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Math Help - A simple problem of convergence...

  1. #1
    Moo
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    A simple problem of convergence...

    ...But I can't get it

    We have a sequence of iid nonnegative rv's (X_n)_{n\in\mathbb{N}}

    Assuming that E[X_1] is finite, we have by the sLLN that \frac{X_1+\dots+X_n}{n} converges to a finite limit.

    But how can I conclude that \lim_{n\to\infty} \frac{X_n}{n}=0 ?


    I tried to relate it to the series \sum \frac{X_n}{n}, but the inequality is the other way round than the one that would be helpful...

    ...so since inequalities don't help me, I hope you guys can help me


    Thanks !


    Note : I also - miserably - tried CesÓro's mean.
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  2. #2
    MHF Contributor matheagle's Avatar
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    By subtraction

    {X_n\over n} ={S_n-S_{n-1}\over n}

    ={S_n\over n}-\left({S_{n-1}\over n-1}\right)\left({n-1\over n}\right)

    Now take your limits
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  3. #3
    Moo
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    I had a doubt about it :

    If Y_n\to a almost surely, then for any continuous function g, g(Y_n)\to g(a) a.s.

    But here, g depends on n ?
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  4. #4
    MHF Contributor matheagle's Avatar
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    I think you can use Borel-Cantelli too since you have a first moment...

    \sum_{n=1}^{\infty}P\{|X|>\epsilon n\}<\infty
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  5. #5
    Moo
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    Quote Originally Posted by matheagle View Post
    I think you can use Borel-Cantelli too since you have a first moment...

    \sum_{n=1}^{\infty}P\{|X|>\epsilon n\}<\infty
    Hum is it okay whether X is a continuous or a discrete rv ? And P(|X|>en) is the cdf of Xn/n, okay, but how does it help actually ?

    Sorry, I'm a bit lost (and tired, it's 2am here )

    And the hint says to use the LLN...
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  6. #6
    MHF Contributor matheagle's Avatar
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    Look at Theorem 2 on page 125 of..........

    Probability theory: independence ... - Google Books
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  7. #7
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    Quote Originally Posted by Moo View Post
    I had a doubt about it :

    If Y_n\to a almost surely, then for any continuous function g, g(Y_n)\to g(a) a.s.

    But here, g depends on n ?
    MathEagle's proof works fine : almost surely,
    - the sequence \Big(\frac{S_n}{n}\Big)_n converges to E[X_1]
    - the sequence \Big(\frac{S_{n-1}}{n-1}\Big)_n converges to E[X_1] (because of the first point)
    - the sequence \Big(\frac{n-1}{n}\Big)_n converges to 1 (obviously),
    and these three events together (I could also have let the third one appart) imply that \Big(\frac{X_n}{n}\Big)_n converges to 0, hence this latter limit holds almost-surely.
    By the way, I'm not sure how to use Borel-Cantelli here.
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  8. #8
    Moo
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    Thank(s to) you, I feel stupid now !
    Last edited by Moo; April 3rd 2010 at 11:20 AM.
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  9. #9
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Moo View Post
    Thank(s to) you, I feel stupid now !
    I'm glad that I could help.
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  10. #10
    MHF Contributor matheagle's Avatar
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    If E|X|<\infty then E\left|{X\over \epsilon}\right|<\infty for all \epsilon>0

    Hence  \sum_{n=1}^{\infty}P(|X_n|>n\epsilon)=\sum_{n=1}^{  \infty}P(|X|>n\epsilon)<\infty

    This follows from

     \sum_{n=1}^{\infty}P(|X|>n^{1/r})\le E|X|^r\le \sum_{n=0}^{\infty}P(|X|>n^{1/r})<\infty

    Thus X_n/n\to 0

    I don't understand why the rvs are nonnegative.
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  11. #11
    Moo
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    Quote Originally Posted by matheagle View Post
    If E|X|<\infty then E\left|{X\over \epsilon}\right|<\infty for all \epsilon>0

    Hence  \sum_{n=1}^{\infty}P(|X_n|>n\epsilon)=\sum_{n=1}^{  \infty}P(|X|>n\epsilon)<\infty

    This follows from

     \sum_{n=1}^{\infty}P(|X|>n^{1/r})\le E|X|^r\le \sum_{n=0}^{\infty}P(|X|>n^{1/r})<\infty

    Thus X_n/n\to 0

    I don't understand why the rvs are nonnegative.
    I was wondering, as I did when I saw your second post, but I'm not sure : doesn't this inequality hold for discrete random variables only ?

    Perhaps they're nonnegative because I'm studying populations, so it can't be negative lol. And it's E[X], not E|X|
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  12. #12
    Junior Member autumn's Avatar
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    Quote Originally Posted by Moo View Post
    I was wondering, as I did when I saw your second post, but I'm not sure : doesn't this inequality hold for discrete random variables only ?

    Perhaps they're nonnegative because I'm studying populations, so it can't be negative lol. And it's E[X], not E|X|
    Let a_n be any positive strictly increasing sequence going to infinity. Let a(x) be any continuous extension of a_n then for all positive random variables X...

     \sum_{n=1}^{\infty}P(X\ge a_n) \le E(a^{-1}(X))\le \sum_{n=0}^{\infty}P(X> a_n)

    If we let f(x)=a^{-1}(x) and

    W=\sum_{k=1}^{\infty}kI(k\le f(X)<k+1)

    Y=\sum_{k=0}^{\infty}(k+1)I(k< f(X)\le k+1)

    we have W\le f(X)\le Y so E(W)\le E(f(X))\le E(Y)

    However

     \sum_{n=1}^{\infty}P(X\ge a_n) \le \sum_{n=1}^{\infty}P(f(X)\ge n)

     =\sum_{n=1}^{\infty}\sum_{k=n}^{\infty}P(k\le f(X)<k+1)

     =\sum_{k=1}^{\infty}kP(k\le f(X)<k+1)=E(W)

    while

     \sum_{n=0}^{\infty}P(X> a_n) \le \sum_{n=0}^{\infty}\sum_{k=n}^{\infty}P(k< f(X)\le k+1)

     =\sum_{k=0}^{\infty}(k+1)P(k< f(X)\le k+1)=E(Y)

    So for any random variable X and positive number r...

     \sum_{n=1}^{\infty}P(|X|\ge n^{1/r})\le E|X|^r\le \sum_{n=0}^{\infty}P(|X|> n^{1/r})
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