How to find a one dimensional sufficient statistic.
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 2 days ago
Maths Barry
286
add a comment |
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 2 days ago
Maths Barry
286
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.
– Lee David Chung Lin
2 days ago
2
No. $prod x_i^{-3}=(prod x_i)^{-3}$.
– StubbornAtom
2 days ago
$prod x_i^{-3}$ is not the same as $x^{-3n}$?
– Maths Barry
yesterday
@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.
– StubbornAtom
yesterday
add a comment |
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 2 days ago
Maths Barry
286
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 2 days ago
Maths Barry
286
asked 2 days ago
Maths Barry
286
edited 2 days ago
asked 2 days ago
Maths Barry
286
asked 2 days ago
Maths Barry
286
asked 2 days ago
Maths Barry
286
286
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.
– Lee David Chung Lin
2 days ago
2
No. $prod x_i^{-3}=(prod x_i)^{-3}$.
– StubbornAtom
2 days ago
$prod x_i^{-3}$ is not the same as $x^{-3n}$?
– Maths Barry
yesterday
@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.
– StubbornAtom
yesterday
add a comment |
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.
– Lee David Chung Lin
2 days ago
2
No. $prod x_i^{-3}=(prod x_i)^{-3}$.
– StubbornAtom
2 days ago
$prod x_i^{-3}$ is not the same as $x^{-3n}$?
– Maths Barry
yesterday
@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.
– StubbornAtom
yesterday
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.
– Lee David Chung Lin
2 days ago
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.
– Lee David Chung Lin
2 days ago
2
2
No. $prod x_i^{-3}=(prod x_i)^{-3}$.
– StubbornAtom
2 days ago
No. $prod x_i^{-3}=(prod x_i)^{-3}$.
– StubbornAtom
2 days ago
$prod x_i^{-3}$ is not the same as $x^{-3n}$?
– Maths Barry
yesterday
$prod x_i^{-3}$ is not the same as $x^{-3n}$?
– Maths Barry
yesterday
@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.
– StubbornAtom
yesterday
@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.
– StubbornAtom
yesterday
add a comment |
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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.
– Lee David Chung Lin
2 days ago
2
No. $prod x_i^{-3}=(prod x_i)^{-3}$.
– StubbornAtom
2 days ago
$prod x_i^{-3}$ is not the same as $x^{-3n}$?
– Maths Barry
yesterday
@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.
– StubbornAtom
yesterday