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if $f(x) = a exp(-a (x-b))$ . Find sufficient statistic for b

Is there a sufficient statistic for shifted exponential distribution?Sufficient Statistic for a ParameterWhat is the sufficient statistic for this function?Sufficient statistic for Bernoulli r.v.Sufficient statistic for Uniform distribution.Sufficient statistic of one parameterOrder statisticminimal sufficient statistic of Cauchy distributionSufficient statistic for a Uniform distribution uniform($itheta$)Sufficient statistic for class of distributions

0

$begingroup$

Intuitively I feel that the answer should be $min(X_1,X_2,ldots,X_n)$ where $X_i$'s are iid, but I don't know how to prove it.

statistics order-statistics

share|cite|improve this question

edited Mar 21 at 18:51

Daniele Tampieri

2,65221022

asked Mar 21 at 17:56

Sambhav KhuranaSambhav Khurana

254

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried?
  $endgroup$
  – clathratus
  Mar 21 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tried to do it by Neyman Factorisation theorum. I found joint distribution to be a^n exp(-a(x- nb)). But I don't know how to proceed.
  $endgroup$
  – Sambhav Khurana
  Mar 21 at 18:25
  

  
  
  
  

add a comment | 

0

$begingroup$

Intuitively I feel that the answer should be $min(X_1,X_2,ldots,X_n)$ where $X_i$'s are iid, but I don't know how to prove it.

statistics order-statistics

share|cite|improve this question

edited Mar 21 at 18:51

Daniele Tampieri

2,65221022

asked Mar 21 at 17:56

Sambhav KhuranaSambhav Khurana

254

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried?
  $endgroup$
  – clathratus
  Mar 21 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tried to do it by Neyman Factorisation theorum. I found joint distribution to be a^n exp(-a(x- nb)). But I don't know how to proceed.
  $endgroup$
  – Sambhav Khurana
  Mar 21 at 18:25
  

  
  
  
  

add a comment | 

$begingroup$

Intuitively I feel that the answer should be $min(X_1,X_2,ldots,X_n)$ where $X_i$'s are iid, but I don't know how to prove it.

statistics order-statistics

share|cite|improve this question

edited Mar 21 at 18:51

Daniele Tampieri

2,65221022

asked Mar 21 at 17:56

Sambhav KhuranaSambhav Khurana

254

$endgroup$

share|cite|improve this question

edited Mar 21 at 18:51

Daniele Tampieri

2,65221022

asked Mar 21 at 17:56

Sambhav KhuranaSambhav Khurana

254

share|cite|improve this question

edited Mar 21 at 18:51

Daniele Tampieri

2,65221022

asked Mar 21 at 17:56

Sambhav KhuranaSambhav Khurana

254

edited Mar 21 at 18:51

Daniele Tampieri

2,65221022

edited Mar 21 at 18:51

Daniele Tampieri

2,65221022

asked Mar 21 at 17:56

Sambhav KhuranaSambhav Khurana

254

asked Mar 21 at 17:56

Sambhav KhuranaSambhav Khurana

254

1 Answer
1

active

oldest

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1

$begingroup$

The domain for this problem is very important. You haven't specified it but I'm assuming the problem is: find the sufficient statistic for
$$f(x) = a, exp[-a(x-b)] , textbf1x > b.$$

Then the joint pdf is $$f(x) = a^n expleft[-a sum_i=1^n x_iright] ,exp[a,b,n] , prod_i=1^ntextbf1x_i>b.$$ Notice all the $x_i$ will be greater than $b$ as long as the minimum is greater than $b$, therefore the likelihood function is $$L(b) =a^n expleft[-a sum_i=1^n X_iright] ,exp[a,b,n] , textbf1X_1,n>b,$$ where $X_1,n = min(X_1,...,X_n)$. Then $X_1,n$ is a sufficient statistic. It is also the MLE.

share|cite|improve this answer

answered Mar 21 at 23:22

astudentastudent

587

$endgroup$

add a comment | 

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1

$begingroup$

The domain for this problem is very important. You haven't specified it but I'm assuming the problem is: find the sufficient statistic for
$$f(x) = a, exp[-a(x-b)] , textbf1x > b.$$

Then the joint pdf is $$f(x) = a^n expleft[-a sum_i=1^n x_iright] ,exp[a,b,n] , prod_i=1^ntextbf1x_i>b.$$ Notice all the $x_i$ will be greater than $b$ as long as the minimum is greater than $b$, therefore the likelihood function is $$L(b) =a^n expleft[-a sum_i=1^n X_iright] ,exp[a,b,n] , textbf1X_1,n>b,$$ where $X_1,n = min(X_1,...,X_n)$. Then $X_1,n$ is a sufficient statistic. It is also the MLE.

share|cite|improve this answer

answered Mar 21 at 23:22

astudentastudent

587

$endgroup$

add a comment | 

1

$begingroup$

The domain for this problem is very important. You haven't specified it but I'm assuming the problem is: find the sufficient statistic for
$$f(x) = a, exp[-a(x-b)] , textbf1x > b.$$

Then the joint pdf is $$f(x) = a^n expleft[-a sum_i=1^n x_iright] ,exp[a,b,n] , prod_i=1^ntextbf1x_i>b.$$ Notice all the $x_i$ will be greater than $b$ as long as the minimum is greater than $b$, therefore the likelihood function is $$L(b) =a^n expleft[-a sum_i=1^n X_iright] ,exp[a,b,n] , textbf1X_1,n>b,$$ where $X_1,n = min(X_1,...,X_n)$. Then $X_1,n$ is a sufficient statistic. It is also the MLE.

share|cite|improve this answer

answered Mar 21 at 23:22

astudentastudent

587

$endgroup$

add a comment | 

$begingroup$

The domain for this problem is very important. You haven't specified it but I'm assuming the problem is: find the sufficient statistic for
$$f(x) = a, exp[-a(x-b)] , textbf1x > b.$$

Then the joint pdf is $$f(x) = a^n expleft[-a sum_i=1^n x_iright] ,exp[a,b,n] , prod_i=1^ntextbf1x_i>b.$$ Notice all the $x_i$ will be greater than $b$ as long as the minimum is greater than $b$, therefore the likelihood function is $$L(b) =a^n expleft[-a sum_i=1^n X_iright] ,exp[a,b,n] , textbf1X_1,n>b,$$ where $X_1,n = min(X_1,...,X_n)$. Then $X_1,n$ is a sufficient statistic. It is also the MLE.

share|cite|improve this answer

answered Mar 21 at 23:22

astudentastudent

587

$endgroup$

share|cite|improve this answer

answered Mar 21 at 23:22

astudentastudent

587

answered Mar 21 at 23:22

astudentastudent

587

answered Mar 21 at 23:22

astudentastudent

587