# Thread: Expected value of the MLE of a binomial distribution

1. ## Expected value of the MLE of a binomial distribution

Hi

I've this question:

X1,...,Xn it's a sample from a binomial distribution b(n,a). Find the maximum likelihood estimator of a and the expected value of the estimator.

I can find the MLE

Ignoring the binomial coefficient i have

$
f(x_1,...,x_n|a)=a^{\sum_{1}^{n}x_i}(1-a)^{n^2-\sum_{1}^{n}x_i}
$

I take logs and differentiate to obtain:

$
\frac{d}{da}logf(x_1,...,x_n|a)=\frac{\sum_{1}^{n} x_i}{a}-\frac{n^2-\sum_{1}^{n}x_i}{1-a}
$

Upon equating to zero and solving i obtain that:

$
a=\frac{\sum_{1}^{n}x_i}{n^2}
$

Now i must find the expected value of a but I'm very confused on how to do this. Any hint?

Thanks

2. Originally Posted by noname
Hi

I've this question:

X1,...,Xn it's a sample from a binomial distribution b(n,a). Find the maximum likelihood estimator of a and the expected value of the estimator.

I can find the MLE

Ignoring the binomial coefficient i have

$
f(x_1,...,x_n|a)=a^{\sum_{1}^{n}x_i}(1-a)^{n^2-\sum_{1}^{n}x_i}
$

I take logs and differentiate to obtain:

$
\frac{d}{da}logf(x_1,...,x_n|a)=\frac{\sum_{1}^{n} x_i}{a}-\frac{n^2-\sum_{1}^{n}x_i}{1-a}
$

Upon equating to zero and solving i obtain that:

$
a=\frac{\sum_{1}^{n}x_i}{n^2}
$

Now i must find the expected value of a but I'm very confused on how to do this. Any hint?

Thanks
$E \left( \sum_{i = 1}^{n} x_i \right) = \sum_{i = 1}^{n} E(x_i)$.

3. Thank you!

So if I'm correct then

$
E(a)=E(\frac{\sum_{i=1}^{n}x_i}{n^2})=\frac{1}{n^2 }\sum_{i=1}^{n}E(x_i)=\frac{1}{n^2}\sum_{i=1}^{n}n a=\frac{1}{n^2}n^2a=a
$

4. Originally Posted by noname
Thank you!

So if I'm correct then

$
E(a)=E(\frac{\sum_{i=1}^{n}x_i}{n^2})=\frac{1}{n^2 }\sum_{i=1}^{n}E(x_i)=\frac{1}{n^2}\sum_{i=1}^{n}n a=\frac{1}{n^2}n^2a=a
$
There's a difference between the two a's ....

Personally I'd use the notation $\hat{a}$ for the MLE of $a$. Then you have $E(\hat{a}) = a$. So the MLE of $a$ is an unbiased estimator of $a$.