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 $a$?


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 $\tilde c_\alpha<c_{\alpha/2}$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.