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Thread: 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 $\displaystyle X_1, X_2, \ldots X_n $ are iid random variable each distributed U[1,0] (uniform distribution) Suppose 0 < a < b <: Show that:
    $\displaystyle 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
    $\displaystyle P[log(a) \leq (X_1, X_2 \ldots X_n)^{\frac{1}{\sqrt{n}}} \leq b] $

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

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

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

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

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

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

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

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

    $\displaystyle \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; Nov 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 $\displaystyle X_1, X_2, \ldots X_n $ are iid random variable each distributed U[1,0] (uniform distribution) Suppose 0 < a < b <: Show that:
    $\displaystyle 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
    $\displaystyle P[a \leq (X_1, X_2 \ldots X_n)^{\frac{1}{\sqrt{n}}} \leq b] $

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

    let $\displaystyle Y = \frac{1}{\sqrt{n}}log(X_i)
    \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: $\displaystyle \log(0)=-\infty$

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

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

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

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

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

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

    $\displaystyle \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 : $\displaystyle \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: $\displaystyle \log\left( (X_1\cdots X_n)^{1/\sqrt{n}}\right) = \frac{1}{\sqrt{n}}(Z_1+\cdots+Z_n)$ where $\displaystyle Z_i=\log X_i$. Since $\displaystyle E[Z_1]=\int_0^1 \log t dt =-1$ exists and is finite, the law of large numbers gives $\displaystyle \frac{Z_1+\cdots+Z_n}{n}\to -1$ a.s.. As a consequence, $\displaystyle \frac{1}{\sqrt{n}}(Z_1+\cdots+Z_n)\to_n -\infty$ a.s., hence, taking exponential, $\displaystyle (X_1\cdots X_n)^{1/\sqrt{n}}\to 0$ a.s., and the probability $\displaystyle 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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