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Consider $n$ iid observations $X_1,X_2,\dots ,X_n$ from a $Uniform(a,b)$ distribution, where $a$ and $b$ are both unknown. How do we construct a joint confidence interval for $(a,b)$?

I would prefer a rectangular shape confidence interval, which can be obtained by using Bonferroni's method, so I guess the question can also be reformed as: How do we get a confidence interval for $a$?

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Let $Y_i:=\frac{X_i-a}{b-a}$, so that the $Y_i$'s are iid from $U(0,1)$, and for the corresponding order statistics one has $X_{(i)}=a+(b-a)Y_{(i)}$. Let $R_n:=X_{(n)}-X_{(1)}$, the ``sample range''. Then, for any real $c>0$ \begin{equation*} \alpha:=P(X_{(1)}>a+cR_n) =P(Y_{(1)}>(Y_{(n)}-Y_{(1)})c)=P(Y_{(n)}<Y_{(1)}\frac{1+c}c). \end{equation*} For $n\ge2$, the joint pdf of $(Z_1,Z_n):=(Y_{(1)},(Y_{(n)})$ equals $n(n-1)(z_n-z_1)^{n-2}$ if $0<z_1<z_n<1$ and $0$ otherwise. So, \begin{equation*} \alpha=\int_0^1 n\,dz_1\int_{z_1}^{1\wedge[z_1(1+c)/c]} dz_n\,(n-1)(z_n-z_1)^{n-2} =\int_0^1 n\,dz_1\, [(1-z_1)\wedge(z_1/c)]^{n-1} \end{equation*} \begin{equation*} =\int_{c/(c+1)}^1 n\,dz_1\, (1-z_1)^{n-1} +\int_0^{c/(c+1)} n\,dz_1\, (z_1/c)^{n-1} =\frac1{(c+1)^{n-1}}, \tag{1} \end{equation*} so that \begin{equation*} c=c_\alpha:=\alpha^{-1/(n-1)}-1 \end{equation*} and \begin{equation*} P(X_{(1)}-c_\alpha R_n<a<X_{(1)}) =P(X_{(1)}-c_\alpha R_n\le a)=1-\alpha. \end{equation*}

That is, $[X_{(1)}-c_\alpha R_n,X_{(1)}]$ is a $(1-\alpha)$-confidence interval for $a$.

Similarly or by symmetry, $[X_{(n)}, X_{(n)}+c_\alpha R_n]$ is a $(1-\alpha)$-confidence interval for $b$.


Consider now the ``joint probability'' \begin{equation*} p(c):=P\big(a\in[X_{(1)}-cR_n,X_{(1)}],b\in[X_{(n)}, X_{(n)}+cR_n]\big), \tag{2} \end{equation*} again for real $c>0$. By a calculation similar to, but a bit more tedious than, (1), one can obtain the following rather simple expression for $p(c)$:
\begin{equation*} p(c)=1 - 2 (1 + c)^{1 - n} + (1 + 2 c)^{1 - n}. \end{equation*} Indeed, letting $\ell(z_1):=1\wedge(\frac{1+c}c\, z_1\vee\frac{1 + c z_1}{1 + c})$, we have \begin{multline*} p(c)=\int_0^1 n\,dz_1\int_{\ell(z_1)}^1 dz_n\,(n-1)(z_n-z_1)^{n-2} \\ =\int_0^{c/(1+2c)} n\,dz_1\, (1-z_1)^{n-1}[1-(1+c)^{1-n}] \\ +\int_{c/(1+2c)}^{c/(1+c)} n\,dz_1\, [(1-z_1)^{n-1}-z_1^{n-1}c^{1-n}] \\ =1 - 2 (1 + c)^{1 - n} + (1 + 2 c)^{1 - n}. \end{multline*}

By (2), $p(c)$ obviously increases from $0$ to $1$ as $c$ increases from $0$ to $\infty$. So, given any natural $n\ge2$ and any real $\alpha\in(0,1)$, one can easily find (numerically) the unique solution, $\tilde c_\alpha$, of the equation \begin{equation*} p(\tilde c_\alpha)=1-\alpha. \end{equation*} Thus, with $c=\tilde c_\alpha$, \begin{equation*} [X_{(1)}-c R_n,X_{(1)}]\times[X_{(n)}, X_{(n)}+c R_n] \end{equation*} is an exact $(1-\alpha)$-confidence rectangle for the pair $(a,b)$.

It appears that $\tilde c_\alpha$ differs rather little from the "Bonferroni" value $c_{\alpha/2}$. E.g., for $n=10$ and $\alpha=0.05$, we have $\tilde c_\alpha\approx0.500243$ vs. $c_{\alpha/2}\approx0.50663$. For $n=100$ and $\alpha=0.05$, we have $\tilde c_\alpha\approx0.0378116$ vs. $c_{\alpha/2}\approx0.0379643$.

Clearly, we always have $\tilde c_\alpha<c_{\alpha/2}$.

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  • $\begingroup$ Is the calculus correct in the last step of the big equation? $\endgroup$ – Matt F. Aug 15 '17 at 0:22
  • $\begingroup$ @MattF. : There was indeed a typo in the second line of the two-line display. It is now fixed. The ultimate expression for $\alpha$ is, anyway, correct. $\endgroup$ – Iosif Pinelis Aug 15 '17 at 2:34
  • $\begingroup$ I have provided details on the calculation of $p(c)$. $\endgroup$ – Iosif Pinelis Aug 15 '17 at 16:09
  • $\begingroup$ This is COOL! Thanks a lot! I was thinking about using conditional probability and a pivotal quantity afterwards to arrive at a CI. I think the result matches what you had in the first part. I never thought about the second part that we can get exact confidence region for $(a,b)$ jointly. Good job! $\endgroup$ – Oliver Aug 15 '17 at 23:42

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