Timeline for How to get the maximum likelihood estimator of $U(\theta,\theta +1)$?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29, 2016 at 18:04 | vote | accept | Onix | ||
| Jan 29, 2016 at 18:03 | comment | added | Onix | I can not upvote right now my reputation is not enough | |
| Jan 29, 2016 at 18:03 | comment | added | Onix | Oh yeah sorry i thought i upvoted it already. | |
| Jan 29, 2016 at 18:02 | comment | added | JohnK | @Abomm You are welcome. If you are satisfied with my answer, you can upvote it and accept it as the correct one by clicking that tick. | |
| Jan 29, 2016 at 18:01 | comment | added | Onix | Ok i got it why the other two are not the mle as they are outside the required interval as you suggested in the answer "you have the freedom of choice then any estimator within this interval will be mle. Thanks for your time. | |
| Jan 29, 2016 at 16:55 | history | edited | JohnK | CC BY-SA 3.0 |
added 74 characters in body
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| Jan 29, 2016 at 16:37 | comment | added | JohnK | @Abomm You don't have to subtract anything. This is just the midpoint of your interval. Recall that we require $y_n - 1<\theta< y_1 $. Then try to see why the other two estimators won't work. | |
| Jan 29, 2016 at 11:02 | comment | added | Onix | And subtracting the factor of 1/2 gives the lower bound of the interval as the interval length is 1? | |
| Jan 29, 2016 at 10:58 | comment | added | JohnK | It is the third one because it is the midpoint of the allowed interval, while the first two lie outside. Recall that in an interval (a,b), the midpoint is given by $\frac{a+b}{2}$. | |
| Jan 29, 2016 at 10:57 | comment | added | Onix | Yeah I agree with that one, (tell me if i am wrong in reasoning) Then the 3rd estimator may be correct as $X_{(1)}\text{and}X_{(2)}$ are minimum and maximum resp. values of the sample they are the overestimate and underestimate of $\theta,\theta+1$ resp. there sum divide by 2 and subtracted by $\frac{1}{2}$ is MLE as here$\frac{\theta+\theta+1}{2}-\frac{1}{2}=\theta$ no? | |
| Jan 29, 2016 at 10:21 | history | answered | JohnK | CC BY-SA 3.0 |