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Consider i.i.d random variables $X_1$, $X_2$, . . . , $X_n$ having pdf

$$f_X(x\mid\theta) = \begin{cases} \theta x^{−2} & x\geq\theta \\ 0 & x\lt\theta \end{cases}$$

where $\theta \gt 0$ and $n\geq 3$

(a) Give the likelihood function, expressing it clearly as a function of $\theta$

(b) Give the MSE of the MLE of $\theta$

My Attempt:

(a) $$L(\theta\mid\vec{x}) = \begin{cases} \theta^n\left(\prod_{i=1}^n x_i\right)^{−2} & x_{(1)}\geq\theta \\ 0 & x_{(1)}\lt\theta \end{cases}$$

(b) Clearly the MLE of $\theta$ is $X_{(1)}$. We have

$$\begin{align*} F_{X_{(1)}}(x) &=\mathsf P(\text{min}{\{X_1,...,X_n}\}\leq x)\\\\ &=1-\mathsf P(\text{min}{\{X_1,...,X_n}\}\gt x)\\\\ &=1-\left(1-F_X(x)\right)^n\\\\ &=1-\left(\frac{\theta}{x}\right)^n \end{align*}$$

so

$$f_{X_{(1)}}(x)=\left(1-\left(\frac{\theta}{x}\right)^n\right)'=\frac{n\theta^n}{x^{n+1}}I_{[\theta,\infty)}(x)$$

It follows that

$$\begin{align*} \mathsf E\left(X_{(1)}^2\right) &=\int_{\theta}^{\infty}\frac{n\theta^n}{x^{n-1}}dx\\\\ &=n\theta^n\left(\frac{x^{-n+2}}{-n+2}\Biggr{|}_{\theta}^{\infty}\right)\\\\ &=\frac{n\theta^2}{n-2} \end{align*}$$

and

$$\begin{align*} \mathsf E\left(X_{(1)}\right) &=\int_{\theta}^{\infty}\frac{n\theta^n}{x^{n}}dx\\\\ &=n\theta^n\left(\frac{x^{-n+1}}{-n+1}\Biggr{|}_{\theta}^{\infty}\right)\\\\ &=\frac{n\theta}{n-1} \end{align*}$$

so

$$\begin{align*} \mathsf{Var}\left(X_{(1)}\right) &=\frac{n\theta^2}{n-2}-\left(\frac{n\theta}{n-1}\right)^2\\\\ &=\theta^2\left(\frac{n}{n-2}-\frac{n^2}{(n-1)^2}\right) \end{align*}$$

We also have

$$\begin{align*} \text{bias}^2\left(\hat{\theta}\right) &=\left(\mathsf E\left(\hat{\theta}\right)-\theta\right)^2\\\\ &=\left(\frac{n\theta}{n-1}-\theta\right)^2\\\\ &=\left(\theta\left(\frac{n}{n-1}-1\right)\right)^2 \end{align*}$$

Finally the MSE is given by

$$\begin{align*} \mathsf{Var}\left(X_{(1)}\right)+\text{bias}^2\left(\hat{\theta}\right) &=\theta^2\left(\frac{n}{n-2}-\frac{n^2}{(n-1)^2}\right)+\left(\theta\left(\frac{n}{n-1}-1\right)\right)^2\\\\ &=\theta^2\left(\frac{n}{n-2}-\frac{n^2}{(n-1)^2}+\left(\frac{n}{n-1}-1\right)^2\right)\\\\ &=\frac{2\theta^2}{(n-1)(n-2)} \end{align*}$$

Are these valid solutions?

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    $\begingroup$ You have followed the correct steps. You can check your computations by noting that the population pdf is a Pareto density. And so is the pdf of $X_{(1)}$. $\endgroup$ – StubbornAtom Nov 8 '18 at 20:01

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