# Thread: How to prove consistency of the biased MLE estimator for sigma^2

1. ## How to prove consistency of the biased MLE estimator for sigma^2

Hi guys,
I have a small math problem
I have a sample of size N iid normal distributed variables. Then I get the MLE for mu and sigma^2(but the biased estimater for sigma^2, the one that is 1/N*(theRest) ). Now I have to prove consistecy of them both. Proving that for the unbiased estimator of mu is easy using Chebyshev.
My problem is proving the consistency of the biased MLE estimator of sigma^2.
Can anybody help me?
Cheers

2. Originally Posted by pimponi
Hi guys,
I have a small math problem
I have a sample of size N iid normal distributed variables. Then I get the MLE for mu and sigma^2(but the biased estimater for sigma^2, the one that is 1/N*(theRest) ). Now I have to prove consistecy of them both. Proving that for the unbiased estimator of mu is easy using Chebyshev.
My problem is proving the consistency of the biased MLE estimator of sigma^2.
Can anybody help me?
Cheers
What exactly do you mean by consistency?

3. Originally Posted by Anonymous1
What exactly do you mean by consistency?
That the ML estimated parametar converges with probability to the real value of the parameter

4. chebyshev is not the way to go.
you would need a fourth moment in that case
You can obtain STRONG consistency by the Strong Law of Large Numbers.

${\sum (X_i-\bar X)^2\over n} ={\sum X_i^2\over n}-(\bar X)^2\to E(X^2)-(\mu)^2=\sigma^2$ almost surely

And almost sure implies convergence in prob, but we do have strong consistency here

AND I used $\bar X\to \mu$ a.s. which doesn't need cheby either.
But I can prove these with cheby, but you will need a fourth moment
And we don't need normality either here.

5. Originally Posted by matheagle
chebyshev is not the way to go.
you would need a fourth moment in that case
You can obtain STRONG consistency by the Strong Law of Large Numbers.

${\sum (X_i-\bar X)^2\over n} ={\sum X_i^2\over n}-(\bar X)^2\to E(X^2)-(\mu)^2=\sigma^2$ almost surely

And almost sure implies convergence in prob, but we do have strong consistency here

AND I used $\bar X\to \mu$ a.s. which doesn't need cheby either.
But I can prove these with cheby, but you will need a fourth moment
And we don't need normality either here.
Thanks a lot.