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Math Help - convergence in distribution

  1. #1
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    Post convergence in distribution

    let X1,X2, . . ., Xn be independent random variables all with uniform distribution on [0,1]. Let 0<=a<=b
    show P(a<=(X1*X2*...*Xn)^(1/Sqr[n]))<=b) tends to a limit as n tends to infinity and find an expression for it?

    i have some problems with my solution and am not sure it is right ..... thanks for any help

    Let Y=(X1*X2*...*Xn)^(1/Sqr[n]))

    c.d.f FY(y) = P({Y<=y})= P({(X1*X2*...*Xn)^(1/Sqr[n]))<=y})
    =P({X1^(n/Sqr[n])<=y}) = P({X1^(Sqr[n])<=y})
    =P({X1<=Sqr[y] to the (Sqr[n])th root})

    =
    0 if (Sqr[y] to the (Sqr[n])th root)<=0
    Sqr[y] to the (Sqr[n])th root if 0<= Sqr[y] to the (Sqr[n])th root <=1
    1 if 1<= Sqr[y] to the (Sqr[n])th root


    but not sure if this is right and then when considering looking at this function as n tends to infinity i got that it tends to 1 for all y eeeeek ?????


    thanks for any help

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  2. #2
    MHF Contributor matheagle's Avatar
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    These rvs are not the same.

    So the statement X_1X_2\cdots X_n\le a

    is not the same as X_1^n\le a
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  3. #3
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    ok i see that now thankssss

    to solve this i want to find c.d.f of Y=(X1*X2*...*Xn)^(1/Sqr(n))

    Fy(y) = P(Y<=y) = P((X1*X2*...*Xn)^(1/Sqr(n))<=y)
    = P(X1*X2*...*Xn<= Sqr[y] to the (1/Sqr(n))th root)

    would i have to do an n length intergral of the joint density function to solve this??? is this right or is there an something better i can do ????
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  4. #4
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    Have you thought of looking at \log Y? I can smell Central Limit theorem here
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  5. #5
    MHF Contributor matheagle's Avatar
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    Log Y is correct, but I'm not sure you even need the CLT.
    The logarithm turns thisthe product into a sum and the log of a uniform is just an exponential.
    The sum of iid exponentials is a gamma...

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    Last edited by matheagle; December 30th 2009 at 05:43 PM.
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  6. #6
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    Sweet so i have

    i have Fy(y) = P(Y<=y) = P((X1*X2*...*Xn)^(1/Sqr(n))<=y))
    = P(log((X1*X2*...*Xn)^(1/Sqr(n))<=log(y))
    = P((1/Sqr(n)log(X1*X2*...*Xn)<=log(y))
    = P(log(X1*X2*...*Xn)<=Sqr(n)*log(y))
    = P(log(X1)+log(X2)+...+log(Xn)<=log(y))

    and the m.d.f

    Flog(X1)(x) =
    = exp[x] if x<0
    = 1 if 0<=0



    where do i go from here, what i have tryed dosnt come out very nice . . .

    thanks
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  7. #7
    MHF Contributor matheagle's Avatar
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    please use tex, this is hard to read
    use [tex] in front and [/math ] no space, at the end of each line
    and you should hit the Thanks link.
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  8. #8
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    Hello,

    Quote Originally Posted by silk23 View Post
    Sweet so i have

    i have Fy(y) = P(Y<=y) = P((X1*X2*...*Xn)^(1/Sqr(n))<=y))
    = P(log((X1*X2*...*Xn)^(1/Sqr(n))<=log(y))
    = P((1/Sqr(n)log(X1*X2*...*Xn)<=log(y))
    = P(log(X1*X2*...*Xn)<=Sqr(n)*log(y))
    = P(log(X1)+log(X2)+...+log(Xn)<=log(y))

    and the m.d.f

    Flog(X1)(x) =
    = exp[x] if x<0
    = 1 if 0<=0



    where do i go from here, what i have tryed dosnt come out very nice . . .

    thanks
    The Xi's are independent. So are the -log(Xi), which follow an exponential distribution, according to the cdf (slight transformation)
    And as oldeagle says, the sum of iid exponential distributions is a Gamma distribution (see here : Exponential distribution - Wikipedia, the free encyclopedia)

    The problem with this method is that it's impossible to get a defined formula for the cdf.

    So the best solution seems to be approximating this by the Central Limit Theorem, since you're interested in the limit !
    For this, you just need to know that \frac{X_1+\dots+X_n-nE[X_1]}{\sqrt{var[X_1]}\cdot\sqrt{n}} converges in distribution to a normal distribution (0,1)
    So the cdf converges too.

    please use tex, this is hard to read
    use [tex] in front and [/math ] no space, at the end of each line
    and you should hit the Thanks link.
    This is way better to read than many other threads.
    And he knows the Thanks link (he thanked 2 posts), but it's up to him to decide if a post deserves one...
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