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Thread: 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 $\displaystyle (X_n)_{n\in\mathbb{N}}$

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

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


    I tried to relate it to the series $\displaystyle \sum \frac{X_n}{n}$, 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

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

    $\displaystyle ={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 $\displaystyle Y_n\to a$ almost surely, then for any continuous function g, $\displaystyle g(Y_n)\to g(a)$ 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...

    $\displaystyle \sum_{n=1}^{\infty}P\{|X|>\epsilon n\}<\infty$
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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...

    $\displaystyle \sum_{n=1}^{\infty}P\{|X|>\epsilon n\}<\infty$
    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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    Quote Originally Posted by Moo View Post
    I had a doubt about it :

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

    But here, g depends on n ?
    MathEagle's proof works fine : almost surely,
    - the sequence $\displaystyle \Big(\frac{S_n}{n}\Big)_n$ converges to $\displaystyle E[X_1]$
    - the sequence $\displaystyle \Big(\frac{S_{n-1}}{n-1}\Big)_n$ converges to $\displaystyle E[X_1]$ (because of the first point)
    - the sequence $\displaystyle \Big(\frac{n-1}{n}\Big)_n$ converges to 1 (obviously),
    and these three events together (I could also have let the third one appart) imply that $\displaystyle \Big(\frac{X_n}{n}\Big)_n$ 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; Apr 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 $\displaystyle E|X|<\infty$ then $\displaystyle E\left|{X\over \epsilon}\right|<\infty$ for all $\displaystyle \epsilon>0$

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

    This follows from

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

    Thus $\displaystyle X_n/n\to 0$

    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 $\displaystyle E|X|<\infty$ then $\displaystyle E\left|{X\over \epsilon}\right|<\infty$ for all $\displaystyle \epsilon>0$

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

    This follows from

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

    Thus $\displaystyle X_n/n\to 0$

    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 $\displaystyle a_n$ be any positive strictly increasing sequence going to infinity. Let $\displaystyle a(x)$ be any continuous extension of $\displaystyle a_n$ then for all positive random variables X...

    $\displaystyle \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 $\displaystyle f(x)=a^{-1}(x)$ and

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

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

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

    However

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

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

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

    while

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

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

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

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