0
$\begingroup$

If $X_1,\dots,X_n$ are iid $\mathrm{Binomial}(3,p)$, then the maximum likelihood estimator of $p$ is

$$\hat p = \frac{1}{n}\sum_i X_i$$

Find the bias, variance and MSE of $\hat p$?

We are asked to provide theoretical answers. Would bias equal $E(\hat p) - p$ ? What about variance and MSE?

I am just looking for some pointers to work this out. Any help would be greatly appreciated.

$\endgroup$
3
  • $\begingroup$ Your thoughts on bias are correct. How would you normally calculate variance? $\endgroup$ Apr 15, 2015 at 14:44
  • 1
    $\begingroup$ for a binomial is it np(1-p) = 3p(1-p)? $\endgroup$ Apr 15, 2015 at 15:27
  • $\begingroup$ Indeed, E(p-hat)=3p, not p. $\endgroup$
    – Did
    Apr 15, 2015 at 15:43

2 Answers 2

1
$\begingroup$

Bias is zero as $E(\hat p = \frac{1}{n}\sum_i X_i)=\frac{1}{n}E(\sum_i X_i)=\frac{1}{n}\sum_i E(X_i)= \frac{np}{n}=p$

Hence, $MSE=Var(\hat p)+bias^2=Var(\hat p)+0$

$Var(\hat p)=Var(\frac{1}{n}\sum_i X_i)=\frac{1}{n^2}Var(\sum_i X_i)=\frac{1}{n^2}\sum_i Var(X_i)=\frac{np(1-p)}{n^2}=\frac{p(1-p)}{n}$

therefore $MSE(\hat p)=Var(\hat p)=\frac{p(1-p)}{n}$

$\endgroup$
0
$\begingroup$

$$MSE=variance+Bias^2$$

Variance is the variance of random variable which is your estimator.

Your calculation of Bias is correct.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .