Existence of MLE












1












$begingroup$


I have a problem with MLE's definition:



Casella Berger in Statistical Inference and Nitis Mukhopadhyay in Probability and Statistics said that MLE for a parameter $thetainTheta$ is respectively $argsup_{thetainTheta}{L(thetamid x)}$ or $argmax_{thetainTheta}{L(thetamid x)}$.



But this estimator is a supremum or a maximum? If it is a supremum why we don't call it supremum likelihood estimator? Conversely if it is a maximum, the likelihood function must be continuous and the parametric space must be compact (sufficient condition for the existence of maximum)



What is the truth or the minimal condition for the existence of MLE?






probability statistics statistical-inference estimation maximum-likelihood







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Do you understand the difference between a "supremum" and a "maximum"? Any set of real numbers, having an upper bound, as a supremum but not necessarily a maximum. A set of real numbers has a maximum if and only if the supremum is in the set in which case the maximum is the supremum.
    $endgroup$
    – user247327
    Jan 2 at 17:47










  • $begingroup$
    Of course I know the difference. My problem is that in an estimation, the estimate parameter must belong to the parametric space but if the parametric space is for example an open set, the supremum may stay in its boundary and this thing may has no sense. So I want to understand which is the correct definition of MLE; there is argsup or argmax? because the second one may not belong to the parametric space. Thanks
    $endgroup$
    – Elia
    Jan 2 at 18:54






  • 1




    $begingroup$
    Then you realize that if the maximum [b]exists[/b], there is no difference between the two. If the maximum does not exist, then you [b]have[/b] to use the supremum.
    $endgroup$
    – user247327
    Jan 3 at 18:12
















1












$begingroup$


I have a problem with MLE's definition:



Casella Berger in Statistical Inference and Nitis Mukhopadhyay in Probability and Statistics said that MLE for a parameter $thetainTheta$ is respectively $argsup_{thetainTheta}{L(thetamid x)}$ or $argmax_{thetainTheta}{L(thetamid x)}$.



But this estimator is a supremum or a maximum? If it is a supremum why we don't call it supremum likelihood estimator? Conversely if it is a maximum, the likelihood function must be continuous and the parametric space must be compact (sufficient condition for the existence of maximum)



What is the truth or the minimal condition for the existence of MLE?






probability statistics statistical-inference estimation maximum-likelihood







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Do you understand the difference between a "supremum" and a "maximum"? Any set of real numbers, having an upper bound, as a supremum but not necessarily a maximum. A set of real numbers has a maximum if and only if the supremum is in the set in which case the maximum is the supremum.
    $endgroup$
    – user247327
    Jan 2 at 17:47










  • $begingroup$
    Of course I know the difference. My problem is that in an estimation, the estimate parameter must belong to the parametric space but if the parametric space is for example an open set, the supremum may stay in its boundary and this thing may has no sense. So I want to understand which is the correct definition of MLE; there is argsup or argmax? because the second one may not belong to the parametric space. Thanks
    $endgroup$
    – Elia
    Jan 2 at 18:54






  • 1




    $begingroup$
    Then you realize that if the maximum [b]exists[/b], there is no difference between the two. If the maximum does not exist, then you [b]have[/b] to use the supremum.
    $endgroup$
    – user247327
    Jan 3 at 18:12














1












1








1


1



$begingroup$


I have a problem with MLE's definition:



Casella Berger in Statistical Inference and Nitis Mukhopadhyay in Probability and Statistics said that MLE for a parameter $thetainTheta$ is respectively $argsup_{thetainTheta}{L(thetamid x)}$ or $argmax_{thetainTheta}{L(thetamid x)}$.



But this estimator is a supremum or a maximum? If it is a supremum why we don't call it supremum likelihood estimator? Conversely if it is a maximum, the likelihood function must be continuous and the parametric space must be compact (sufficient condition for the existence of maximum)



What is the truth or the minimal condition for the existence of MLE?






probability statistics statistical-inference estimation maximum-likelihood







share|cite|improve this question











$endgroup$




I have a problem with MLE's definition:



Casella Berger in Statistical Inference and Nitis Mukhopadhyay in Probability and Statistics said that MLE for a parameter $thetainTheta$ is respectively $argsup_{thetainTheta}{L(thetamid x)}$ or $argmax_{thetainTheta}{L(thetamid x)}$.



But this estimator is a supremum or a maximum? If it is a supremum why we don't call it supremum likelihood estimator? Conversely if it is a maximum, the likelihood function must be continuous and the parametric space must be compact (sufficient condition for the existence of maximum)



What is the truth or the minimal condition for the existence of MLE?






probability statistics statistical-inference estimation maximum-likelihood




probability statistics statistical-inference estimation maximum-likelihood






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 17:42









StubbornAtom

5,51211138




5,51211138










asked Jan 2 at 16:20









EliaElia

295




295








  • 2




    $begingroup$
    Do you understand the difference between a "supremum" and a "maximum"? Any set of real numbers, having an upper bound, as a supremum but not necessarily a maximum. A set of real numbers has a maximum if and only if the supremum is in the set in which case the maximum is the supremum.
    $endgroup$
    – user247327
    Jan 2 at 17:47










  • $begingroup$
    Of course I know the difference. My problem is that in an estimation, the estimate parameter must belong to the parametric space but if the parametric space is for example an open set, the supremum may stay in its boundary and this thing may has no sense. So I want to understand which is the correct definition of MLE; there is argsup or argmax? because the second one may not belong to the parametric space. Thanks
    $endgroup$
    – Elia
    Jan 2 at 18:54






  • 1




    $begingroup$
    Then you realize that if the maximum [b]exists[/b], there is no difference between the two. If the maximum does not exist, then you [b]have[/b] to use the supremum.
    $endgroup$
    – user247327
    Jan 3 at 18:12














  • 2




    $begingroup$
    Do you understand the difference between a "supremum" and a "maximum"? Any set of real numbers, having an upper bound, as a supremum but not necessarily a maximum. A set of real numbers has a maximum if and only if the supremum is in the set in which case the maximum is the supremum.
    $endgroup$
    – user247327
    Jan 2 at 17:47










  • $begingroup$
    Of course I know the difference. My problem is that in an estimation, the estimate parameter must belong to the parametric space but if the parametric space is for example an open set, the supremum may stay in its boundary and this thing may has no sense. So I want to understand which is the correct definition of MLE; there is argsup or argmax? because the second one may not belong to the parametric space. Thanks
    $endgroup$
    – Elia
    Jan 2 at 18:54






  • 1




    $begingroup$
    Then you realize that if the maximum [b]exists[/b], there is no difference between the two. If the maximum does not exist, then you [b]have[/b] to use the supremum.
    $endgroup$
    – user247327
    Jan 3 at 18:12








2




2




$begingroup$
Do you understand the difference between a "supremum" and a "maximum"? Any set of real numbers, having an upper bound, as a supremum but not necessarily a maximum. A set of real numbers has a maximum if and only if the supremum is in the set in which case the maximum is the supremum.
$endgroup$
– user247327
Jan 2 at 17:47




$begingroup$
Do you understand the difference between a "supremum" and a "maximum"? Any set of real numbers, having an upper bound, as a supremum but not necessarily a maximum. A set of real numbers has a maximum if and only if the supremum is in the set in which case the maximum is the supremum.
$endgroup$
– user247327
Jan 2 at 17:47












$begingroup$
Of course I know the difference. My problem is that in an estimation, the estimate parameter must belong to the parametric space but if the parametric space is for example an open set, the supremum may stay in its boundary and this thing may has no sense. So I want to understand which is the correct definition of MLE; there is argsup or argmax? because the second one may not belong to the parametric space. Thanks
$endgroup$
– Elia
Jan 2 at 18:54




$begingroup$
Of course I know the difference. My problem is that in an estimation, the estimate parameter must belong to the parametric space but if the parametric space is for example an open set, the supremum may stay in its boundary and this thing may has no sense. So I want to understand which is the correct definition of MLE; there is argsup or argmax? because the second one may not belong to the parametric space. Thanks
$endgroup$
– Elia
Jan 2 at 18:54




1




1




$begingroup$
Then you realize that if the maximum [b]exists[/b], there is no difference between the two. If the maximum does not exist, then you [b]have[/b] to use the supremum.
$endgroup$
– user247327
Jan 3 at 18:12




$begingroup$
Then you realize that if the maximum [b]exists[/b], there is no difference between the two. If the maximum does not exist, then you [b]have[/b] to use the supremum.
$endgroup$
– user247327
Jan 3 at 18:12










1 Answer
1






active

oldest

votes


















0












$begingroup$

Sufficient conditions for the MLE to exist are $L$ continuous in $theta$ over a compact support (extreme value theorem).

We can built an example where the maximum in $theta$ does not exist, but the supremum does. The main open difficulty is whether the "supremum likelihood" is meaningful...

If $U$ is uniformly distributed on $theta=(a,b)$, where $a$ and $b$ are parameters, then the likelihood (for one observation) is $$ L(theta)=frac{1}{b-a} hspace{4mm} text{ if } a<u_i<b $$ and $0$ otherwise. Then the MLE does not exist, but the SLE does (but is not unique…). See for instance the lecture notes of Songfeng Zheng (p. 5 and 6) for a discussion:
http://people.missouristate.edu/songfengzheng/Teaching/MTH541/Lecture%20notes/MLE.pdf






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much: I've searching this example and now I've understand the problem thanks to the discussion in the lecture notes.
    $endgroup$
    – Elia
    Jan 8 at 20:41











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$begingroup$

Sufficient conditions for the MLE to exist are $L$ continuous in $theta$ over a compact support (extreme value theorem).

We can built an example where the maximum in $theta$ does not exist, but the supremum does. The main open difficulty is whether the "supremum likelihood" is meaningful...

If $U$ is uniformly distributed on $theta=(a,b)$, where $a$ and $b$ are parameters, then the likelihood (for one observation) is $$ L(theta)=frac{1}{b-a} hspace{4mm} text{ if } a<u_i<b $$ and $0$ otherwise. Then the MLE does not exist, but the SLE does (but is not unique…). See for instance the lecture notes of Songfeng Zheng (p. 5 and 6) for a discussion:
http://people.missouristate.edu/songfengzheng/Teaching/MTH541/Lecture%20notes/MLE.pdf






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much: I've searching this example and now I've understand the problem thanks to the discussion in the lecture notes.
    $endgroup$
    – Elia
    Jan 8 at 20:41
















0












$begingroup$

Sufficient conditions for the MLE to exist are $L$ continuous in $theta$ over a compact support (extreme value theorem).

We can built an example where the maximum in $theta$ does not exist, but the supremum does. The main open difficulty is whether the "supremum likelihood" is meaningful...

If $U$ is uniformly distributed on $theta=(a,b)$, where $a$ and $b$ are parameters, then the likelihood (for one observation) is $$ L(theta)=frac{1}{b-a} hspace{4mm} text{ if } a<u_i<b $$ and $0$ otherwise. Then the MLE does not exist, but the SLE does (but is not unique…). See for instance the lecture notes of Songfeng Zheng (p. 5 and 6) for a discussion:
http://people.missouristate.edu/songfengzheng/Teaching/MTH541/Lecture%20notes/MLE.pdf






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much: I've searching this example and now I've understand the problem thanks to the discussion in the lecture notes.
    $endgroup$
    – Elia
    Jan 8 at 20:41














0












0








0





$begingroup$

Sufficient conditions for the MLE to exist are $L$ continuous in $theta$ over a compact support (extreme value theorem).

We can built an example where the maximum in $theta$ does not exist, but the supremum does. The main open difficulty is whether the "supremum likelihood" is meaningful...

If $U$ is uniformly distributed on $theta=(a,b)$, where $a$ and $b$ are parameters, then the likelihood (for one observation) is $$ L(theta)=frac{1}{b-a} hspace{4mm} text{ if } a<u_i<b $$ and $0$ otherwise. Then the MLE does not exist, but the SLE does (but is not unique…). See for instance the lecture notes of Songfeng Zheng (p. 5 and 6) for a discussion:
http://people.missouristate.edu/songfengzheng/Teaching/MTH541/Lecture%20notes/MLE.pdf






share|cite|improve this answer











$endgroup$



Sufficient conditions for the MLE to exist are $L$ continuous in $theta$ over a compact support (extreme value theorem).

We can built an example where the maximum in $theta$ does not exist, but the supremum does. The main open difficulty is whether the "supremum likelihood" is meaningful...

If $U$ is uniformly distributed on $theta=(a,b)$, where $a$ and $b$ are parameters, then the likelihood (for one observation) is $$ L(theta)=frac{1}{b-a} hspace{4mm} text{ if } a<u_i<b $$ and $0$ otherwise. Then the MLE does not exist, but the SLE does (but is not unique…). See for instance the lecture notes of Songfeng Zheng (p. 5 and 6) for a discussion:
http://people.missouristate.edu/songfengzheng/Teaching/MTH541/Lecture%20notes/MLE.pdf







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 10 at 8:19

























answered Jan 6 at 19:41









BertrandBertrand

963




963












  • $begingroup$
    Thank you so much: I've searching this example and now I've understand the problem thanks to the discussion in the lecture notes.
    $endgroup$
    – Elia
    Jan 8 at 20:41


















  • $begingroup$
    Thank you so much: I've searching this example and now I've understand the problem thanks to the discussion in the lecture notes.
    $endgroup$
    – Elia
    Jan 8 at 20:41
















$begingroup$
Thank you so much: I've searching this example and now I've understand the problem thanks to the discussion in the lecture notes.
$endgroup$
– Elia
Jan 8 at 20:41




$begingroup$
Thank you so much: I've searching this example and now I've understand the problem thanks to the discussion in the lecture notes.
$endgroup$
– Elia
Jan 8 at 20:41


















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