Estimation, bias, and mean square error

Question

Let $X$ be a continuous random variable with pdf $f(x) =\frac{1}{2}(1+ \theta x)$, for $-1 < x < 1$, and $-1 < \theta < 1$

(a) Show that $E(X) = \frac{\theta}{3}$.

(b) Given a random sample of size $n$ from a population with pdf $f(X)$, consider the estimator $\hat{\theta} = 3\bar{X}$. Find the variance, bias, and mean squared error of $\hat{\theta}$.

For (a) I simply integrate $x f(x)$

For (b) I'm not sure it is related to part (a) or the pdf at all. I'm not quiet sure how to proceed.

I know that $\operatorname{Var}(\hat{\theta})=E((\hat{\theta}-E[\theta])^2)$

$\operatorname{Bias}(\hat{\theta})=E(\hat{\theta}-\theta))$, and

$\operatorname{MSE}=\sqrt{\operatorname{Var}(\hat{\theta})}$

Do I just need to manipulate the definitions?

Answer 1

$\newcommand{\var}{\operatorname{var}}\newcommand{\E}{\operatorname{E}}$ \begin{align} \E(\bar X) & = \E\left( \frac {X_1+\cdots+X_n} n \right) \\[8pt] & = \frac 1 n \E(X_1+\cdots+X_n) = \frac 1 n (\E(X_1)+\cdots+\E(X_n)) \\[8pt] & = \frac 1 n \Big( n\E(X_1) \Big) = \E(X_1). \\[25pt] \var( \bar X ) & = \var \left( \frac {X_1+\cdots+X_n} n \right) \\[8pt] & = \frac 1 {n^2} \var(X_1+\cdots+X_n) = \frac 1 {n^2} (\var(X_1)+\cdots+\var(X_n)) \\[8pt] & = \frac 1 {n^2} \Big( n \var(X_1) \Big) = \frac 1 n \var(X_1). \end{align}

Based on the above, you should be able to find the bias and the mean squared error.

Answer 2

$$\operatorname{Var}\left(\hat\theta\right)=\frac{3(1-θ)}{n}$$

$$\operatorname{Bias}\left(\hat\theta\right)=E(\hat\theta\right-theta)=0

$$\operatorname{Bias}^2\left(\hat\theta\right)=\frac{27(1-\theta)+4n\theta^2}{9n}$$