Let $X_1,\cdots,X_n \sim f(x|\theta)=\frac{\theta}{x^2}, x> \theta$ be a random sample where $\theta>0$ is unknown. I want to use $\frac{\theta}{X_{(1)}}$ as a pivotal quantity. How can I use this pivotal quantity to find the shortest $100(1-\alpha)$% confidence interval for $\theta$?

my work:

Let $Y=X_{(1)}$. Then, $f_Y(y)=nf_x(y)[1-F_x(y)]^{n-1}=n\frac{\theta}{y^2}(\frac{\theta}{y})^{n-1}=\frac{n\theta^n}{y^{n+1}}, y > \theta$.

However, how can I find the shortest confidence interval? I know that I need to find the distribution of $\frac{\theta}{Y}$, but I do not know how.

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Find the distribution function of $\theta/X_{(1)}$. For $t\in(0,1)$, one should end up with

\begin{align} P\left[\frac{\theta}{X_{(1)}}\le t\right]&=P\left[X_{(1)}\ge \frac{\theta}{t}\right] \\&=\left\{P\left[X_1\ge \frac{\theta}{t}\right]\right\}^n \\&=t^n \end{align}

Now there exists $(\ell_1,\ell_2)$ with $0\le \ell_1<\ell_2\le 1$ such that $$P_{\theta}\left[\ell_1<\frac{\theta}{X_{(1)}}<\ell_2\right]=P_{\theta}\left[\ell_1X_{(1)}<\theta<\ell_2X_{(1)}\right]=1-\alpha\quad\,\forall\,\theta>0 \tag{1}$$

This gives a confidence interval $(\ell_1X_{(1)},\ell_2X_{(1)})$ for $\theta$. Expected length of this interval is


Since $E[X_{(1)}]$ is a constant, need only to minimize $\ell_2-\ell_1=f$ (say) subject to $(1)$, that is $$\ell_2^n-\ell_1^n=1-\alpha \tag{2}$$

This constrained optimization problem can be solved by usual calculus methods.

Differentiating both sides of $(2)$ with respect to $\ell_2$, we have


Or, $$\frac{d\ell_1}{d\ell_2}=\left(\frac{\ell_2}{\ell_1}\right)^{n-1}$$

Therefore differentiating $f$ we get


Hence $f$ is decreasing in $\ell_2$, so its minimum occurs when $\ell_2$ is maximum.

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  • $\begingroup$ Thank you for your answer. How do you know that $0 < l_1 < l_2 < 1$? Additionally, how did you reduce to $l^n_2 - l_1^n=1 - \alpha$? $\endgroup$ – Ron Snow Apr 2 at 16:43
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    $\begingroup$ The support of $\theta/X_{(1)}$ is $(0,1)$ (it is a Beta distribution). Did you read the answer carefully? $\endgroup$ – StubbornAtom Apr 2 at 19:25
  • $\begingroup$ Ah, I see. I was wondering if it was from $t \in (0,1)$, but that didn't make much sense to me. Thank you! $\endgroup$ – Ron Snow Apr 2 at 19:46
  • $\begingroup$ I just solved it to the point you had; I am currently working through solving the optimization problem. $\endgroup$ – Ron Snow Apr 2 at 21:22
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    $\begingroup$ One of them should depend on $\alpha$. $\endgroup$ – StubbornAtom Apr 3 at 15:37

To find the distribution of $Z=\frac{\theta}{Y}$ you can do the following:

$F_Z(z) = P(Z\leq z) = P(\frac{\theta}{Y}\leq z) = P(\frac{\theta}{z} \leq Y)$ because $Y=X_{(1)}>0$, so

$F_Z(z) = 1-F_Y(\frac{\theta}{z})$.

Hope this helps!

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  • $\begingroup$ This is very helpful- thank you! $\endgroup$ – Ron Snow Apr 2 at 21:22

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