1. ## Convergence it Distribution

Let the r.v Yn have a dist that is b(n,p)

(a) Prove that Yn / n converges in probability p. This result is one form of the weak law of large numbers.

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

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

2. Okay, the first is just chebyshev's
P(|Y/n-p|>epsilon)=P(|Y-np|>n epsilon)
Use Chebyshev's we have
<V(Y)/(n^2 Epsilon^2)
=(npq)/(n^2 Epsilon^2)
=(pq)/(n Epsilon^2)
which goes to zero as n->infinity
We can quote result someone else asked me last week.
If The variance is finite and the stat is unbiased then it consistent.
Thats what I just did, here.

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I'm going to make dinner.
I'll do parts b,c later.

By the way this is not convergence in distribution, this is a stronger mode of convergence, convergence in probability.
Convergence in distribution is where a sequence of cdfs converges to a cdf, like in the central limit theorem.
LOL, see
http://www.statisticalengineering.com/convergence.htm
"Convergence in probability" is not quite the same as convergence in distribution.
this is quite useful...
http://en.wikipedia.org/wiki/Converg...ndom_variables

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Part b can be done two ways.
It's basically the same as a, just switch what is a success and a failure
That changes p and q and you're done.
However, let epsilon>0...
P(|(1-Y/n)-(1-p)|>epsilon)
=P(|-Y/n+p|>epsilon)
=P(|Y/n-p|>epsilon) since |-1|=1
=P(|Y-np|>n epsilon)...
the rest is as b4.

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Part c follows from...
If x_n->a and y_n->b (both in probability),
then (x_n)(y_n) ->ab, in probability.
I'm trying to prove your problem directly via the triangle inequality.
Let epsilon>0...
P(|(Y/n)(1-Y/n)-p(1-p)|>epsilon)
=P(|(Y/n)-(Y/n)^2-p+p^2)|>epsilon)
=P(|[(Y/n)-p] + [p^2-(Y/n)^2]|>epsilon) now the Triangle
<=P(|(Y/n)-p|>epsilon/2)+P(|(Y/n)^2-p^2|>epsilon/2).
Since the sum is greater than epsilon, at least one of them in abs value must be greater than epsilon/2.
Now we've already shown the first term goes to zero.
The second term goes to zero since the square of a sequence going to p has their terms going to p^2 in probability too.
See, since Y/n->p in probability, the square does too.
In general if x_n->a in prob and g(.) is any continuous function
then g(x_n)->g(a) in prob
That can be found at...
http://planetmath.org/encyclopedia/C...ormations.html