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Math Help - limiting distribution

  1. #1
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    limiting distribution

    Hi!! Um I'm really lost with limiting distributions, so can someone please explain to me slowly how to solve this problem, or just help me get a start on it?

    Find the limiting distribution of Yn = nXn(1)
    f(x) = 1/ (1+x)^2

    THank you.. !
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  2. #2
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    Quote Originally Posted by gconfused View Post
    Hi!! Um I'm really lost with limiting distributions, so can someone please explain to me slowly how to solve this problem, or just help me get a start on it?

    Find the limiting distribution of Yn = nXn(1)
    f(x) = 1/ (1+x)^2

    THank you.. !
    Is Xn(1) defined to be \min \{X_1, \, X_2, \, .... \, X_n \} where X_1, \, X_2, \, .... \, X_n are i.i.d. random variables with pdf given by f(x)?
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    Thank you for replying. but ummm well the question only says "Let X1, X2, .., Xn be a random sample from a Pareto distribution. Find the Limiting distribution". So I'm really not sure what I'm supposed to do, coz my textbook doesnt say anything about limiting distributions..
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  4. #4
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    Quote Originally Posted by gconfused View Post
    Thank you for replying. but ummm well the question only says "Let X1, X2, .., Xn be a random sample from a Pareto distribution. Find the Limiting distribution". So I'm really not sure what I'm supposed to do, coz my textbook doesnt say anything about limiting distributions..
    Limiting distribution of what?

    Neither of your two posts in this thread make any sense to me. Please post the complete and entire question, exactly as it's written in your textbook or wherever it is that you got it from.
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  5. #5
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    umm well the whole question is:

    "Let X1, X2, .. Xn be a random sample from a Pareto distribution which has probability density function

    f(x) = 1/(1+x)^2, x>0

    Find the limiting distribution of Yn = nX(1)."
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  6. #6
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    Quote Originally Posted by gconfused View Post
    umm well the whole question is:

    "Let X1, X2, .. Xn be a random sample from a Pareto distribution which has probability density function

    f(x) = 1/(1+x)^2, x>0

    Find the limiting distribution of Yn = nX(1)."
    I'm guessing that X(1) is \min \{X_1, \, X_2, \, .... \, X_n \}. To get the pdf of nX(1):

    \Pr(n X(1) \leq x) = \Pr\left( X(1) \leq \frac{x}{n} \right) = 1 - \Pr\left( X(1) > \frac{x}{n} \right)

     = 1 - \Pr\left( X_1 > \frac{x}{n} \right) \cdot \Pr\left( X_1 > \frac{x}{n} \right) \cdot .... \cdot \Pr\left( X_n > \frac{x}{n} \right)

     = 1 - \left[1 - F\left( \frac{x}{n}\right) \right]^n.

    But F\left( \frac{x}{n}\right) = \int_0^{x/n} \frac{1}{(1 + u)^2} \, du = 1 - \frac{n}{x + n}.

    Therefore \Pr(n X(1) \leq x) = 1 - \left[\frac{n}{x + n} \right]^n.

    Differentiation of this gives the pdf of n X(1): \left[\frac{n}{x + n}\right]^{n+1}.
    Last edited by mr fantastic; May 4th 2009 at 11:43 PM. Reason: Popped the membrane of a runaway 2 and brought it down a bit.
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  7. #7
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by mr fantastic View Post
    I'm guessing that X(1) is \min \{X_1, \, X_2, \, .... \, X_n \}. To get the pdf of nX(1):

    \Pr(n X(1) \leq x) = \Pr\left( X(1) \leq \frac{x}{n} \right) = 1 - \Pr\left( X(1) > \frac{x}{n} \right)

     = 1 - \Pr\left( X_1 > \frac{x}{n} \right) \cdot \Pr\left( X_1 > \frac{x}{n} \right) \cdot .... \cdot \Pr\left( X_n > \frac{x}{n} \right)

     = 1 - \left[1 - F\left( \frac{x}{n}\right) \right]^n.

    But F\left( \frac{x}{n}\right) = \int_0^{x/n} \frac{1}{1 + u}^2 \, du = 1 - \frac{n}{x + n}.

    Therefore \Pr(n X(1) \leq x) = 1 - \left[\frac{n}{x + n} \right]^n.
    the limit is... \Pr(n X(1) \leq x) = 1 - \left[\frac{n}{x + n} \right]^n\to 1-e^{-x}.

    And whats with that floating 2 in that integrand. It looks suspicious.
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  8. #8
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    Quote Originally Posted by matheagle View Post
    the limit is... \Pr(n X(1) \leq x) = 1 - \left[\frac{n}{x + n} \right]^n\to 1-e^{-x}.

    And whats with that floating 2 in that integrand. It looks suspicious.
    I guess it was filled with helium gas .....
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    I'm guessing that X(1) is \min \{X_1, \, X_2, \, .... \, X_n \}. To get the pdf of nX(1):

    \Pr(n X(1) \leq x) = \Pr\left( X(1) \leq \frac{x}{n} \right) = 1 - \Pr\left( X(1) > \frac{x}{n} \right)

     = 1 - \Pr\left( X_1 > \frac{x}{n} \right) \cdot \Pr\left( X_1 > \frac{x}{n} \right) \cdot .... \cdot \Pr\left( X_n > \frac{x}{n} \right)

     = 1 - \left[1 - F\left( \frac{x}{n}\right) \right]^n.

    But F\left( \frac{x}{n}\right) = \int_0^{x/n} \frac{1}{1 + u}^2 \, du = 1 - \frac{n}{x + n}.

    Therefore \Pr(n X(1) \leq x) = 1 - \left[\frac{n}{x + n} \right]^n.

    Differentiation of this gives the pdf of n X(1): \left[\frac{n}{x + n}\right]^{n+1}.


    So is that the answer?
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  10. #10
    MHF Contributor matheagle's Avatar
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    I posted it, look up^^^^^
    This converges in distribution to an exponential with mean 1.

    hot air 4 sure
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  11. #11
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    Quote Originally Posted by matheagle View Post
    I posted it, look up^^^^^
    This converges in distribution to an exponential with mean 1.

    hot air 4 sure
    We make a good team - I do the heavy lifting and baldeagle writes down the answer.

    Thread closed before it drifts (like a helium filled 2) off-topic (sorry, I like to get the last word ).
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