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Math Help - Determining Limiting Distributions

  1. #1
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    Determining Limiting Distributions

    1. Let Xn denote the mean of a random sample of size n from a distribution that is N(\mu, \sigma ^2). Find the Limiting distribution of Xn.

    2. Let Y1 denote the first order statistic of a random sample of size n from a distribution that has the PDF f(x) = e^{-(x - \theta)} \ \ \theta < x . Let Z_n = n(Y_1 - \theta). Find the distribution of Zn.



    1.

    So to find the limiting distribution, I need to take the limit of the PDF as n approaches infinity.

    X_n = \frac{X_1 + X_2 + ... + X_n}{n} = \frac{nX_n}{n} = X_n

    Kinda stuck here.

    Any help would be appreciated.
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  2. #2
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    Can't manipulate RV like that when it comes to distributions. Best way is to go through the moment generating function. You should expect to get another normal distribution <br />
N(\mu, \sigma ^2/n)<br />
.
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  3. #3
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    Quote Originally Posted by statmajor View Post
    1. Let Xn denote the mean of a random sample of size n from a distribution that is N(\mu, \sigma ^2). Find the Limiting distribution of Xn.

    2. Let Y1 denote the first order statistic of a random sample of size n from a distribution that has the PDF f(x) = e^{-(x - \theta)} \ \ \theta < x . Let Z_n = n(Y_1 - \theta). Find the distribution of Zn.



    1.

    So to find the limiting distribution, I need to take the limit of the PDF as n approaches infinity.

    X_n = \frac{X_1 + X_2 + ... + X_n}{n} = \frac{nX_n}{n} = X_n

    Kinda stuck here.

    Any help would be appreciated.
    1. It makes absolutely no sense to use the symbol X_n twice in the same equation but to mean two different things.

    2. The distribution of U = X_1 + X_2 + ... + X_n is not n X_n. Review how to find the distribution of a sum of iid normal random variables.
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  4. #4
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    So: E[e^{Ut}]

    where  U = X_1 + X_2 + ... + X_n?
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  5. #5
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    Quote Originally Posted by statmajor View Post
    So: E[e^{Ut}]

    where  U = X_1 + X_2 + ... + X_n?
    No.

    Post #2 told you the distribution. Your job is to understand why this is the distribution. Your job is also to review how to get the moment generating function of a sum of independent random variables.

    Then you have to take the limit of the distribution as n --> +oo. The result is unsurprising (Google Dirac delta function).

    This question has uncovered several apparent fundamental gaps in your knowledge and understanding. Some of these gaps have been pointed out and you must now take the necessary steps to address this.
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  6. #6
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    Quote Originally Posted by statmajor View Post
    So: E[e^{Ut}]

    where  U = X_1 + X_2 + ... + X_n?
    Okay, but you lack 1/n factor for U. You also need to do further job than that. Evaluate that expected value for a start and then do what the guy above suggested.
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  7. #7
    MHF Contributor matheagle's Avatar
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    The sum of n random variables is not n times one of them.
    THIS is what you meant...


    \overline X_n = \frac{X_1 + X_2 + ... + X_n}{n}\ne X_n
    Last edited by matheagle; October 19th 2009 at 10:49 PM.
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  8. #8
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    Let me give this another try:

    E(e^{tX_n}) = E(e^{t \frac{X_1 + ... + X_n}{n}}) = \prod E(e^{t\frac{X_i}{n}})= (e^{\mu t + 0.5 \frac{\sigma^2}{n}t^2})^n

    and I would take the limit of (e^{\mu t + 0.5 \frac{\sigma^2}{n}t^2})^n as n approaches infinity (which would equal infinity)?

    or did I make another dumb mistake somewhere?
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  9. #9
    MHF Contributor matheagle's Avatar
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    The sample mean converges almost sure to \mu almost surely.
    All you need is a finite first moment and in this case you even have a finite second moment.
    Normality is not necessary.
    Last edited by matheagle; October 20th 2009 at 03:31 PM.
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  10. #10
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    but is <br />
(e^{\mu t + 0.5 \frac{\sigma^2}{n}t^2})^n<br />
correct?
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  11. #11
    MHF Contributor matheagle's Avatar
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    substitute t/n for t in the MGF
    that will fix this

    Quote Originally Posted by statmajor View Post
    Let me give this another try:

    E(e^{tX_n}) = E(e^{t \frac{X_1 + ... + X_n}{n}}) = \prod E(e^{{t\over n}X_i})= (e^{\mu {t\over n} + 0.5 \frac{\sigma^2}{n^2}t^2})^n\to e^{\mu t}



    or did I make another dumb mistake somewhere?
    Last edited by matheagle; October 20th 2009 at 03:30 PM.
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  12. #12
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    That makes senses.

    Thank you (and everyone else) for all their help
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