1. ## Consistent Estimator

If y has a binomial distribution with n trials and success probability p, show that Y/n is a consistent estimator of p.

Can someone show how to show this. I appreciate it any and all help. thanks.

2. Hello,

Yn=X1+...+Xn, where the Xi's are independent Bernoulli distributions B(p).
Law of large numbers : Yn/n converges to E[X1] almost surely, and hence in probability -> consistent.

3. Originally Posted by jzellt
If y has a binomial distribution with n trials and success probability p, show that Y/n is a consistent estimator of p.

Can someone show how to show this. I appreciate it any and all help. thanks.
To show that Y/n is a consistent estimator of p, it is sufficient to show that $\lim_{n\to\infty}E(\frac{Y}{n})=p$ and $\lim_{n\to\infty}Var(\frac{Y}{n})=0$.

4. Originally Posted by Sambit
To show that Y/n is a consistent estimator of p, it is sufficient to show that $\lim_{n\to\infty}E(\frac{Y}{n})=p$ and $\lim_{n\to\infty}Var(\frac{Y}{n})=0$.
But here it may be much easier to prove convergence in probability, since we obviously have the law of large numbers ?

5. Originally Posted by Moo
But here it may be much easier to prove convergence in probability, since we obviously have the law of large numbers ?
Basically both the procedures need same labour which is very little indeed. Since $Y\sim Bin(n,p)$, Obviously $E(Y)=np$ which gives $E(\frac{Y}{n})=p$ and $Var(Y)=np(1-p)$ , so $Var(\frac{Y}{n})=\frac{np(1-p)}{n^2} = \frac{p(1-p)}{n}\to\0$ as $n\to\infty$. No tough job

6. there are a lot of meanings to consistency, strong, weak...
I was curious as to what was meant here.