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Help with finding the expectation

  1. Jan 6, 2010 #1
    1. The problem statement, all variables and given/known data

    Let X1,...,Xn denote a random sample from a [tex]N(\mu , \sigma)[/tex] distribution. Let [tex]Y = \Sigma \frac{(X_i - \overline{X})^2}{n}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    How would I find E(Y)?

    Any help would be greately appreciated.
     
  2. jcsd
  3. Jan 6, 2010 #2

    statdad

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    Homework Helper

    You need to be sure you can justify steps and perform the omitted work

    [tex]
    (X_i - \bar X)^2 = X_1^2 - 2X_1 \bar X + {\bar X}^2
    [/tex]

    When you compute [tex] E[(X_i - \bar X)^2][/tex]

    [tex]
    E[X_i^2]
    [/tex]

    should be easily found, and doesn't depend on i.

    [tex]
    2E[X_i \bar X] = \frac 2 n E[X_i (X_1 + X_2 + \dots + X_n)]
    [/tex]

    should be easily determined (and will not depend on i). Also,

    [tex]
    E[{\bar X}^2]
    [/tex]

    should be easy to find (since you know the distribution of the sample mean).

    Work these out individually, then combine them.
     
  4. Jan 6, 2010 #3
    [tex]X^2_i[/tex] is a chi-square distribution with n degrees of freedom (since there are n Xi) and it's expectation would be n

    [tex]\overline{X} -> N(\mu, \frac{\sigma^2}{n})[/tex] is a noncentral chisquare distribution [tex]\frac{X^2_i}{\sigma^2 /n}[/tex] and it's expectation is n*sigma^2

    Kinda stuck on this one: [tex]2E[X_i \bar X] = \frac 2 n E[X_i (X_1 + X_2 + \dots + X_n)][/tex]

    I was just wondering, would I be able to use the following property of the chi square distribution:

    2b87c537781cd265449ee4541fabf8ae.png
     
  5. Jan 6, 2010 #4

    statdad

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    Homework Helper

    It would be a non-central Chi-square - but you don't need that. For any random variable what do you know about an expression for [tex] E[X^2] [/tex] in terms of the first two moments?
    I would make a comment similar to the first one here: you know the distribution of the sample mean, what do you know about the expectation of its square in terms of the first two moments?
    Write it as
    [tex]
    \frac 2 n \left( E[X_i^2] + \sum_{j \ne i} E[X_i X_j]\right)
    [/tex]
    and remember that for [tex] i \ne j [/tex] the Xs are independent.
    You could - IF you have already obtained that result elsewhere in your class.
     
  6. Jan 6, 2010 #5
    Is this what you wre talking about:

    [tex]E((X_i - \overline{X})^2)= \sigma^2 = E(X)^2 - E(X^2)[/tex]

    not sure how that helps
     
  7. Jan 6, 2010 #6

    statdad

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    Homework Helper

    Actually, just the second part:
    If
    [tex]
    \sigma^2=E[X^2] - \left(E[X]\right)^2
    [/tex]

    what does [tex] E[X^2] [/tex] itself equal?
     
  8. Jan 6, 2010 #7
    [tex]E(X^2_i) = \sigma^2 + E(X_i)^2 = \sigma^2 + (n \mu)^2[/tex]

    Would the expectation for Xbar2 be equal to the expectation of of Xi?
     
  9. Jan 8, 2010 #8

    statdad

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    Homework Helper

    Why do you have

    [tex]
    E(X_i)^2 = (n\mu)^2
    [/tex]

    Should the [tex] n [/tex] really be there?

    For [tex] E[\bar X^2] [/tex], remember that the sample mean is normally distributed with mean [tex] \mu [/tex] and variance [tex] \frac{\sigma^2} n [/tex].
     
  10. Jan 8, 2010 #9
    Isn't it because [tex]\Sigma E[X_i] = E[X_1] + ... + E[X_n] = n \mu[/tex] and each E[X] = mu, so there are n of them?
     
    Last edited: Jan 8, 2010
  11. Jan 8, 2010 #10

    statdad

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    Homework Helper

    I asked because you didn't have the sum. My point is that for any random variable that has a variance,

    [tex]
    E[X^2] = \sigma^2_X + \mu^2_x
    [/tex]

    that is - the expectation of the square equals the variance plus the square of the mean.
     
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