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Some time ago I wrote a question about what I think/thought (up to my understanding) is an ambiguity of the common definition of sufficient statistics :

Conditioning in the definition of sufficient statistics

I was wondering if there is a Bayesian definition of sufficiency. In such a setting a parameter $\theta$ can be a random variable. And we could define a statistics $T$ as sufficient if:

$$p(\theta|x_1,..,x_n,t)=p(\theta|t) \ [1]$$

(in other words, $\theta$ is conditional independent on the sample, given $t$).

This would remove some ambiguities of the standard definition. But could such a definition be equivalent to the standard one in some sense ? Since in a Bayesian setting we need to specify a prior over $\theta$ to define it as a random variable, I wonder if the validity of condition [1] may depend on the value of the prior on $\theta$.

Is there some approach to define sufficient statistics in a Bayesian setting ? Are they trivially equivalent ? Or sufficient statistics is a concept belonging only to non-Bayesian statistics ? To me the concept of conditional independence looks closely related, at an intuitive level...

Thomas
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  • Some related posts: https://stats.stackexchange.com/questions/369445/is-there-a-difference-between-bayesian-and-classical-sufficiency, https://stats.stackexchange.com/questions/339075/how-does-bayesian-sufficiency-relate-to-frequentist-sufficiency, – kjetil b halvorsen Jan 02 '22 at 04:38
  • Thank you! I will read it carefully and if I find the answer to my doubts will delete this question. Happy new year ! – Thomas Jan 02 '22 at 09:10
  • I am not sure about Taylor's answer. He says that the frequentist definition of sufficiency is $p(x|\theta,t)=p(x|t)$ and derives an equivalence between the Bayesian definition and the frequentist one. But in a Bayesian setting the right member is, in a more explicit form : $p(x|t)=\int d\theta p(\theta|t)p(x|\theta,t)$, i.e. is a mean over different values of $\theta$. Isn't this at variance with the frequentist definition, where the right member would be evaluated with a fixed value of $\theta$ ? – Thomas Jan 02 '22 at 10:10

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