Consider a simple random sample $X_{1},X_{2},\ldots,X_{n}$ whose distribution is given by $X\sim U(0,\theta)$. Moreover, consider the estimators $\hat{\theta}_{1} = c_{1}\overline{X}$ and $\hat{\theta}_{2} = c_{2}X_{(n)} = c_{2}\max{X_{1},X_{2},\ldots,X_{n})}$.

(a) Find $c_{1}$ and $c_{2}$ such that $\hat{\theta}_{1}$ and $\hat{\theta}_{2}$ becomes unbiased.

(b) Find both Mean Squared Errors.


The estimator $\hat{\theta}_{1}$ is unbiased iff $\textbf{E}(\hat{\theta}_{1}) = \theta_{1}$, which is equivalent to \begin{align*} \textbf{E}(\hat{\theta}_{1}) = \textbf{E}(c_{1}\overline{X}) = c_{1}\textbf{E}(\overline{X}) = \frac{c_{1}\theta}{2} = \theta \Longleftrightarrow c_{1} = 2 \end{align*}

As to the second case, I am unable to work with $X_{(n)}$. Could someone help me?

(b) Since $\hat{\theta}_{1}$ is unbiased for $c_{1} = 2$, we conclude that $\text{MSE}(\hat{\theta}_{1}) = \text{Var}(\hat{\theta}_{1})$. Consequently, \begin{align*} \text{Var}(\hat{\theta}_{1}) = \text{Var}(2\overline{X}) = 4\text{Var}(\overline{X}) = 4\times\frac{\theta^{2}}{12n} = \frac{\theta^{2}}{3n} \end{align*}

The same problem applies to $\text{MSE}(\hat{\theta}_{2})$. Could someone help me with this as well?

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  • 2
    $\begingroup$ Please add the self-study tag. Can you find the CDF of $X_{(n)}$? $\endgroup$ – StubbornAtom Mar 26 '19 at 23:47


$F_{X_{(n)}}(x)=P(X_{(n)}\leq x)=P(\max \{X_1,\cdots, X_n\} \leq x)=P(X_1 \leq x,\cdots ,X_n\leq x)=(P(X_1\leq x))^n=(F_{X_1}(x))^n=(\frac{x}{\theta})^n$

so $f_{X_{(n)}}(x)=\frac{nx^{n-1}}{\theta^n} \hspace{1cm} 0\leq x \leq \theta$

now can you calculate $E(c X_{(n)})$ and $Var(c X_{(n)})$ ?

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  • $\begingroup$ Thank you for the answer! I think I can handle the problem now :) $\endgroup$ – user1337 Apr 1 '19 at 22:51
  • $\begingroup$ you are welcome $\endgroup$ – Masoud Apr 1 '19 at 22:53

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