# How can I use this pivotal quantity to find the shortest length confidence interval for $\theta$?

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.

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

$$E[\ell_2X_{(1)}-\ell_1X_{(1)}]=(\ell_2-\ell_1)E[X_{(1)}]$$

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

$$n\ell_2^{n-1}-n\ell_1^{n-1}\frac{d\ell_1}{d\ell_2}=0$$

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

Therefore differentiating $$f$$ we get

$$\frac{df}{d\ell_2}=1-\frac{d\ell_1}{d\ell_2}=1-\left(\frac{\ell_2}{\ell_1}\right)^{n-1}<0$$

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

• Thank you for your answer. How do you know that $0 < l_1 < l_2 < 1$? Additionally, how did you reduce to $l^n_2 - l_1^n=1 - \alpha$? – Ron Snow Apr 2 at 16:43
• The support of $\theta/X_{(1)}$ is $(0,1)$ (it is a Beta distribution). Did you read the answer carefully? – StubbornAtom Apr 2 at 19:25
• 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! – Ron Snow Apr 2 at 19:46
• I just solved it to the point you had; I am currently working through solving the optimization problem. – Ron Snow Apr 2 at 21:22
• One of them should depend on $\alpha$. – 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!

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