Timeline for How can one construct a confidence interval for the mean of a uniformly distributed random variable using a sample of size 2
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Mar 16 at 19:51 | answer | added | Will Sawin | timeline score: 5 | |
| Mar 14 at 0:14 | comment | added | Will Sawin | @MichaelHardy No, I do not, deny this, and this was clear already from the first sentence of my first comment which gives a formula for the confidence interval depending on $X$ and $Y$. (I guess it's not obvious from this formula that $a$ is a constant and not a function of $M,L$ but indeed $a$ is a constant and was made explicit in my later comments.) | |
| Mar 13 at 21:41 | comment | added | Michael Hardy | @WillSawin $\qquad \uparrow \qquad$ | |
| Mar 11 at 5:18 | comment | added | Michael Hardy | Later I'll look more closely, but can you tell me whether you are denying that the confidence interval can depend on $(X,Y,M,L)$ ONLY through $(X,Y)\text{?}$ | |
| Mar 11 at 1:16 | comment | added | Will Sawin | @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. | |
| Mar 11 at 1:06 | comment | added | Aaron Hill | Thank you for these helpful comments! | |
| Mar 10 at 21:55 | comment | added | Michael Hardy | @WillSawin : Remember that the confidence interval must depend on the quadruple $(X,Y,M,L)$ ONLY through the pair $(X,Y).$ It seems as if you missed that. | |
| Mar 10 at 21:51 | comment | added | Michael Hardy | @WillSawin : You're entirely wrong about that. The original poster was clear that $L$ was to be estimated based on the data. You say certain things are invariant under affine transformations, but the fact is you cannot know WHICH affine transformation is the right one for the occasion without knowing the value of $L$. | |
| Mar 10 at 3:18 | comment | added | Will Sawin | @IosifPinelis Indeed, and then we take $a=44.5$ I guess. Or said more elegantly $\left[ \frac{X+Y - 99 |X-Y|}{2}, \frac{X+Y+99|X-Y|}{2}\right]$. | |
| Mar 10 at 2:50 | comment | added | Iosif Pinelis | Previous comment continued: Then the confidence probability will be $\frac{2a}{1+2a}$ for $a>0$. | |
| Mar 10 at 2:43 | comment | added | Iosif Pinelis | @WillSawin : I think you meant $\frac{X+Y}2$ in place of $\frac{X-Y}2$. | |
| Mar 9 at 21:39 | comment | added | Will Sawin | @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 $(1/2,1)$. | |
| Mar 9 at 20:55 | comment | added | Michael Hardy | The term "true mean" is often used, but I think it is better to use the term "population mean." Both the population mean and the sample mean are truly the mean of something. | |
| Mar 9 at 20:53 | comment | added | Michael Hardy | @WillSawin : How can you do that if you don't know the value of $L\text{ ?} \qquad$ | |
| Mar 9 at 20:40 | comment | added | Will Sawin | 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 $a$. | |
| Mar 9 at 20:39 | comment | added | Michael Hardy | $\ldots\,$what to do when $L$ is not known is not clear to me yet. | |
| Mar 9 at 20:39 | comment | added | Michael Hardy | 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$\,\ldots\qquad$ | |
| Mar 9 at 20:18 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 19 characters in body
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| Mar 9 at 6:45 | history | asked | Aaron Hill | CC BY-SA 4.0 |