Mixing
Sufficient statistic for a Uniform distribution uniform($itheta$)
$begingroup$
Let $X_1, X_2, ..., X_n$ be $n$ independent random variables, where $X_i$~uniform($itheta$), $i=1,2,...,n$. Find a sufficient statistics for $theta$.
My Attempt
The conditional distribution :
$$ f( x_1 ,ldots ,x_n mid itheta ) = frac 1 {(itheta)^n}, text{ where } ( x_i leq itheta , i=1,2,ldots,n).$$
Rewriting the conditional distribution as:
$$ f( x_1 ,ldots ,x_n mid itheta ) = {(itheta)^{-n}} quad ( x_i leq itheta , i=1,2,ldots,n) tag 1$$
So,
$$ f( x_1 ,ldots ,x_n mid theta ) = {(itheta)^{-n}} quad ( max(x_i) leq itheta)tag 1$$
Thus, the sufficient statistic $max(x_i/i)%$ is taken.
My question is, how to I take into account uniform($itheta$) for the discrete case as the question implies?
statistics statistical-inference
asked Jan 23 at 18:59
LadyLady
1198
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add a comment |
$begingroup$
Let $X_1, X_2, ..., X_n$ be $n$ independent random variables, where $X_i$~uniform($itheta$), $i=1,2,...,n$. Find a sufficient statistics for $theta$.
My Attempt
The conditional distribution :
$$ f( x_1 ,ldots ,x_n mid itheta ) = frac 1 {(itheta)^n}, text{ where } ( x_i leq itheta , i=1,2,ldots,n).$$
Rewriting the conditional distribution as:
$$ f( x_1 ,ldots ,x_n mid itheta ) = {(itheta)^{-n}} quad ( x_i leq itheta , i=1,2,ldots,n) tag 1$$
So,
$$ f( x_1 ,ldots ,x_n mid theta ) = {(itheta)^{-n}} quad ( max(x_i) leq itheta)tag 1$$
Thus, the sufficient statistic $max(x_i/i)%$ is taken.
My question is, how to I take into account uniform($itheta$) for the discrete case as the question implies?
statistics statistical-inference
asked Jan 23 at 18:59
LadyLady
1198
$endgroup$
$begingroup$
You mean, $max(x_i/i)$?
$endgroup$
– Did
Jan 23 at 19:12
$begingroup$
@Did, yes! I made the correction. But is the steps right?
$endgroup$
– Lady
Jan 23 at 19:17
add a comment |
$begingroup$
Let $X_1, X_2, ..., X_n$ be $n$ independent random variables, where $X_i$~uniform($itheta$), $i=1,2,...,n$. Find a sufficient statistics for $theta$.
My Attempt
The conditional distribution :
$$ f( x_1 ,ldots ,x_n mid itheta ) = frac 1 {(itheta)^n}, text{ where } ( x_i leq itheta , i=1,2,ldots,n).$$
Rewriting the conditional distribution as:
$$ f( x_1 ,ldots ,x_n mid itheta ) = {(itheta)^{-n}} quad ( x_i leq itheta , i=1,2,ldots,n) tag 1$$
So,
$$ f( x_1 ,ldots ,x_n mid theta ) = {(itheta)^{-n}} quad ( max(x_i) leq itheta)tag 1$$
Thus, the sufficient statistic $max(x_i/i)%$ is taken.
My question is, how to I take into account uniform($itheta$) for the discrete case as the question implies?
statistics statistical-inference
asked Jan 23 at 18:59
LadyLady
1198
$endgroup$
Let $X_1, X_2, ..., X_n$ be $n$ independent random variables, where $X_i$~uniform($itheta$), $i=1,2,...,n$. Find a sufficient statistics for $theta$.
My Attempt
The conditional distribution :
$$ f( x_1 ,ldots ,x_n mid itheta ) = frac 1 {(itheta)^n}, text{ where } ( x_i leq itheta , i=1,2,ldots,n).$$
Rewriting the conditional distribution as:
$$ f( x_1 ,ldots ,x_n mid itheta ) = {(itheta)^{-n}} quad ( x_i leq itheta , i=1,2,ldots,n) tag 1$$
So,
$$ f( x_1 ,ldots ,x_n mid theta ) = {(itheta)^{-n}} quad ( max(x_i) leq itheta)tag 1$$
Thus, the sufficient statistic $max(x_i/i)%$ is taken.
My question is, how to I take into account uniform($itheta$) for the discrete case as the question implies?
statistics statistical-inference
statistics statistical-inference
asked Jan 23 at 18:59
LadyLady
1198
asked Jan 23 at 18:59
LadyLady
1198
edited Jan 23 at 19:16
Lady
asked Jan 23 at 18:59
LadyLady
1198
asked Jan 23 at 18:59
LadyLady
1198
asked Jan 23 at 18:59
LadyLady
1198
1198
$begingroup$
You mean, $max(x_i/i)$?
$endgroup$
– Did
Jan 23 at 19:12
$begingroup$
@Did, yes! I made the correction. But is the steps right?
$endgroup$
– Lady
Jan 23 at 19:17
add a comment |
$begingroup$
You mean, $max(x_i/i)$?
$endgroup$
– Did
Jan 23 at 19:12
$begingroup$
@Did, yes! I made the correction. But is the steps right?
$endgroup$
– Lady
Jan 23 at 19:17
$begingroup$
You mean, $max(x_i/i)$?
$endgroup$
– Did
Jan 23 at 19:12
$begingroup$
You mean, $max(x_i/i)$?
$endgroup$
– Did
Jan 23 at 19:12
$begingroup$
@Did, yes! I made the correction. But is the steps right?
$endgroup$
– Lady
Jan 23 at 19:17
$begingroup$
@Did, yes! I made the correction. But is the steps right?
$endgroup$
– Lady
Jan 23 at 19:17
add a comment |
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$begingroup$
You mean, $max(x_i/i)$?
$endgroup$
– Did
Jan 23 at 19:12
$begingroup$
@Did, yes! I made the correction. But is the steps right?
$endgroup$
– Lady
Jan 23 at 19:17