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Convergence of Probability

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Let the random variable Yn have the distribution b(n,p).

    a)Prove that Yn/n converges in probability p.

    b)Prove that 1 - Yn/n converges to 1 - p.

    c)Prove that (Yn/n)(1 - Yn/n) converges in probability to p(1-p)

    2. Relevant equations



    3. The attempt at a solution

    Note: when lim -> the limit of as n approaches infinity.

    a) lim Yn/n = lim Yn * lim 1/n.

    But the lim 1/n = 0 => lim Yn/n = 0. But it's supposed to converge to p.

    Where did I make the mistake?
     
  2. jcsd
  3. Sep 30, 2009 #2
    You can't take the limit of the 1/n part independently of the Yn part.

    That is like saying that

    1 goes to 0 as n goes to infinity since

    [tex]1=\frac{n}{n} = n \cdot \frac{1}{n}[/tex] and that 1/n goes to 0!
     
  4. Sep 30, 2009 #3
    I assume you mean Yn/n converges in probability to p.

    What is your defn of "converges in probability"? Are you supposed to show that for each [tex]\varepsilon>0[/tex],

    [tex]P(|Y_n/n - p|\ge \varepsilon)\to 0[/tex] as [tex]n\to\infty[/tex]?

    Have you learned Chebyshev's inequality?
     
  5. Sep 30, 2009 #4
    So I tried doing it with Yn, and I tried "splitting" the limits up:

    lim n!/n! * lim(n-x!)-1 * lim px * lim(1-p)n * lim(1-p)-x * lim n-1

    lim n!/n! = 1 so:

    lim(n-x!)-1 * lim px * lim(1-p)n * lim(1-p)-x * lim n-1

    Still stuck though.

    Didn't see your message.

    Chebyshev's inequality: [tex]P(|X - \mu |\geq k \sigma) \leq 1/k^2[/tex]

    would mu be np (because this is a binomial distribution)?
     
    Last edited: Sep 30, 2009
  6. Sep 30, 2009 #5
    This is not even close to the correct method. See my post above (which got posted while you were writing).
     
  7. Sep 30, 2009 #6
    [tex]P(|\frac{Y_n}{n} - np |\geq p \sqrt{np(1-p)}) \leq \frac{1}{p^2}[/tex]
     
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