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Let $X_1, \dots ,X_n$ be a random sample of size $n$ from the continuous distribution with pdf $f_X(x\mid\theta) = \dfrac{2\theta^2}{x^3} I(x)_{(\theta;\infty)}$ where $\theta \in \Theta = (0, \infty)$.

(1) Show that $X_{(1)}$ is sufficient for $\theta$.

(2) Show directly that the pdf for $X_{(1)}$ is $f_{X(1)}(x\mid\theta) = \dfrac{2n\theta^{2n}}{x^{2n+1}} I(x)(\theta,\infty)$.

(3) When $\Theta = (0, \infty)$, the probability distribution of $X_{(1)}$ is complete. In this case, find the best unbiased estimator for $\theta$.

(4) Suppose that $\Theta = (0; 1]$. Show that the probability distribution of $X_{(1)}$ is not complete in this setting by considering the function $g(X_{(1)}) = \Big [ X_{(1)} - \frac{2n}{2n-1} \Bigr] I(X_{(1)})_{(1,\infty)}$.

For (1), this was pretty easy to show using Factorization Theorem.

For (2), I think I am integrating my pdf wrong because I can't seem to arrive at the answer.

For (3), I am trying to use a Theorem that states "If T is a complete and sufficient statistic for $\theta$ and $\phi(T)$ is any estimator based only on T, then $\phi(T)$ is the unique best unbiased estimator of its expected values", but I can't seem to simplify the expected value to get $\theta$.

For (4), I am getting stuck trying to show $P(g(X_{(1)})=0) = 1$ using the given function.

Any assistance is greatly appreciated.

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(1) Awesome!

(2) Did you try $P(X_{(1)} \leq x) = 1 - P(X_{(1)} > x) = 1 - P(X_{1} > x)^{n}$. No integration is necessary :D

(3) Can you find an unbiased estimator? What is $E[X_{1}]$? Let $U$ be an unbiased estimator, can you find $E[U|X_{(1)}]$? How does that help?

(4) You have to show that $E[g(X_{(1)})] = 0$. Not what you wrote.

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For Part (1), great!

For Part (2), I'm unsure about that one.

For Part (3), Note: that the original distribution $f_{X}(x|\beta) = \frac{2*\theta^{2}}{x^{3}}*I_{(\theta,\infty)}(x)$ resembles a famous distribution, but this famous distribution has 2 parameters, $\alpha$ and $\beta$, where the value of $\beta = 2$, and your $\alpha = \theta$.

Side Note: May I ask what you got for Part (3) when you integrated? (If you did integrate?)

it's the Pareto distribution

and once you know the right distribution, then finding the expected value for the BUE is easier than having to integrate.

For Part(4), all you have to do is show that its Expected value of that function is not equal to 0, and, therefore, the function is not complete.

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