Sufficient statistic

  • Thread starter EvLer
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  • #1
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If T(x1...xn) is a sufficient statistic and is paired up with another statistic S(x1...xn) then the (T,S) is still a sufficient statistic i think, correct?
But how would I "formally" explain that?
thanks.
 

Answers and Replies

  • #2
ssd
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I am not sure... just giving my thoughts..

Let, T(x) be sufficient for some parameter 'theta' say. Then T(x) contains all information about 'theta'. Now the question is how are you 'pairing up' S(x) with T(x) .... by addition or multiplication or through some other relation? If the process of 'pairing' alters the information in T(x) then the combined statistic may not be sufficient.
As for example, consider the normal distribution N(theta,1). Let x1,x2,...,xn be a random sample from it drawn independently of each other. Then T(x)= (x1+x2....+xn)/n is sufficient (minimal) for theta. Consider
S(x)= (x1-x2-x3...-xn)/n. Then if you pair up by addition, ie, form a new statistic T(x)+S(x), it is not sufficient for theta.
 
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  • #3
EnumaElish
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I think what EvLer means is "if a scalar statistic T is sufficient then a vector (pair) of statistics (T,S) is also sufficient." The heuristic answer is "because even if one were to ignore S, they would still have sufficiency by virtue of having T."
 
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  • #4
ssd
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I think what EvLer means is "if a scalar statistic T is sufficient then a vector (pair) of statistics (T,S) is also sufficient." The heuristic answer is "because even if one were to ignore S, they would still have sufficiency by virtue of having T."
If T is scalar then it can be sufficient for a parameter 'theta' which is also scalar valued. Then you mean that, a vector valued statistic (T,S) can be sufficient for a scalar valued parameter 'theta'?.... The idea is unknown to me.
 
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