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Does L^2 Convergance Imply Convergance of L^2 norms?

  1. Aug 16, 2013 #1
    The answer seems to obviously be yes. But it's not so obvious to show it.

    I'm working with random variables. So the [itex]L^2[/itex] norm of X is [itex]E(X^2)^{1/2}[/itex], where E is the expected value. Thus, we want to show: if [itex]E((X_n-X)^2)\to0[/itex], then [itex]E(X_n^2)\to E(X^2)[/itex].

    From [itex]E((X_n^2-X)^2)\to0[/itex], we get
    [tex]E(X_n^2)\to2E(X_nX)-E(X^2).[/tex]

    I think it should be true that [itex]2E(X_nX)\to2E(X^2)[/itex], which would prove the result, but I'm not sure how to prove that.

    Any help?

    Or a reference? Is the result in a book?
     
  2. jcsd
  3. Aug 16, 2013 #2

    micromass

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    Use the Cauchy-Bunyakovski-Schwarz inequality

    [tex]|E(X_n X) - E(X^2)| = |E( (X_n - X) X)|\leq \sqrt{ E((X_n - X)^2) E(X^2)}[/tex]
     
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