0
$\begingroup$

So I am struggling with the following question. It asks to show that the MSE($\hat{p}_1$)=$\frac{4-16p+16p^2+np-np^2}{(n+4)^2}$ Where $\hat{p}_1$= $\frac{X+2}{n+4}$, $E(X)=np$ and $Var(X)=np(1-p)$ i.e. Binomial Distribution

I am aware of the formula for calculating the MSE, but I don't understand what $E($$\hat{p}_1$) and $Var($$\hat{p}_1$) is going to equal. Or am I totally looking at this the wrong way?

Improve this question
$\endgroup$
2
  • $\begingroup$ Need to specify what parameter or function of parameters you try to estimate by $\hat p_1$. In addition, you have $E(X)$, and could not find $E(\frac {X+2}{n+4})$? $\endgroup$
    – user158565
    Commented May 7, 2017 at 8:51
  • $\begingroup$ Sorry for being a bit vague, X follows a binomial distribution with parameters n and p. $\hat p_1$ is an estimator for p. $E(\frac {X+2}{n+4})$ and $Var(\frac {X+2}{n+4})$ is what I am having difficulty calculating. $\endgroup$
    – user158031
    Commented May 7, 2017 at 9:22

1 Answer 1

Reset to default
0
$\begingroup$

$E(\frac{X+2}{n+4}) = E(\frac{X}{m+4} + \frac{2}{n+4})$ $ = E(\frac{X}{n+4}) + E(\frac{2}{n+4}) = \frac{E(X)}{n+4} + \frac{2}{n+4}$ $ = \frac{np}{n+4} + \frac{2}{n+4}$

$V(\frac{X+2}{n+4}) = V(\frac{X}{m+4} + \frac{2}{n+4})$ $ = V(\frac{X}{n+4}) + V(\frac{2}{n+4}) = \frac{V(X)}{(n+4)^2} + 0$ $ = \frac{np(1-p)}{(n+4)^2}$

MSE = $\left(\frac{np}{n+4} + \frac{2}{n+4} - p\right)^2 + \frac{np(1-p)}{(n+4)^2}$

The rest is your job.

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.