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Limit definition examples

  1. Mar 10, 2005 #1
    First 3 definitions:
    \left( 1 \right)\mathop {\lim }\limits_{n \to \infty } P\left( {\left| {X_n - X} \right| > \varepsilon } \right) = 0 \\
    \left( 2 \right)P\left( {\mathop {\lim }\limits_{n \to \infty } X_n = X} \right) = 1 \\
    \left( 3 \right)\mathop {\lim }\limits_{n \to \infty } E\left[ {\left( {X_n - X} \right)^2 } \right] = 0 \\

    I need to find:
    a. an example that (1) does not give (3).
    b. an example that (1) does not give (2).
  2. jcsd
  3. Mar 28, 2005 #2
    Your stuff seems quite hard...I don't know where to begin...are you sure there are no things like 2=>1, or other logical implications ??

    Have you tried some series like : X_n=1/ln(n) ?
  4. Mar 28, 2005 #3
    let X_1 =1 on [0,1]
    let X_2=1 on [0,1/2] , 0 otherwise
    let x_3=1 on [1/2,1] , 0 ow
    let x_4=1 on[0,1/3] ,0 ow
    let x_5=1 on [1/3,2/3] , 0 ow

    can you show that X_n converges in probability (1)
    but x_n does not converge a.s. (2)
  5. Mar 28, 2005 #4
    Hm..Zone Ranger interpreted the x_n as functions....I took them as numbers.

    If X_n are functions, then it's easy to find what you want:

    Let X_n be functions over [0;1], with

    X_n(x)=g(n) if x in [0;1/n], 0 otherwise, with a stricly increasing function g(n)


    1) P(|X_n|>e)<=1/n, hence the limit gives 0
    b: 2) X_n does not tend towards the 0 function, since X_n(0)>0 forall n
    a: E(X_n^2)=g(n)^2/n...here choose g(n)=Sqrt(n)....you get

    limit n->infty E((X_n-0)^2)=1

    But if the X_n are numbers, I don't know how to solve this...

    Thanks to Zone Ranger.
  6. Mar 28, 2005 #5
    the X_n have to be random variables (measurable functions).

    you are correct that for your choice of X_n, X_n(0)>0. but still X_n->0 a.s.
    so with your X_n (2) still holds.
  7. Mar 28, 2005 #6
    Assuming the X_n's do not have be a random sample from X, then you could define the RV's as discrete, each taking on a value with probability 1 and all other values with probability 0. It's easy to set up examples that way.
    Last edited: Mar 28, 2005
  8. Mar 28, 2005 #7
    I don't know what u mean with a.s. (maybe converges uniformly ???)
  9. Mar 28, 2005 #8
  10. Apr 7, 2005 #9
    I'm sorry for not answering sooner, but after 10 days and no answer I thought noone was interested. :smile:

    I know that (2) gives (1) and also (3) gives (1), these two are not hard to prove.
    I'm going to try your suggestions now.
  11. Apr 7, 2005 #10
    Taking this example, if I understand correctly, X_n does not converge almost surely (2) since it does not converge to a single value but rather keeps dividing the interval [0,1] into halfs, and "jumps back and forth" within the interval.
    X_n does not converge in mean square to X (3) because of the same reason, namely the expected value of [tex](X_n-X)^2[/tex] kepps changing as [tex]{n \to \infty }[/tex].
    However, I don't know how to show that X_n converges in probability.
  12. May 2, 2005 #11

    [tex]X_n[/tex] does converge in mean square to X (X=0)

    the expected value of [tex](X_n-X)^2[/tex] goes to 0 as [tex]{n \to \infty }[/tex].
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