Proving that $max_{i in {1, ldots, n }}(X_i)$ is a minimal statistics for uniform distribution
$begingroup$
Let $X_1, ldots, X_n$ be independent random variables with uniform distribution on $[0, theta] text{ }, theta in [0, infty)$.
I am to prove that $max_{i in {1, ldots, n }}(X_i)$ is a minimal sufficient statistics.
It's easy to show that it is a sufficient statistics (using factorization theorem).
I have problems with showing that it is moreover a minimal statistics.
How can I show that if the ratio
$$frac{f(x|theta)}{f(y|theta)}$$
does not depend on $theta$ then $max_{i in {1, ldots, n }}(X_i) = max_{i in {1, ldots, n }}(Y_i)$?
statistics statistical-inference
edited Dec 5 '18 at 14:58
mathcounterexamples.net
26k21955
asked Dec 5 '18 at 14:57
HendrraHendrra
1,163516
$endgroup$
add a comment |
$begingroup$
Let $X_1, ldots, X_n$ be independent random variables with uniform distribution on $[0, theta] text{ }, theta in [0, infty)$.
I am to prove that $max_{i in {1, ldots, n }}(X_i)$ is a minimal sufficient statistics.
It's easy to show that it is a sufficient statistics (using factorization theorem).
I have problems with showing that it is moreover a minimal statistics.
How can I show that if the ratio
$$frac{f(x|theta)}{f(y|theta)}$$
does not depend on $theta$ then $max_{i in {1, ldots, n }}(X_i) = max_{i in {1, ldots, n }}(Y_i)$?
statistics statistical-inference
edited Dec 5 '18 at 14:58
mathcounterexamples.net
26k21955
asked Dec 5 '18 at 14:57
HendrraHendrra
1,163516
$endgroup$
1
$begingroup$
You might consider the contrapositive that if the two maxima are different then the ratio does depend on $theta$ in the sense that it will be different in the case $theta$ between the two maxima compared with the case where $theta$ is greater than both maxima
$endgroup$
– Henry
Dec 5 '18 at 15:39
$begingroup$
@Henry You're right! That might work!
$endgroup$
– Hendrra
Dec 5 '18 at 15:43
add a comment |
$begingroup$
Let $X_1, ldots, X_n$ be independent random variables with uniform distribution on $[0, theta] text{ }, theta in [0, infty)$.
I am to prove that $max_{i in {1, ldots, n }}(X_i)$ is a minimal sufficient statistics.
It's easy to show that it is a sufficient statistics (using factorization theorem).
I have problems with showing that it is moreover a minimal statistics.
How can I show that if the ratio
$$frac{f(x|theta)}{f(y|theta)}$$
does not depend on $theta$ then $max_{i in {1, ldots, n }}(X_i) = max_{i in {1, ldots, n }}(Y_i)$?
statistics statistical-inference
edited Dec 5 '18 at 14:58
mathcounterexamples.net
26k21955
asked Dec 5 '18 at 14:57
HendrraHendrra
1,163516
$endgroup$
Let $X_1, ldots, X_n$ be independent random variables with uniform distribution on $[0, theta] text{ }, theta in [0, infty)$.
I am to prove that $max_{i in {1, ldots, n }}(X_i)$ is a minimal sufficient statistics.
It's easy to show that it is a sufficient statistics (using factorization theorem).
I have problems with showing that it is moreover a minimal statistics.
How can I show that if the ratio
$$frac{f(x|theta)}{f(y|theta)}$$
does not depend on $theta$ then $max_{i in {1, ldots, n }}(X_i) = max_{i in {1, ldots, n }}(Y_i)$?
statistics statistical-inference
statistics statistical-inference
edited Dec 5 '18 at 14:58
mathcounterexamples.net
26k21955
asked Dec 5 '18 at 14:57
HendrraHendrra
1,163516
edited Dec 5 '18 at 14:58
mathcounterexamples.net
26k21955
asked Dec 5 '18 at 14:57
HendrraHendrra
1,163516
edited Dec 5 '18 at 14:58
mathcounterexamples.net
26k21955
edited Dec 5 '18 at 14:58
mathcounterexamples.net
26k21955
edited Dec 5 '18 at 14:58
mathcounterexamples.net
26k21955
26k21955
asked Dec 5 '18 at 14:57
HendrraHendrra
1,163516
asked Dec 5 '18 at 14:57
HendrraHendrra
1,163516
asked Dec 5 '18 at 14:57
HendrraHendrra
1,163516
1,163516
1
$begingroup$
You might consider the contrapositive that if the two maxima are different then the ratio does depend on $theta$ in the sense that it will be different in the case $theta$ between the two maxima compared with the case where $theta$ is greater than both maxima
$endgroup$
– Henry
Dec 5 '18 at 15:39
$begingroup$
@Henry You're right! That might work!
$endgroup$
– Hendrra
Dec 5 '18 at 15:43
add a comment |
1
$begingroup$
You might consider the contrapositive that if the two maxima are different then the ratio does depend on $theta$ in the sense that it will be different in the case $theta$ between the two maxima compared with the case where $theta$ is greater than both maxima
$endgroup$
– Henry
Dec 5 '18 at 15:39
$begingroup$
@Henry You're right! That might work!
$endgroup$
– Hendrra
Dec 5 '18 at 15:43
1
1
$begingroup$
You might consider the contrapositive that if the two maxima are different then the ratio does depend on $theta$ in the sense that it will be different in the case $theta$ between the two maxima compared with the case where $theta$ is greater than both maxima
$endgroup$
– Henry
Dec 5 '18 at 15:39
$begingroup$
You might consider the contrapositive that if the two maxima are different then the ratio does depend on $theta$ in the sense that it will be different in the case $theta$ between the two maxima compared with the case where $theta$ is greater than both maxima
$endgroup$
– Henry
Dec 5 '18 at 15:39
$begingroup$
@Henry You're right! That might work!
$endgroup$
– Hendrra
Dec 5 '18 at 15:43
$begingroup$
@Henry You're right! That might work!
$endgroup$
– Hendrra
Dec 5 '18 at 15:43
add a comment |
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$begingroup$
You might consider the contrapositive that if the two maxima are different then the ratio does depend on $theta$ in the sense that it will be different in the case $theta$ between the two maxima compared with the case where $theta$ is greater than both maxima
$endgroup$
– Henry
Dec 5 '18 at 15:39
$begingroup$
@Henry You're right! That might work!
$endgroup$
– Hendrra
Dec 5 '18 at 15:43