# Mean Squared Error Question

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?

• 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 $E(\frac {X+2}{n+4})$? – user158565 May 7 '17 at 8:51
• 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. – user158031 May 7 '17 at 9:22

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