How to find a one dimensional sufficient statistic.
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Please could you help check my approach to finding a one dimensional sufficient statistic.
Here is the problem.
$X_1,...,X_n$ is a random sample with each $X_i$ having the pdf $f(x;theta)=2theta^2x^{-3} $ and $0<thetaleq x<infty$
Approach
$L(theta;x)= prod_{i=1}^n 2theta^2x^{-3}_i $
$L(theta;x)= 2^n theta^{2n}x^{-3n}_i $
$L(theta;x)= [2^n] [theta^{2n}][x^{-3n}_i] $
So by factorisation theorem, $x^{-3n}_i$ is a sufficient statistic.
Is this right?
statistics statistical-inference
asked Jan 3 at 16:38
Maths BarryMaths Barry
439
$endgroup$
add a comment |
$begingroup$
Please could you help check my approach to finding a one dimensional sufficient statistic.
Here is the problem.
$X_1,...,X_n$ is a random sample with each $X_i$ having the pdf $f(x;theta)=2theta^2x^{-3} $ and $0<thetaleq x<infty$
Approach
$L(theta;x)= prod_{i=1}^n 2theta^2x^{-3}_i $
$L(theta;x)= 2^n theta^{2n}x^{-3n}_i $
$L(theta;x)= [2^n] [theta^{2n}][x^{-3n}_i] $
So by factorisation theorem, $x^{-3n}_i$ is a sufficient statistic.
Is this right?
statistics statistical-inference
asked Jan 3 at 16:38
Maths BarryMaths Barry
439
$endgroup$
$begingroup$
A friendly advice: title is NOT the first sentence of your question. In particular, see the last bullet: the question post should be comprehensible without the title, even though one should make good use of the title to provide extra info.
$endgroup$
– Lee David Chung Lin
Jan 3 at 16:43
2
$begingroup$
No. $prod x_i^{-3}=(prod x_i)^{-3}$.
$endgroup$
– StubbornAtom
Jan 3 at 16:44
$begingroup$
$prod x_i^{-3}$ is not the same as $x^{-3n}$?
$endgroup$
– Maths Barry
Jan 4 at 9:31
1
$begingroup$
@MathsBarry Your sufficient statistic is $prod_{i=1}^n x_i^{-3}=x_1^{-3}ldots x_n^{-3}$; the $x_i$'s are not constant as you seem to think.
$endgroup$
– StubbornAtom
Jan 4 at 13:43
add a comment |
$begingroup$
Please could you help check my approach to finding a one dimensional sufficient statistic.
Here is the problem.
$X_1,...,X_n$ is a random sample with each $X_i$ having the pdf $f(x;theta)=2theta^2x^{-3} $ and $0<thetaleq x<infty$
Approach
$L(theta;x)= prod_{i=1}^n 2theta^2x^{-3}_i $
$L(theta;x)= 2^n theta^{2n}x^{-3n}_i $
$L(theta;x)= [2^n] [theta^{2n}][x^{-3n}_i] $
So by factorisation theorem, $x^{-3n}_i$ is a sufficient statistic.
Is this right?
statistics statistical-inference
asked Jan 3 at 16:38
Maths BarryMaths Barry
439
$endgroup$
Please could you help check my approach to finding a one dimensional sufficient statistic.
Here is the problem.
$X_1,...,X_n$ is a random sample with each $X_i$ having the pdf $f(x;theta)=2theta^2x^{-3} $ and $0<thetaleq x<infty$
Approach
$L(theta;x)= prod_{i=1}^n 2theta^2x^{-3}_i $
$L(theta;x)= 2^n theta^{2n}x^{-3n}_i $
$L(theta;x)= [2^n] [theta^{2n}][x^{-3n}_i] $
So by factorisation theorem, $x^{-3n}_i$ is a sufficient statistic.
Is this right?
statistics statistical-inference
statistics statistical-inference
asked Jan 3 at 16:38
Maths BarryMaths Barry
439
asked Jan 3 at 16:38
Maths BarryMaths Barry
439
edited Jan 3 at 17:02
Maths Barry
asked Jan 3 at 16:38
Maths BarryMaths Barry
439
asked Jan 3 at 16:38
Maths BarryMaths Barry
439
asked Jan 3 at 16:38
Maths BarryMaths Barry
439
439
$begingroup$
A friendly advice: title is NOT the first sentence of your question. In particular, see the last bullet: the question post should be comprehensible without the title, even though one should make good use of the title to provide extra info.
$endgroup$
– Lee David Chung Lin
Jan 3 at 16:43
2
$begingroup$
No. $prod x_i^{-3}=(prod x_i)^{-3}$.
$endgroup$
– StubbornAtom
Jan 3 at 16:44
$begingroup$
$prod x_i^{-3}$ is not the same as $x^{-3n}$?
$endgroup$
– Maths Barry
Jan 4 at 9:31
1
$begingroup$
@MathsBarry Your sufficient statistic is $prod_{i=1}^n x_i^{-3}=x_1^{-3}ldots x_n^{-3}$; the $x_i$'s are not constant as you seem to think.
$endgroup$
– StubbornAtom
Jan 4 at 13:43
add a comment |
$begingroup$
A friendly advice: title is NOT the first sentence of your question. In particular, see the last bullet: the question post should be comprehensible without the title, even though one should make good use of the title to provide extra info.
$endgroup$
– Lee David Chung Lin
Jan 3 at 16:43
2
$begingroup$
No. $prod x_i^{-3}=(prod x_i)^{-3}$.
$endgroup$
– StubbornAtom
Jan 3 at 16:44
$begingroup$
$prod x_i^{-3}$ is not the same as $x^{-3n}$?
$endgroup$
– Maths Barry
Jan 4 at 9:31
1
$begingroup$
@MathsBarry Your sufficient statistic is $prod_{i=1}^n x_i^{-3}=x_1^{-3}ldots x_n^{-3}$; the $x_i$'s are not constant as you seem to think.
$endgroup$
– StubbornAtom
Jan 4 at 13:43
$begingroup$
A friendly advice: title is NOT the first sentence of your question. In particular, see the last bullet: the question post should be comprehensible without the title, even though one should make good use of the title to provide extra info.
$endgroup$
– Lee David Chung Lin
Jan 3 at 16:43
$begingroup$
A friendly advice: title is NOT the first sentence of your question. In particular, see the last bullet: the question post should be comprehensible without the title, even though one should make good use of the title to provide extra info.
$endgroup$
– Lee David Chung Lin
Jan 3 at 16:43
2
2
$begingroup$
No. $prod x_i^{-3}=(prod x_i)^{-3}$.
$endgroup$
– StubbornAtom
Jan 3 at 16:44
$begingroup$
No. $prod x_i^{-3}=(prod x_i)^{-3}$.
$endgroup$
– StubbornAtom
Jan 3 at 16:44
$begingroup$
$prod x_i^{-3}$ is not the same as $x^{-3n}$?
$endgroup$
– Maths Barry
Jan 4 at 9:31
$begingroup$
$prod x_i^{-3}$ is not the same as $x^{-3n}$?
$endgroup$
– Maths Barry
Jan 4 at 9:31
1
1
$begingroup$
@MathsBarry Your sufficient statistic is $prod_{i=1}^n x_i^{-3}=x_1^{-3}ldots x_n^{-3}$; the $x_i$'s are not constant as you seem to think.
$endgroup$
– StubbornAtom
Jan 4 at 13:43
$begingroup$
@MathsBarry Your sufficient statistic is $prod_{i=1}^n x_i^{-3}=x_1^{-3}ldots x_n^{-3}$; the $x_i$'s are not constant as you seem to think.
$endgroup$
– StubbornAtom
Jan 4 at 13:43
add a comment |
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$begingroup$
A friendly advice: title is NOT the first sentence of your question. In particular, see the last bullet: the question post should be comprehensible without the title, even though one should make good use of the title to provide extra info.
$endgroup$
– Lee David Chung Lin
Jan 3 at 16:43
2
$begingroup$
No. $prod x_i^{-3}=(prod x_i)^{-3}$.
$endgroup$
– StubbornAtom
Jan 3 at 16:44
$begingroup$
$prod x_i^{-3}$ is not the same as $x^{-3n}$?
$endgroup$
– Maths Barry
Jan 4 at 9:31
1
$begingroup$
@MathsBarry Your sufficient statistic is $prod_{i=1}^n x_i^{-3}=x_1^{-3}ldots x_n^{-3}$; the $x_i$'s are not constant as you seem to think.
$endgroup$
– StubbornAtom
Jan 4 at 13:43