product of n unifom distribution as n-> infinty

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November 11th 2009, 09:04 AM
rosh3000

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
November 11th 2009, 09:36 AM
Laurent

Quote:

Originally Posted by rosh3000

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.
November 11th 2009, 10:35 AM
rosh3000

thanks you make it sound so simple

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