Sufficient Statistic: T, S Pairing Explained

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A sufficient statistic T(x1...xn) retains all information about a parameter 'theta', and when paired with another statistic S(x1...xn), the sufficiency of the pair (T,S) depends on how they are combined. If the pairing alters the information contained in T(x), the resulting statistic may not remain sufficient. For instance, combining T(x) with S(x) through addition can lead to a loss of sufficiency, as shown in the example with the normal distribution. However, if T is a scalar sufficient statistic, then the vector (T,S) can still be sufficient for the same parameter. The key takeaway is that the sufficiency of the pair relies on the nature of the combination and the information retained.
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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.
 
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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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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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EnumaElish said:
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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The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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