{\sigma_y} \right)[/tex] for proving that correlation coefficient has absolute magnitude <= 1?
     
    Last edited: Sep 29, 2007
  13. Sep 29, 2007 #12

    D H

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    The variance of any random variable is tautologically non-negative. Look at the definition of variance.
     
  14. Sep 29, 2007 #13

    EnumaElish

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    You mean [tex] Var(X) = \sigma_x^2, Var(Y) = \sigma_y^2[/tex]

    The idea behind correlation is to standardize variables X and Y by dividing each by its standard deviation before finding their correlation.
     
  15. Oct 1, 2007 #14

    Simfish

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    Okay I see. Thanks. :)
     
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