Convergence of Random Variables in Probability

[cse63146]

Homework Statement

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

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

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

c)Prove that (Y_n/n)(1 - Y_n/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 Y_n/n = lim Y_n * lim 1/n.

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

Where did I make the mistake?

 

[kermy]

You can't take the limit of the 1/n part independently of the Y_n part.

That is like saying that

1 goes to 0 as n goes to infinity since

1=\frac{n}{n} = n \cdot \frac{1}{n} and that 1/n goes to 0!

 

[Billy Bob]

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

I assume you mean Y_n/n converges in probability to p.

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

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

Have you learned Chebyshev's inequality?

 

[cse63146]

So I tried doing it with Y_n, and I tried "splitting" the limits up:

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

lim n!/n! = 1 so:

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

Still stuck though.

Didn't see your message.

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

would mu be np (because this is a binomial distribution)?

 

[Billy Bob]

So I tried doing it with Y_n, and I tried "splitting" the limits up:

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

lim n!/n! = 1 so:

lim(n-x!)^-1 * lim p^x * 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).

 

[cse63146]

P(|\frac{Y_n}{n} - np |\geq p \sqrt{np(1-p)}) \leq \frac{1}{p^2}