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The expected value of the square of the sample mean?

  1. Aug 23, 2015 #1
    1. The problem statement, all variables and given/known data
    In my notes I keep stumbling upon this equation:

    Equation 1: E(X-bar^2) = (σ^2)/n + μ^2

    I was wondering why the above equation is true and how it is derived.


    3. The attempt at a solution
    E(X-bar^2)

    ##Summations are from i/j=1 to n

    = E[(Σx_i/n)^2)]

    = E[(Σx_i/n)(Σx_j/n)]

    =(1/n^2)E[(Σx_i)(Σx_j)]

    =(1/n^2)ΣΣE(x_i*x_j)

    =E(x_1^2) + E( x_2^2 + ... + E(x_n^2)
    + E(x_1*x_2) + E(x_1*x_3) + ... : Equation 2

    I'm stuck at this point. Could someone show me how you get to Equation 1 from Equation 2? Assuming that my derivation to Equation 2 is correct.
     
    Last edited: Aug 23, 2015
  2. jcsd
  3. Aug 23, 2015 #2

    Ray Vickson

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    If you mean
    [tex] E \overline{X}^2 = \frac{\sigma^2}{n} + \mu [/tex]
    then that cannot possibly be right! For one thing, the "dimensions" don't match. For example, if ##X## is measured in meters, then ##\sigma^2## is measured in ##\text{meters}^2##, while ##\mu## is in meters.

    The easiest approach is to recognize that the mean square and variance of any random variable ##Y## are connected by the equation
    [tex] \text{Var}(Y) = E(Y^2) - (E Y)^2, \; \text{or} \; E(Y^2) = \text{Var}(Y) + (EY)^2 [/tex]

    Apply this to ##Y = \bar{X} = \sum X_i / n##. Assuming the ##X_i## are independent and identically distributed there are easy and well-known formulas for ##EY## and ##\text{Var} (Y)## in terms of ##\mu##, ##\sigma## and ##n##. That will tell you the value of ##E \overline{X}^2##.

    On the other hand, if you meant that you want to know
    [tex] E \overline{X^2} = E \sum X_i^2 / n, [/tex]
    then you have a different set of formulas to use. However, you can approach it again using the relationship between mean square and variance.
     
    Last edited: Aug 23, 2015
  4. Aug 23, 2015 #3

    SteamKing

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    I think your Equation 1 may contain a slight error, as Ray pointed out

    I think the correct equation is E(x-bar)2 = (σ2 / n) + μ2
     
  5. Aug 23, 2015 #4
    Yes, you are both correct, the correct equation was actually

    Equation 1: E(X-bar^2) = (σ^2)/n + μ^2
    The mu had to be squared.

    In fact, I can see that I have made a huge blunder in how I was looking at this problem. Thank you both for the insight!
     
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