# Thread: Minimum Variance Unbiased Estimator

1. ## Minimum Variance Unbiased Estimator

We have a r.v. $X$~ $Bin (12,\theta)$

And I have worked out that the Cramer Rao Lower Bound for the var of any unbiased est of $\theta$ to be

$\frac{1}{\frac{1}{12}(\frac{1}{\theta} + \frac{1}{1-\theta})}$

Now, I'm given that $T = \frac{\sum X_{i}}{12n}$

(in case there's any confusion, the sum is from i=1 to n, I'm not sure how to do that on LaTex)

How do I show that $T$ is a minimum variance unbiased estimator of $\theta$?

2. Originally Posted by Zenter
We have a r.v. $X$~ $Bin (12,\theta)$

And I have worked out that the Cramer Rao Lower Bound for the var of any unbiased est of $\theta$ to be

$\frac{1}{\frac{1}{12}(\frac{1}{\theta} + \frac{1}{1-\theta})}$

Now, I'm given that $T = \frac{\sum X_{i}}{12n}$

(in case there's any confusion, the sum is from i=1 to n, I'm not sure how to do that on LaTex)

How do I show that $T$ is a minimum variance unbiased estimator of $\theta$?
First calculate its mean and show its equal to $\theta$ (so its unbiased). Now calculate its variance and if it is equal to the CRLB you are done.

(there is something strange about your CRLB, it seems to be independedent of sample size amoung other things)

CB

3. Oops, I made an error when typing out the CRLB, it's actually

$\frac{1}{12(\frac{1}{\theta} + \frac{1}{1-\theta})}$

It's the CRLB for one observation of X, and we know n = 12, so I subbed it in for any n in the $i(\theta)$ equation. Was this wrong?

4. Originally Posted by Zenter
Oops, I made an error when typing out the CRLB, it's actually

$\frac{1}{12(\frac{1}{\theta} + \frac{1}{1-\theta})}$

It's the CRLB for one observation of X, and we know n = 12, so I subbed it in for any n in the $i(\theta)$ equation. Was this wrong?
Your estimator looks like it is using n sets of 12 (that is a total of 12n Bernoulli trials). It is the sum of n (presumably independent) RVs ~Bin(12,theta)

(Other than the above you will find that the variance of your estimator is equal to the CRLB, it would also help if you simplify what is inside your brackets)

CB

5. Oh, so then my value for the CRLB should actually be (after simplifying the brackets and correcting my error):

$\frac{\theta(1-\theta)}{12n}$

Right?

6. Originally Posted by Zenter
Oh, so then my value for the CRLB should actually be (after simplifying the brackets and correcting my error):

$\frac{\theta(1-\theta)}{12n}$

Right?
I think so, and that is also the variance of your estimator, as expected!

CB

7. Yea, I just worked it out, and it is Thanks!