Example of a maximum likelihood estimator that is not a sufficient statistic
$begingroup$
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
asked Dec 20 '18 at 11:00
P. QuintonP. Quinton
1,8161213
$endgroup$
add a comment |
$begingroup$
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
asked Dec 20 '18 at 11:00
P. QuintonP. Quinton
1,8161213
$endgroup$
add a comment |
$begingroup$
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
asked Dec 20 '18 at 11:00
P. QuintonP. Quinton
1,8161213
$endgroup$
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
asked Dec 20 '18 at 11:00
P. QuintonP. Quinton
1,8161213
asked Dec 20 '18 at 11:00
P. QuintonP. Quinton
1,8161213
asked Dec 20 '18 at 11:00
P. QuintonP. Quinton
1,8161213
asked Dec 20 '18 at 11:00
P. QuintonP. Quinton
1,8161213
asked Dec 20 '18 at 11:00
P. QuintonP. Quinton
1,8161213
1,8161213
add a comment |
add a comment |
1 Answer
1
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oldest
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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 $T$.
answered Dec 20 '18 at 13:56
StubbornAtomStubbornAtom
6,00811238
$endgroup$
$begingroup$
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
add a comment |
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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 $T$.
answered Dec 20 '18 at 13:56
StubbornAtomStubbornAtom
6,00811238
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $T$.
answered Dec 20 '18 at 13:56
StubbornAtomStubbornAtom
6,00811238
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $T$.
answered Dec 20 '18 at 13:56
StubbornAtomStubbornAtom
6,00811238
$endgroup$
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 $T$.
answered Dec 20 '18 at 13:56
StubbornAtomStubbornAtom
6,00811238
edited Dec 20 '18 at 14:16
answered Dec 20 '18 at 13:56
StubbornAtomStubbornAtom
6,00811238
answered Dec 20 '18 at 13:56
StubbornAtomStubbornAtom
6,00811238
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
add a comment |
$begingroup$
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
add a comment |
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