Example of a maximum likelihood estimator that is not a sufficient statistic












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I am currently researching on providing some bounds on estimation using some information theoretic tools (I won't expend on that here for now, I may make a post about it later) and turns out that given a phenomenon $X$, an observation $Y$, then $hat{x}(Y)$, the maximum likelihood estimator of $X$ based on $Y$, may apparently not be a sufficient statistic and this is a something I would like to study, the answer to this post states that such an example exists when $Y$ consists of samples that are not i.i.d but fails to provide such an example and I haven't found anything about it. Have anyone seen something of the sort ?






statistics estimation maximum-likelihood data-sufficiency







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am currently researching on providing some bounds on estimation using some information theoretic tools (I won't expend on that here for now, I may make a post about it later) and turns out that given a phenomenon $X$, an observation $Y$, then $hat{x}(Y)$, the maximum likelihood estimator of $X$ based on $Y$, may apparently not be a sufficient statistic and this is a something I would like to study, the answer to this post states that such an example exists when $Y$ consists of samples that are not i.i.d but fails to provide such an example and I haven't found anything about it. Have anyone seen something of the sort ?






    statistics estimation maximum-likelihood data-sufficiency







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am currently researching on providing some bounds on estimation using some information theoretic tools (I won't expend on that here for now, I may make a post about it later) and turns out that given a phenomenon $X$, an observation $Y$, then $hat{x}(Y)$, the maximum likelihood estimator of $X$ based on $Y$, may apparently not be a sufficient statistic and this is a something I would like to study, the answer to this post states that such an example exists when $Y$ consists of samples that are not i.i.d but fails to provide such an example and I haven't found anything about it. Have anyone seen something of the sort ?






      statistics estimation maximum-likelihood data-sufficiency







      share|cite|improve this question









      $endgroup$




      I am currently researching on providing some bounds on estimation using some information theoretic tools (I won't expend on that here for now, I may make a post about it later) and turns out that given a phenomenon $X$, an observation $Y$, then $hat{x}(Y)$, the maximum likelihood estimator of $X$ based on $Y$, may apparently not be a sufficient statistic and this is a something I would like to study, the answer to this post states that such an example exists when $Y$ consists of samples that are not i.i.d but fails to provide such an example and I haven't found anything about it. Have anyone seen something of the sort ?






      statistics estimation maximum-likelihood data-sufficiency




      statistics estimation maximum-likelihood data-sufficiency






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Dec 20 '18 at 11:00









      P. QuintonP. Quinton

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      1,8161213






















          1 Answer
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          1












          $begingroup$


          If $T$ is a sufficient statistic for $theta$ and a unique MLE of $theta$ exists, then the MLE must be a function of $T$.




          So if you can find a situation where there can be several maximum likelihood estimators, there remains a possibility that you can choose one MLE that might not be a function of a sufficient statistic alone.



          A simple example to consider is the $U(theta,theta+1)$ distribution.



          Consider i.i.d random variables $X_1,X_2,ldots,X_n$ having the above distribution.



          Then the likelihood function given the sample $(x_1,ldots,x_n)$ is



          $$L(theta)=prod_{i=1}^n mathbf1_{theta<x_i<theta+1}=mathbf1_{theta<x_{(1)},x_{(n)}<theta+1}quad,,thetainmathbb R$$



          A sufficient statistic for $theta$ is $$T(X_1,ldots,X_n)=(X_{(1)},X_{(n)})$$



          And an MLE of $theta$ is any $hattheta$ satisfying $$hattheta<X_{(1)},X_{(n)}<hattheta+1$$



          or equivalently, $$X_{(n)}-1<hattheta<X_{(1)} tag{1}$$



          Choose $$hattheta'= (sin^2 X_1)(X_{(n)}-1) + (cos^2 X_1)(X_{(1)})$$



          Then $hattheta'$ satisfies $(1)$ but it does not depend on the sample only through $T$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I am a bit curious, do you think there is always a element in the set of MLE that is a sufficient statistic ? (in this case yes, if we take $(X_{(1)}+X_{(n)})/2$) and is it unique ? Also I am interested in other measures of error, do you know if there are some examples of minimum MSE estimator that is not a sufficient statistic too ?
            $endgroup$
            – P. Quinton
            Dec 20 '18 at 16:40












          • $begingroup$
            @P.Quinton There is of course an MLE that is a function of $T$. Any MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ is a function of the sufficient statistic $T$ where $alphain(0,1)$ is a constant.
            $endgroup$
            – StubbornAtom
            Dec 20 '18 at 18:10












          • $begingroup$
            Yes but that's not really what I asked, in my case I was wondering if $theta - hattheta - (X_1, dots, X_n)$ is a Markov chain (supposing that $theta$ is a random variable), I think that in your formula, if we know alpha then we can find $X_{(1)}$ and $X_{(n)}$ back, so it's a sufficient statistic, is that correct ?
            $endgroup$
            – P. Quinton
            Dec 20 '18 at 18:16












          • $begingroup$
            @P.Quinton Well that is out of my league to answer. I only aimed to give you an MLE that is not a function of sufficient statistic, as asked in your post. An MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ satisfies condition $(1)$ in my answer. If we choose $alphain(0,1)$ to be a constant, then the MLE depends on the sample only through $T$.
            $endgroup$
            – StubbornAtom
            Dec 20 '18 at 18:19













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          $begingroup$


          If $T$ is a sufficient statistic for $theta$ and a unique MLE of $theta$ exists, then the MLE must be a function of $T$.




          So if you can find a situation where there can be several maximum likelihood estimators, there remains a possibility that you can choose one MLE that might not be a function of a sufficient statistic alone.



          A simple example to consider is the $U(theta,theta+1)$ distribution.



          Consider i.i.d random variables $X_1,X_2,ldots,X_n$ having the above distribution.



          Then the likelihood function given the sample $(x_1,ldots,x_n)$ is



          $$L(theta)=prod_{i=1}^n mathbf1_{theta<x_i<theta+1}=mathbf1_{theta<x_{(1)},x_{(n)}<theta+1}quad,,thetainmathbb R$$



          A sufficient statistic for $theta$ is $$T(X_1,ldots,X_n)=(X_{(1)},X_{(n)})$$



          And an MLE of $theta$ is any $hattheta$ satisfying $$hattheta<X_{(1)},X_{(n)}<hattheta+1$$



          or equivalently, $$X_{(n)}-1<hattheta<X_{(1)} tag{1}$$



          Choose $$hattheta'= (sin^2 X_1)(X_{(n)}-1) + (cos^2 X_1)(X_{(1)})$$



          Then $hattheta'$ satisfies $(1)$ but it does not depend on the sample only through $T$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I am a bit curious, do you think there is always a element in the set of MLE that is a sufficient statistic ? (in this case yes, if we take $(X_{(1)}+X_{(n)})/2$) and is it unique ? Also I am interested in other measures of error, do you know if there are some examples of minimum MSE estimator that is not a sufficient statistic too ?
            $endgroup$
            – P. Quinton
            Dec 20 '18 at 16:40












          • $begingroup$
            @P.Quinton There is of course an MLE that is a function of $T$. Any MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ is a function of the sufficient statistic $T$ where $alphain(0,1)$ is a constant.
            $endgroup$
            – StubbornAtom
            Dec 20 '18 at 18:10












          • $begingroup$
            Yes but that's not really what I asked, in my case I was wondering if $theta - hattheta - (X_1, dots, X_n)$ is a Markov chain (supposing that $theta$ is a random variable), I think that in your formula, if we know alpha then we can find $X_{(1)}$ and $X_{(n)}$ back, so it's a sufficient statistic, is that correct ?
            $endgroup$
            – P. Quinton
            Dec 20 '18 at 18:16












          • $begingroup$
            @P.Quinton Well that is out of my league to answer. I only aimed to give you an MLE that is not a function of sufficient statistic, as asked in your post. An MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ satisfies condition $(1)$ in my answer. If we choose $alphain(0,1)$ to be a constant, then the MLE depends on the sample only through $T$.
            $endgroup$
            – StubbornAtom
            Dec 20 '18 at 18:19


















          1












          $begingroup$


          If $T$ is a sufficient statistic for $theta$ and a unique MLE of $theta$ exists, then the MLE must be a function of $T$.




          So if you can find a situation where there can be several maximum likelihood estimators, there remains a possibility that you can choose one MLE that might not be a function of a sufficient statistic alone.



          A simple example to consider is the $U(theta,theta+1)$ distribution.



          Consider i.i.d random variables $X_1,X_2,ldots,X_n$ having the above distribution.



          Then the likelihood function given the sample $(x_1,ldots,x_n)$ is



          $$L(theta)=prod_{i=1}^n mathbf1_{theta<x_i<theta+1}=mathbf1_{theta<x_{(1)},x_{(n)}<theta+1}quad,,thetainmathbb R$$



          A sufficient statistic for $theta$ is $$T(X_1,ldots,X_n)=(X_{(1)},X_{(n)})$$



          And an MLE of $theta$ is any $hattheta$ satisfying $$hattheta<X_{(1)},X_{(n)}<hattheta+1$$



          or equivalently, $$X_{(n)}-1<hattheta<X_{(1)} tag{1}$$



          Choose $$hattheta'= (sin^2 X_1)(X_{(n)}-1) + (cos^2 X_1)(X_{(1)})$$



          Then $hattheta'$ satisfies $(1)$ but it does not depend on the sample only through $T$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I am a bit curious, do you think there is always a element in the set of MLE that is a sufficient statistic ? (in this case yes, if we take $(X_{(1)}+X_{(n)})/2$) and is it unique ? Also I am interested in other measures of error, do you know if there are some examples of minimum MSE estimator that is not a sufficient statistic too ?
            $endgroup$
            – P. Quinton
            Dec 20 '18 at 16:40












          • $begingroup$
            @P.Quinton There is of course an MLE that is a function of $T$. Any MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ is a function of the sufficient statistic $T$ where $alphain(0,1)$ is a constant.
            $endgroup$
            – StubbornAtom
            Dec 20 '18 at 18:10












          • $begingroup$
            Yes but that's not really what I asked, in my case I was wondering if $theta - hattheta - (X_1, dots, X_n)$ is a Markov chain (supposing that $theta$ is a random variable), I think that in your formula, if we know alpha then we can find $X_{(1)}$ and $X_{(n)}$ back, so it's a sufficient statistic, is that correct ?
            $endgroup$
            – P. Quinton
            Dec 20 '18 at 18:16












          • $begingroup$
            @P.Quinton Well that is out of my league to answer. I only aimed to give you an MLE that is not a function of sufficient statistic, as asked in your post. An MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ satisfies condition $(1)$ in my answer. If we choose $alphain(0,1)$ to be a constant, then the MLE depends on the sample only through $T$.
            $endgroup$
            – StubbornAtom
            Dec 20 '18 at 18:19
















          1












          1








          1





          $begingroup$


          If $T$ is a sufficient statistic for $theta$ and a unique MLE of $theta$ exists, then the MLE must be a function of $T$.




          So if you can find a situation where there can be several maximum likelihood estimators, there remains a possibility that you can choose one MLE that might not be a function of a sufficient statistic alone.



          A simple example to consider is the $U(theta,theta+1)$ distribution.



          Consider i.i.d random variables $X_1,X_2,ldots,X_n$ having the above distribution.



          Then the likelihood function given the sample $(x_1,ldots,x_n)$ is



          $$L(theta)=prod_{i=1}^n mathbf1_{theta<x_i<theta+1}=mathbf1_{theta<x_{(1)},x_{(n)}<theta+1}quad,,thetainmathbb R$$



          A sufficient statistic for $theta$ is $$T(X_1,ldots,X_n)=(X_{(1)},X_{(n)})$$



          And an MLE of $theta$ is any $hattheta$ satisfying $$hattheta<X_{(1)},X_{(n)}<hattheta+1$$



          or equivalently, $$X_{(n)}-1<hattheta<X_{(1)} tag{1}$$



          Choose $$hattheta'= (sin^2 X_1)(X_{(n)}-1) + (cos^2 X_1)(X_{(1)})$$



          Then $hattheta'$ satisfies $(1)$ but it does not depend on the sample only through $T$.






          share|cite|improve this answer











          $endgroup$




          If $T$ is a sufficient statistic for $theta$ and a unique MLE of $theta$ exists, then the MLE must be a function of $T$.




          So if you can find a situation where there can be several maximum likelihood estimators, there remains a possibility that you can choose one MLE that might not be a function of a sufficient statistic alone.



          A simple example to consider is the $U(theta,theta+1)$ distribution.



          Consider i.i.d random variables $X_1,X_2,ldots,X_n$ having the above distribution.



          Then the likelihood function given the sample $(x_1,ldots,x_n)$ is



          $$L(theta)=prod_{i=1}^n mathbf1_{theta<x_i<theta+1}=mathbf1_{theta<x_{(1)},x_{(n)}<theta+1}quad,,thetainmathbb R$$



          A sufficient statistic for $theta$ is $$T(X_1,ldots,X_n)=(X_{(1)},X_{(n)})$$



          And an MLE of $theta$ is any $hattheta$ satisfying $$hattheta<X_{(1)},X_{(n)}<hattheta+1$$



          or equivalently, $$X_{(n)}-1<hattheta<X_{(1)} tag{1}$$



          Choose $$hattheta'= (sin^2 X_1)(X_{(n)}-1) + (cos^2 X_1)(X_{(1)})$$



          Then $hattheta'$ satisfies $(1)$ but it does not depend on the sample only through $T$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 20 '18 at 14:16

























          answered Dec 20 '18 at 13:56









          StubbornAtomStubbornAtom

          6,00811238




          6,00811238












          • $begingroup$
            I am a bit curious, do you think there is always a element in the set of MLE that is a sufficient statistic ? (in this case yes, if we take $(X_{(1)}+X_{(n)})/2$) and is it unique ? Also I am interested in other measures of error, do you know if there are some examples of minimum MSE estimator that is not a sufficient statistic too ?
            $endgroup$
            – P. Quinton
            Dec 20 '18 at 16:40












          • $begingroup$
            @P.Quinton There is of course an MLE that is a function of $T$. Any MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ is a function of the sufficient statistic $T$ where $alphain(0,1)$ is a constant.
            $endgroup$
            – StubbornAtom
            Dec 20 '18 at 18:10












          • $begingroup$
            Yes but that's not really what I asked, in my case I was wondering if $theta - hattheta - (X_1, dots, X_n)$ is a Markov chain (supposing that $theta$ is a random variable), I think that in your formula, if we know alpha then we can find $X_{(1)}$ and $X_{(n)}$ back, so it's a sufficient statistic, is that correct ?
            $endgroup$
            – P. Quinton
            Dec 20 '18 at 18:16












          • $begingroup$
            @P.Quinton Well that is out of my league to answer. I only aimed to give you an MLE that is not a function of sufficient statistic, as asked in your post. An MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ satisfies condition $(1)$ in my answer. If we choose $alphain(0,1)$ to be a constant, then the MLE depends on the sample only through $T$.
            $endgroup$
            – StubbornAtom
            Dec 20 '18 at 18:19




















          • $begingroup$
            I am a bit curious, do you think there is always a element in the set of MLE that is a sufficient statistic ? (in this case yes, if we take $(X_{(1)}+X_{(n)})/2$) and is it unique ? Also I am interested in other measures of error, do you know if there are some examples of minimum MSE estimator that is not a sufficient statistic too ?
            $endgroup$
            – P. Quinton
            Dec 20 '18 at 16:40












          • $begingroup$
            @P.Quinton There is of course an MLE that is a function of $T$. Any MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ is a function of the sufficient statistic $T$ where $alphain(0,1)$ is a constant.
            $endgroup$
            – StubbornAtom
            Dec 20 '18 at 18:10












          • $begingroup$
            Yes but that's not really what I asked, in my case I was wondering if $theta - hattheta - (X_1, dots, X_n)$ is a Markov chain (supposing that $theta$ is a random variable), I think that in your formula, if we know alpha then we can find $X_{(1)}$ and $X_{(n)}$ back, so it's a sufficient statistic, is that correct ?
            $endgroup$
            – P. Quinton
            Dec 20 '18 at 18:16












          • $begingroup$
            @P.Quinton Well that is out of my league to answer. I only aimed to give you an MLE that is not a function of sufficient statistic, as asked in your post. An MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ satisfies condition $(1)$ in my answer. If we choose $alphain(0,1)$ to be a constant, then the MLE depends on the sample only through $T$.
            $endgroup$
            – StubbornAtom
            Dec 20 '18 at 18:19


















          $begingroup$
          I am a bit curious, do you think there is always a element in the set of MLE that is a sufficient statistic ? (in this case yes, if we take $(X_{(1)}+X_{(n)})/2$) and is it unique ? Also I am interested in other measures of error, do you know if there are some examples of minimum MSE estimator that is not a sufficient statistic too ?
          $endgroup$
          – P. Quinton
          Dec 20 '18 at 16:40






          $begingroup$
          I am a bit curious, do you think there is always a element in the set of MLE that is a sufficient statistic ? (in this case yes, if we take $(X_{(1)}+X_{(n)})/2$) and is it unique ? Also I am interested in other measures of error, do you know if there are some examples of minimum MSE estimator that is not a sufficient statistic too ?
          $endgroup$
          – P. Quinton
          Dec 20 '18 at 16:40














          $begingroup$
          @P.Quinton There is of course an MLE that is a function of $T$. Any MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ is a function of the sufficient statistic $T$ where $alphain(0,1)$ is a constant.
          $endgroup$
          – StubbornAtom
          Dec 20 '18 at 18:10






          $begingroup$
          @P.Quinton There is of course an MLE that is a function of $T$. Any MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ is a function of the sufficient statistic $T$ where $alphain(0,1)$ is a constant.
          $endgroup$
          – StubbornAtom
          Dec 20 '18 at 18:10














          $begingroup$
          Yes but that's not really what I asked, in my case I was wondering if $theta - hattheta - (X_1, dots, X_n)$ is a Markov chain (supposing that $theta$ is a random variable), I think that in your formula, if we know alpha then we can find $X_{(1)}$ and $X_{(n)}$ back, so it's a sufficient statistic, is that correct ?
          $endgroup$
          – P. Quinton
          Dec 20 '18 at 18:16






          $begingroup$
          Yes but that's not really what I asked, in my case I was wondering if $theta - hattheta - (X_1, dots, X_n)$ is a Markov chain (supposing that $theta$ is a random variable), I think that in your formula, if we know alpha then we can find $X_{(1)}$ and $X_{(n)}$ back, so it's a sufficient statistic, is that correct ?
          $endgroup$
          – P. Quinton
          Dec 20 '18 at 18:16














          $begingroup$
          @P.Quinton Well that is out of my league to answer. I only aimed to give you an MLE that is not a function of sufficient statistic, as asked in your post. An MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ satisfies condition $(1)$ in my answer. If we choose $alphain(0,1)$ to be a constant, then the MLE depends on the sample only through $T$.
          $endgroup$
          – StubbornAtom
          Dec 20 '18 at 18:19






          $begingroup$
          @P.Quinton Well that is out of my league to answer. I only aimed to give you an MLE that is not a function of sufficient statistic, as asked in your post. An MLE of the form $alpha(X_{(n)}-1)+(1-alpha)X_{(1)}$ satisfies condition $(1)$ in my answer. If we choose $alphain(0,1)$ to be a constant, then the MLE depends on the sample only through $T$.
          $endgroup$
          – StubbornAtom
          Dec 20 '18 at 18:19




















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          Mont Emei

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          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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