How to find a one dimensional sufficient statistic.












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 17:02







Maths Barry

















asked Jan 3 at 16:38









Maths BarryMaths Barry

438




438












  • $begingroup$
    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






  • 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










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