Convergence of Probability

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Homework Statement



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)

Homework Equations





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?
 

Answers and Replies

  • #2
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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!
 
  • #3
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a)Prove that Yn/n converges in probability p.
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?
 
  • #4
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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:
  • #5
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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.
This is not even close to the correct method. See my post above (which got posted while you were writing).
 
  • #6
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[tex]P(|\frac{Y_n}{n} - np |\geq p \sqrt{np(1-p)}) \leq \frac{1}{p^2}[/tex]
 

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