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Sufficient statistic equivalence

Show a statistic is not sufficientSufficient statistic.Show that $T$ is not a sufficient statisticHow do i know what's the sufficient statistic/estimator?What is a sufficient statistic of this distribution?Transformation of a sufficient statisticSufficient statistic with…Finding complete sufficient statisticMinimal sufficient statistic criterionThe natural sufficient statistic is minimal sufficient

1

$begingroup$

Let $theta'$, $theta in Theta$ such that $theta' neq theta$. I want to prove that $T$ is a sufficient statistic if and only if $$frac{f(x,theta')}{f(x,theta)}$$
is a function dependent only on $T(x)$.

I tried to use factorization theorem but for different parameters functions can be different and it leads nowhere.

statistics statistical-inference

share|cite|improve this question

asked Mar 12 at 23:06

treskovtreskov

147110

$endgroup$

add a comment | 

1

$begingroup$

Let $theta'$, $theta in Theta$ such that $theta' neq theta$. I want to prove that $T$ is a sufficient statistic if and only if $$frac{f(x,theta')}{f(x,theta)}$$
is a function dependent only on $T(x)$.

I tried to use factorization theorem but for different parameters functions can be different and it leads nowhere.

statistics statistical-inference

share|cite|improve this question

asked Mar 12 at 23:06

treskovtreskov

147110

$endgroup$

add a comment | 

$begingroup$

Let $theta'$, $theta in Theta$ such that $theta' neq theta$. I want to prove that $T$ is a sufficient statistic if and only if $$frac{f(x,theta')}{f(x,theta)}$$
is a function dependent only on $T(x)$.

I tried to use factorization theorem but for different parameters functions can be different and it leads nowhere.

statistics statistical-inference

share|cite|improve this question

asked Mar 12 at 23:06

treskovtreskov

147110

$endgroup$

share|cite|improve this question

asked Mar 12 at 23:06

treskovtreskov

147110

share|cite|improve this question

asked Mar 12 at 23:06

treskovtreskov

147110

asked Mar 12 at 23:06

treskovtreskov

147110

asked Mar 12 at 23:06

treskovtreskov

147110

1 Answer
1

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oldest

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1

$begingroup$

From the factorisation theorem we know that $T$ is sufficient if and only if:

$$f_theta(x) = h(x) g_theta(T(x)).$$

So we then have:

$$R_{theta', theta}(x) equiv frac{f_theta'(x)}{f_theta(x)} = frac{h(x) g_theta'(T(x))}{h(x) g_theta(T(x))} = frac{g_theta'(T(x))}{g_theta(T(x))},$$

which depends on $x$ only through $T(x)$.

share|cite|improve this answer

answered Mar 13 at 9:16

BenBen

1,815215

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1

$begingroup$

From the factorisation theorem we know that $T$ is sufficient if and only if:

$$f_theta(x) = h(x) g_theta(T(x)).$$

So we then have:

$$R_{theta', theta}(x) equiv frac{f_theta'(x)}{f_theta(x)} = frac{h(x) g_theta'(T(x))}{h(x) g_theta(T(x))} = frac{g_theta'(T(x))}{g_theta(T(x))},$$

which depends on $x$ only through $T(x)$.

share|cite|improve this answer

answered Mar 13 at 9:16

BenBen

1,815215

$endgroup$

add a comment | 

1

$begingroup$

From the factorisation theorem we know that $T$ is sufficient if and only if:

$$f_theta(x) = h(x) g_theta(T(x)).$$

So we then have:

$$R_{theta', theta}(x) equiv frac{f_theta'(x)}{f_theta(x)} = frac{h(x) g_theta'(T(x))}{h(x) g_theta(T(x))} = frac{g_theta'(T(x))}{g_theta(T(x))},$$

which depends on $x$ only through $T(x)$.

share|cite|improve this answer

answered Mar 13 at 9:16

BenBen

1,815215

$endgroup$

add a comment | 

$begingroup$

From the factorisation theorem we know that $T$ is sufficient if and only if:

$$f_theta(x) = h(x) g_theta(T(x)).$$

So we then have:

$$R_{theta', theta}(x) equiv frac{f_theta'(x)}{f_theta(x)} = frac{h(x) g_theta'(T(x))}{h(x) g_theta(T(x))} = frac{g_theta'(T(x))}{g_theta(T(x))},$$

which depends on $x$ only through $T(x)$.

share|cite|improve this answer

answered Mar 13 at 9:16

BenBen

1,815215

$endgroup$

share|cite|improve this answer

answered Mar 13 at 9:16

BenBen

1,815215

answered Mar 13 at 9:16

BenBen

1,815215

answered Mar 13 at 9:16

BenBen

1,815215