Timeline for Calculating a confidence interval for the min/max of a distribution when sample may not reflect the underlying distribution
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 22, 2017 at 1:24 | history | edited | G_B | CC BY-SA 3.0 |
added 1861 characters in body
|
| Mar 22, 2017 at 0:28 | comment | added | G_B | Specifying a continuous distribution isn't enough to address this issue. I will update my response to explain why not. | |
| Mar 21, 2017 at 5:44 | comment | added | HXSP1947 | Rereading your response, I believe that I wasn't clear enough that I know that my distribution is continuous. I've updated my question so that it is more clear. That said, I should also add that I know that my continuous distribution is bounded between 0 and 1 (and in fact more tightly than this but I don't know exactly how tight). I have added this as well | |
| Mar 21, 2017 at 5:41 | comment | added | HXSP1947 | Two comments for your first example. First, is this distribution even possible? Did you mean to say x1 = 0 with probability 1-d and x2=0 for d where d is extremely small? Since this is a discrete distribution sampling will either result in you picking the exact minimum or always picking the max. With a continuous distribution I have the same problem but if I generate enough numbers I'm wondering if I will be able to say the minimum is within a certain range of at least one of my samples. | |
| Mar 21, 2017 at 0:26 | history | answered | G_B | CC BY-SA 3.0 |