Timeline for Confidence Intervals: how to formally deal with $P(L(\textbf{X}) \leq \theta, U(\textbf{X})\geq\theta) = 1-\alpha$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Mar 27, 2018 at 19:52 | history | bounty ended | gioxc88 | ||
| S Mar 27, 2018 at 19:52 | history | notice removed | gioxc88 | ||
| Mar 23, 2018 at 6:23 | answer | added | Ben | timeline score: 1 | |
| Mar 21, 2018 at 15:30 | answer | added | Sextus Empiricus | timeline score: 6 | |
| Mar 20, 2018 at 23:37 | comment | added | Sextus Empiricus | What I do not get about your question is why you are dealing with a joint pdf of L(Y) and U(Y) or what is actually the problem. (in your second question do you mean, by using the term 'infinite' that 'there are infinitely many possible intervals' or 'the intervals are infinite size'?) | |
| Mar 20, 2018 at 23:33 | comment | added | Sextus Empiricus | For discrete distributions you can not always compute confidence intervals with exactly every $\alpha$. But you can ignore that comment. It does not really matter, I see now. | |
| Mar 20, 2018 at 22:46 | comment | added | gioxc88 | Honestly I don't follow you. What I gave is the standard definition of CI. If $\alpha = 5\%$ then you CI will cover the true parameter $95\%$ of the times | |
| Mar 20, 2018 at 12:37 | comment | added | Sextus Empiricus | If you define $L(X)$ and $U(X)$ such that for every possible hypothetical $\theta$ you have $$P(Y) \text{ with } \lbrace L(Y)<\theta < U(Y) \rbrace \geq 1-\alpha$$. Then no matter what your true $\theta$ is you will only estimate wrong $L(X)$ and $U(X)$ at most a fraction $\alpha$ of the time. | |
| S Mar 20, 2018 at 2:07 | history | bounty started | gioxc88 | ||
| S Mar 20, 2018 at 2:07 | history | notice added | gioxc88 | Draw attention | |
| Jan 29, 2018 at 13:01 | history | asked | gioxc88 | CC BY-SA 3.0 |