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Cauchy-Schwarz inequality: cov(X,Y)]^2 ≤ var(X) var(Y)

  1. Oct 10, 2009 #1
    Suppose that X andy Y are (scalar) random variables. Show that
    [cov(X,Y)]^2 ≤ var(X) var(Y). (Cauchy-Schwarz inequality)

    Sow that equality holds if and only if there is a relationship of the form

    m.s.
    c=aX+bY (i.e. c is equal to aX+bY in "mean square").

    =========================

    Proof:
    E[(X+tY)^2] is quadratic in t and is ≥0. This means that the graph of it has at most one real root, i.e. discriminant ≤ 0. Setting discriminant ≤ 0 gives gives [E(XY)]^2 ≤ E(X^2) E(Y^2).

    Now I am stuck with the second part. I know that equality can only possibly happen when the graph of E[(X+tY)^2] has exactly one real root, i.e. discriminant = 0, but how can I prove that c is equal to aX+bY in "mean square"?

    Any help is appreciated!:)
     
    Last edited: Oct 10, 2009
  2. jcsd
  3. Oct 15, 2009 #2
    Something is missing in your expressions for variance and covariance?
     
  4. Oct 17, 2009 #3
    [tex]|Cov(X,Y)|^2=|E(X-\mu)(Y-\nu)|^2=|\langle X-\mu, Y-\nu\rangle|^2[/tex]

    [tex]= E(X-\mu)^2E(Y-\nu)^2[/tex]

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

    The Cauchy-Schwarz inequality is:[tex] E|(XY)|^2\leq E(X^2)E(Y^2)[/tex]
     
    Last edited: Oct 17, 2009
  5. Oct 18, 2009 #4
    But now the trouble is when the EQUALITY hold in the inequality [cov(X,Y)]^2 ≤ var(X) var(Y)...!?
     
  6. Oct 19, 2009 #5
    [tex]|Cov(X,Y)|^2\leq \langle X-\mu,X-\mu \rangle \langle Y-\nu,Y-\nu\rangle[/tex]

    http://en.wikipedia.org/wiki/Cauchy–Schwarz_inequality
     
    Last edited: Oct 19, 2009
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