Maximum likelihood estimator of uniform $(a,b)$
0
$begingroup$
Let $X_1, dots, X_n$ be a sample of independent random variables with uniform distribution on $(a,b)$. Derive the maximum likelihood estimators of $a$ and $b$.
Given it is a sample of independent variables we can write:
$$
L = text{product of marginal density} = n^{-n} quadtext{or}quad (b-a+1)^{-n}
$$
then
$$
dfrac{mathrm{d}L}{mathrm{d}a} = -n(b-a+1)^{n-1} = 0
quadtext{then}quad
a = b+1
$$
and
$$
dfrac{mathrm{d}L}{mathrm{d}b} = n(b-a+1)^{n-1} = 0
quadtext{then}quad
b = a-1
$$
statistics
edited Jan 21 at 18:52
Björn Friedrich
2,67961831
asked Jan 21 at 17:53
T. BrunT. Brun
1
$endgroup$
add a comment |
0
$begingroup$
Let $X_1, dots, X_n$ be a sample of independent random variables with uniform distribution on $(a,b)$. Derive the maximum likelihood estimators of $a$ and $b$.
Given it is a sample of independent variables we can write:
$$
L = text{product of marginal density} = n^{-n} quadtext{or}quad (b-a+1)^{-n}
$$
then
$$
dfrac{mathrm{d}L}{mathrm{d}a} = -n(b-a+1)^{n-1} = 0
quadtext{then}quad
a = b+1
$$
and
$$
dfrac{mathrm{d}L}{mathrm{d}b} = n(b-a+1)^{n-1} = 0
quadtext{then}quad
b = a-1
$$
statistics
edited Jan 21 at 18:52
Björn Friedrich
2,67961831
asked Jan 21 at 17:53
T. BrunT. Brun
1
$endgroup$
$begingroup$
This site works better if you tell us in the question what you have tried already
$endgroup$
– Henry
Jan 21 at 17:54
$begingroup$
Given it is a sample of independant variables we can write : L=product of marginal density = n^(-n) or (b-a+1)^(-n) then dL/da = -n(b-a+1)^(n-1) = 0 then a= b+1 and dL/db = n(b-a+1)^(n-1) = 0 then b= a-1
$endgroup$
– T. Brun
Jan 21 at 18:14
2
$begingroup$
I have put your comment into the question - you might check whether it says what you want. It is not correct, partly because you should probably have said the likelihood was $(b-a)^{-n}$ and partly because you have not used any information from $X_1, dots X_n$ in your calculations of the likelihood: in particular all the observations must be in the interval $[a,b]$. As an extra hint, the maximum likelihood is not always associated with its derivative being zero
$endgroup$
– Henry
Jan 21 at 18:40
$begingroup$
If you write down the likelihood correctly including proper support of the joint distribution, it will be clear that the likelihood is not differentiable everywhere and hence MLE cannot be derived using differentiation.
$endgroup$
– StubbornAtom
Jan 21 at 20:03
add a comment |
0
0
0
$begingroup$
Let $X_1, dots, X_n$ be a sample of independent random variables with uniform distribution on $(a,b)$. Derive the maximum likelihood estimators of $a$ and $b$.
Given it is a sample of independent variables we can write:
$$
L = text{product of marginal density} = n^{-n} quadtext{or}quad (b-a+1)^{-n}
$$
then
$$
dfrac{mathrm{d}L}{mathrm{d}a} = -n(b-a+1)^{n-1} = 0
quadtext{then}quad
a = b+1
$$
and
$$
dfrac{mathrm{d}L}{mathrm{d}b} = n(b-a+1)^{n-1} = 0
quadtext{then}quad
b = a-1
$$
statistics
edited Jan 21 at 18:52
Björn Friedrich
2,67961831
asked Jan 21 at 17:53
T. BrunT. Brun
1
$endgroup$
Let $X_1, dots, X_n$ be a sample of independent random variables with uniform distribution on $(a,b)$. Derive the maximum likelihood estimators of $a$ and $b$.
Given it is a sample of independent variables we can write:
$$
L = text{product of marginal density} = n^{-n} quadtext{or}quad (b-a+1)^{-n}
$$
then
$$
dfrac{mathrm{d}L}{mathrm{d}a} = -n(b-a+1)^{n-1} = 0
quadtext{then}quad
a = b+1
$$
and
$$
dfrac{mathrm{d}L}{mathrm{d}b} = n(b-a+1)^{n-1} = 0
quadtext{then}quad
b = a-1
$$
statistics
statistics
edited Jan 21 at 18:52
Björn Friedrich
2,67961831
asked Jan 21 at 17:53
T. BrunT. Brun
1
edited Jan 21 at 18:52
Björn Friedrich
2,67961831
asked Jan 21 at 17:53
T. BrunT. Brun
1
edited Jan 21 at 18:52
Björn Friedrich
2,67961831
edited Jan 21 at 18:52
Björn Friedrich
2,67961831
edited Jan 21 at 18:52
Björn Friedrich
2,67961831
2,67961831
asked Jan 21 at 17:53
T. BrunT. Brun
1
asked Jan 21 at 17:53
T. BrunT. Brun
1
asked Jan 21 at 17:53
T. BrunT. Brun
1
1
$begingroup$
This site works better if you tell us in the question what you have tried already
$endgroup$
– Henry
Jan 21 at 17:54
$begingroup$
Given it is a sample of independant variables we can write : L=product of marginal density = n^(-n) or (b-a+1)^(-n) then dL/da = -n(b-a+1)^(n-1) = 0 then a= b+1 and dL/db = n(b-a+1)^(n-1) = 0 then b= a-1
$endgroup$
– T. Brun
Jan 21 at 18:14
2
$begingroup$
I have put your comment into the question - you might check whether it says what you want. It is not correct, partly because you should probably have said the likelihood was $(b-a)^{-n}$ and partly because you have not used any information from $X_1, dots X_n$ in your calculations of the likelihood: in particular all the observations must be in the interval $[a,b]$. As an extra hint, the maximum likelihood is not always associated with its derivative being zero
$endgroup$
– Henry
Jan 21 at 18:40
$begingroup$
If you write down the likelihood correctly including proper support of the joint distribution, it will be clear that the likelihood is not differentiable everywhere and hence MLE cannot be derived using differentiation.
$endgroup$
– StubbornAtom
Jan 21 at 20:03
add a comment |
$begingroup$
This site works better if you tell us in the question what you have tried already
$endgroup$
– Henry
Jan 21 at 17:54
$begingroup$
Given it is a sample of independant variables we can write : L=product of marginal density = n^(-n) or (b-a+1)^(-n) then dL/da = -n(b-a+1)^(n-1) = 0 then a= b+1 and dL/db = n(b-a+1)^(n-1) = 0 then b= a-1
$endgroup$
– T. Brun
Jan 21 at 18:14
2
$begingroup$
I have put your comment into the question - you might check whether it says what you want. It is not correct, partly because you should probably have said the likelihood was $(b-a)^{-n}$ and partly because you have not used any information from $X_1, dots X_n$ in your calculations of the likelihood: in particular all the observations must be in the interval $[a,b]$. As an extra hint, the maximum likelihood is not always associated with its derivative being zero
$endgroup$
– Henry
Jan 21 at 18:40
$begingroup$
If you write down the likelihood correctly including proper support of the joint distribution, it will be clear that the likelihood is not differentiable everywhere and hence MLE cannot be derived using differentiation.
$endgroup$
– StubbornAtom
Jan 21 at 20:03
$begingroup$
This site works better if you tell us in the question what you have tried already
$endgroup$
– Henry
Jan 21 at 17:54
$begingroup$
This site works better if you tell us in the question what you have tried already
$endgroup$
– Henry
Jan 21 at 17:54
$begingroup$
Given it is a sample of independant variables we can write : L=product of marginal density = n^(-n) or (b-a+1)^(-n) then dL/da = -n(b-a+1)^(n-1) = 0 then a= b+1 and dL/db = n(b-a+1)^(n-1) = 0 then b= a-1
$endgroup$
– T. Brun
Jan 21 at 18:14
$begingroup$
Given it is a sample of independant variables we can write : L=product of marginal density = n^(-n) or (b-a+1)^(-n) then dL/da = -n(b-a+1)^(n-1) = 0 then a= b+1 and dL/db = n(b-a+1)^(n-1) = 0 then b= a-1
$endgroup$
– T. Brun
Jan 21 at 18:14
2
2
$begingroup$
I have put your comment into the question - you might check whether it says what you want. It is not correct, partly because you should probably have said the likelihood was $(b-a)^{-n}$ and partly because you have not used any information from $X_1, dots X_n$ in your calculations of the likelihood: in particular all the observations must be in the interval $[a,b]$. As an extra hint, the maximum likelihood is not always associated with its derivative being zero
$endgroup$
– Henry
Jan 21 at 18:40
$begingroup$
I have put your comment into the question - you might check whether it says what you want. It is not correct, partly because you should probably have said the likelihood was $(b-a)^{-n}$ and partly because you have not used any information from $X_1, dots X_n$ in your calculations of the likelihood: in particular all the observations must be in the interval $[a,b]$. As an extra hint, the maximum likelihood is not always associated with its derivative being zero
$endgroup$
– Henry
Jan 21 at 18:40
$begingroup$
If you write down the likelihood correctly including proper support of the joint distribution, it will be clear that the likelihood is not differentiable everywhere and hence MLE cannot be derived using differentiation.
$endgroup$
– StubbornAtom
Jan 21 at 20:03
$begingroup$
If you write down the likelihood correctly including proper support of the joint distribution, it will be clear that the likelihood is not differentiable everywhere and hence MLE cannot be derived using differentiation.
$endgroup$
– StubbornAtom
Jan 21 at 20:03
add a comment |
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$begingroup$
This site works better if you tell us in the question what you have tried already
$endgroup$
– Henry
Jan 21 at 17:54
$begingroup$
Given it is a sample of independant variables we can write : L=product of marginal density = n^(-n) or (b-a+1)^(-n) then dL/da = -n(b-a+1)^(n-1) = 0 then a= b+1 and dL/db = n(b-a+1)^(n-1) = 0 then b= a-1
$endgroup$
– T. Brun
Jan 21 at 18:14
2
$begingroup$
I have put your comment into the question - you might check whether it says what you want. It is not correct, partly because you should probably have said the likelihood was $(b-a)^{-n}$ and partly because you have not used any information from $X_1, dots X_n$ in your calculations of the likelihood: in particular all the observations must be in the interval $[a,b]$. As an extra hint, the maximum likelihood is not always associated with its derivative being zero
$endgroup$
– Henry
Jan 21 at 18:40
$begingroup$
If you write down the likelihood correctly including proper support of the joint distribution, it will be clear that the likelihood is not differentiable everywhere and hence MLE cannot be derived using differentiation.
$endgroup$
– StubbornAtom
Jan 21 at 20:03