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Two numbers will be randomly (and independently) selected from a uniform distribution on an interval with length $L$ and center $M.$

It is very easy to estimate $M$ (just take the average of the two values), but I'm not sure how to construct a statistically reliable confidence interval for $M$ based on the two values.

One the other hand, while it is not as obvious how to estimate $L$ from the two values, I found a couple of different ways to construct statistically reliable confidence intervals for $L,$ using the sample statistic $X-Y.$ Indeed, here are two $99\%$ confidence intervals for $L.$

  • $\left[\frac{10}{9} |X-Y|, \infty\right)$
  • $\left[|X-Y|, \frac{|X-Y|}{1 - \sqrt{.99}}\right]$
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  • $\begingroup$ In one sense $(\min\{X,Y\}, \max\{X,Y\})$ is a $50\%$ confidence interval for $M.$ If we knew the value of $L,$ then it would be absurd to use that interval in the way in which confidence intervals are normally used, since if the length of this interval is a tiny fraction of $L$ then it would be very unlikely to contain $M,$ whereas if it were nearly as big as $L,$ then it would be very unlikely to fail to contain $M.$ But Fisher's technique of conditioning on an ancillary statistic can rescue us from this situation and give us a reasonable $50\%$ confidence interval. But$\,\ldots\qquad$ $\endgroup$
    – Michael Hardy
    Commented Mar 9 at 20:39
  • $\begingroup$ $\ldots\,$what to do when $L$ is not known is not clear to me yet. $\endgroup$
    – Michael Hardy
    Commented Mar 9 at 20:39
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    $\begingroup$ If we choose the interval of the form $ [ \frac{X-Y}{2} - a |X-Y|, \frac{X+Y}{2} + a |X-Y|]$ then the true mean will be in the confidence interval if $(X,Y)$ lies in a certain union of two quadrilaterals, meeting at the point $(M,M)$. Computing the probability the true mean is in the confidence interval is equivalent to computing the area of these quadrilaterals, which is some kind of elementary plane geometry, and then you can solve for a probability of $.99$ in terms of $a$ to find the optimal value of $a$. $\endgroup$
    – Will Sawin
    Commented Mar 9 at 20:40
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    $\begingroup$ @MichaelHardy Everything is invariant under affine transformations, which can turn any pair $M,L$ into any other, so it suffices to calculate for one pair, maybe $(0,1)$ or $(1/2,1)$. $\endgroup$
    – Will Sawin
    Commented Mar 9 at 21:39
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    $\begingroup$ @MichaelHardy No, I am not incorrect. The goal stated in the question is to estimate $M$. The formula given in my comments clearly does depend only on $X$ and $Y$, not on $L$. The definition of the $99\%$ confidence interval is that for each value of the parameters $M$ and $L$ the probability that the confidence interval contains the population mean $M$ is at least $99\%$. To check this condition, you may use an affine transformation in $M$ and $L$, and when you do this, you clearly know what $M$ and $L$ are. $\endgroup$
    – Will Sawin
    Commented Mar 11 at 1:16

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`Let's answer a slightly more general version, where the distribution is still uniform but the sample has size $n$. Let $\max$ be the maximum sample and $\min$ be the minimum sample. Then I claim a valid $99\%$ confidence interval is

$$\left[\max - \frac{ \max-\min}{2 (.01)^{\frac{1}{n-1}}}, \min + \frac{ \max-\min}{2 (.01)^{\frac{1}{n-1}} } \right].$$

In particular in the $n=2$ case we get

$$\left [ \max - 50 (\max-\min), \min + 50 (\max-\min)\right]$$ which agrees with what I wrote in the comments (thanks to Isoif Pinelis for help in the comments).

To check this interval is valid, i.e. that the population mean $M$ is contained in it with probability (at least $99\%$), consider the sample which is farthest from the population mean $M$. This sample is either $\max$ or $\min$. It is $\max$ with probability $\frac{1}{2}$ and $\min$ with probability $\frac{1}{2}$. It suffices to show that, conditional on either of these events, the probability that $M$ lies in the confidence interval is $99\%$. Everything is symmetric so let us focus on the first case.

Conditional on $\max$ being the furthest from $M$, and conditioning on a particular value of $\max$, every other sample lies in $[ 2M - \max, \max]$. I claim we can treat the other samples as $n-1$ uniform samples from the interval $[2M-\max, \max]$. This is just because we are conditioning on the $i$'th (say) sample taking a particular value $\max$ and all other samples lying in the interval $[ 2M - \max, \max]$, and conditioning a uniform distribution on lying in a smaller interval always gives a uniform distribution on that interval.

Thus the probability that $\min> u$ is the probability that $n-1$ samples from a uniform distribution on $[2M-\max, \max]$ are all greater than $u$, which is $$\left (\frac{ \max-u}{2 (\max-M)}\right)^{n-1}.$$

Hence the probability that $$\min > \max-\hspace{5pt} 2(\max-M) (.01)^{ \frac{1}{n-1}}$$ is $.01$.

So with $99\%$ probability we have

$$\min \leq \max-\hspace{5pt} 2(\max-M) (.01)^{ \frac{1}{n-1}}$$ $$ 2 (\max-M) (.01)^{ \frac{1}{n-1}} \leq \max-\min$$ $$ \max - M \leq \frac{ \max - \min }{ 2(.01)^{\frac{1}{n-1}}}$$ $$ M \geq \max - \frac{ \max - \min }{2 (.01)^{\frac{1}{n-1}}}$$ so $M$ lies above the lower bound of the confidence interval.

And, on the other hand, since $\min \geq 2M-\max $ we have $$ M \leq \frac{\max + \min}{2} = \min - \frac{\max -\min}{2} \leq \frac{ \max - \min}{2 (.01)^{\frac{1}{n-1}}}$$ so $M$ lies below the lower bound as well, hence within the interval, as desired.

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