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Math Help - product of n unifom distribution as n-> infinty

  1. #1
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    product of n unifom distribution as n-> infinty

    Queshion:
    The problem statement, all variables and given/known data[/b]
    Suppose  X_1, X_2, \ldots X_n are iid random variable each distributed U[1,0] (uniform distribution) Suppose 0 < a < b <: Show that:
     P((X_1X_2 \ldots X_n)^{\frac{1}{\sqrt{n}}} \in [a,b]) tends to a limit as n tends to infinty and find an expression for it

    Attempt at solution
     P[log(a) \leq (X_1, X_2 \ldots X_n)^{\frac{1}{\sqrt{n}}} \leq b]

    take natural logs
     = P[log(a) \leq \frac{1}{\sqrt{n}}log(X_1, X_2 \ldots, X_n) \leq log(b)]

    let  Y = \frac{1}{\sqrt{n}}log(X_i)<br />
\Rightarrow f_y(x) = \sqrt{n}e^{\sqrt{n}y} \mbox{ for } x \in [0, log[0])

    let  W = \frac{1}{\sqrt{n}}(log[X_1] + log[X_2], \ldots + log[x_n])

     W = \Sigma^n_{i = 1} (Y_i)

    let Moment generating function of W and Y =  \Phi_{y_i}(t) and  \Phi_w(t) respectively

    \Phi_{w} = \Pi^n_{i=0}(\Phi_{y_i}(t)) by independence.

    \Phi_{y} = \mathbb{E}[e^{xt}] = \frac{\sqrt{n}}{t + \sqrt{n}}

     \Phi_{w} = (\frac{\sqrt{n}}{t + \sqrt{n}})^n

     \stackrel{Lim}{n \rightarrow \infty}[ (\frac{\sqrt{n}}{t + \sqrt{n}})^n ] = e^{-2t} = e^{t\frac{1}{2}}
    Converges to MGF of degenerate distribution with parameter 1/2.
    as MGF converges => distribuiton converges

    the anwser doesn't seem right
    Last edited by mr fantastic; November 11th 2009 at 06:24 PM. Reason: Fixed latex
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  2. #2
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    Quote Originally Posted by rosh3000 View Post
    Queshion:
    The problem statement, all variables and given/known data[/b]
    Suppose  X_1, X_2, \ldots X_n are iid random variable each distributed U[1,0] (uniform distribution) Suppose 0 < a < b <: Show that:
     P((X_1X_2 \ldots X_n)^{\frac{1}{\sqrt{n}}} \in [a,b]) (LaTeX error corrected : \X_n instead of X_n) tends to a limit as n tends to infinty and find an expression for it

    Attempt at solution
     P[a \leq (X_1, X_2 \ldots X_n)^{\frac{1}{\sqrt{n}}} \leq b]

    take natural logs
     = P[log(a) \leq \frac{1}{\sqrt{n}}log(X_1, X_2 \ldots, X_n) \leq log(b)]

    let  Y = \frac{1}{\sqrt{n}}log(X_i)<br />
\Rightarrow f_y(x) = \sqrt{n}e^{\sqrt{n}y} \mbox{ for } x \in [0, log[0]) (problem with variable names, and sign in the exponent; nb: \log(0)=-\infty

    let  W = \frac{1}{\sqrt{n}}(log[X_1] + log[X_2], \ldots + log[x_n])

     W = \Sigma^n_{i = 1} (Y_i)

    let Moment generating function of W and Y =  \Phi_{y_i}(t) and  \Phi_w(t) respectively

    \Phi_{w} = \Pi^n_{i=0}(\Phi_{y_i}(t)) by independence.

    \Phi_{y} = \mathbb{E}[e^{xt}] = \frac{\sqrt{n}}{t + \sqrt{n}}

     \Phi_{w} = (\frac{\sqrt{n}}{t + \sqrt{n}})^n

     \stackrel{Lim}{n \rightarrow \infty}[ (\frac{\sqrt{n}}{t + \sqrt{n}})^n ] = e^{-2t} = e^{t\frac{1}{2}} Wrong
    Converges to MGF of degenerate distribution with parameter 1/2.
    as MGF converges => distribuiton converges

    the anwser doesn't seem right
    The final limit is : \stackrel{Lim}{n \rightarrow \infty}[ (\frac{\sqrt{n}}{t + \sqrt{n}})^n ] = 0 which is not a MGF (except if infinite values are allowed).

    A good reason why this isn't surprising: \log\left( (X_1\cdots X_n)^{1/\sqrt{n}}\right) = \frac{1}{\sqrt{n}}(Z_1+\cdots+Z_n) where Z_i=\log X_i. Since E[Z_1]=\int_0^1 \log t dt =-1 exists and is finite, the law of large numbers gives \frac{Z_1+\cdots+Z_n}{n}\to -1 a.s.. As a consequence, \frac{1}{\sqrt{n}}(Z_1+\cdots+Z_n)\to_n -\infty a.s., hence, taking exponential,  (X_1\cdots X_n)^{1/\sqrt{n}}\to 0 a.s., and the probability  P((X_1X_2 \ldots X_n)^{\frac{1}{\sqrt{n}}} \in [a,b]) tends to 0.
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  3. #3
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    thanks you make it sound so simple

    (the first part of the question was find E(Z))
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