# Thread: Asymptotic distributions of MLEs

1. ## Asymptotic distributions of MLEs

Hello all! Having a problem with general MLE concepts. See if you can help me with this.

Let X1,...Xn be a random sample of size n where Xi~BIN(1,p). How do I find the Asyptotic distribution of the MLE of p?

Thanks for any help!

2. the MLE is $\hat P$ which is an sample mean of Bernoulli's

So ${\hat P-P\over \sqrt{{pq\over n}}}\to N(0,1)$

and by Slutsky's Theorem

${\hat P-P\over \sqrt{{\hat p\hat q\over n}}}\to N(0,1)$ as well.

3. Originally Posted by matheagle
the MLE is $\hat P$ which is an sample mean of Bernoulli's

So ${\hat P-P\over \sqrt{{pq\over n}}}\to N(0,1)$

and by Slutsky's Theorem

${\hat P-P\over \sqrt{{\hat p\hat q\over n}}}\to N(0,1)$ as well.
That looks like the formula $\sqrt{n}(\hat{\theta}-\theta)\sim{N}(0,I_Y^{-1}(\theta))$ adapted to Bin(1,p) for n observations, or, $\sqrt{I_Y(\theta)}(\hat{\theta}-\theta)\sim{N}(0,1)$ for large n

(with MLE= $\hat{p}$ and Fisher information $I_Y=\frac{1}{p(1-p)}$ for one observation, so $I_Y=\frac{n}{p(1-p)}$ for n observations)

So, does the former ( $\sqrt{n}(\hat{\theta}-\theta)->N(0,I_Y^{-1}(\theta))$) comes from Slutsky theorem? I just never seen this name before, and I was also given the above assymptotic formula without explaining where it comes from.

4. The Central Limit Theorem gives you

${\hat P-P\over \sqrt{{pq\over n}}}\to N(0,1)$

replacing the p's and q's with their sample versions also converges in distribution to
a standard normal since

$\hat P\to P$ and $1-\hat P\to 1-P$ by either law of large numbers

all you need is convergence in probability http://en.wikipedia.org/wiki/Slutsky's_theorem

5. Ah! then, that assymptotic formula with Fisher information in it looks just like a version of the Central Limit Theorem. (edited: but of course it would, because it originates from CLT, I just didn't see the proof so I didn't see that it does. Sorry for off-top).