Show that E[S2] = sigma^2

Discussion in 'CT3' started by n111kus, Apr 21, 2015.

  1. n111kus

    n111kus Member

    Hi,

    In the attachment, can someone explain how you get from the top circles terms to the bottom ones...

    Thanks

    Nik
     

    Attached Files:

    n111kus, Apr 21, 2015
    #1
  2. Hemant Rupani

    Hemant Rupani Senior Member

    a(b+c)=ab+ac......you know.

    First n vanished.
     
    Hemant Rupani, Apr 22, 2015
    #2
  3. n111kus

    n111kus Member

    Thanks for your reply..

    I actually circled the wrong term, I wanted to know how you get from nE[x^2] to the second term..

    Thanks

    Nikesh
     
    n111kus, Apr 22, 2015
    #3
  4. Hemant Rupani

    Hemant Rupani Senior Member

    ohhh! I should have understand that.:eek:


    Actually Expected Value of sample mean is population mean
    and Variance of sample mean is (population variance)/n, see
    and Var(a)=E(a^2)-E(a)
     
    Hemant Rupani, Apr 22, 2015
    #4
  5. John Lee

    John Lee ActEd Tutor Staff Member

    Hemant is correct.

    We are using \(E(\bar X^2) = var(\bar X) + E^2(\bar X)\)

    And from the central limit theorem we know that \(E(\bar X) = \mu\) and \(var(\bar X) = \sigma^2/n\).
     
    John Lee, Apr 22, 2015
    #5

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