"Bayes Sufficiency" is defined in two ways. Are they equivalent?


A statistical experiment $S$ is a triplet $\left(\left(\Theta,\mathcal{F}\right),\left(\Omega,\mathcal{A}\right),P\right)$, where $\left(\Theta,\mathcal{F}\right)$, $\left(\Omega,\mathcal{A}\right)$ are measurable spaces and $P:\Theta\times\mathcal{A}\rightarrow\left[0,1\right]$ is a stochastic kernel. $\left(\Theta,\mathcal{F}\right)$ is called the parameter space, $\left(\Omega,\mathcal{A}\right)$ - the sample space and $P$ - the family of sampling distributions (indexed by $\theta\in\Theta$).

Every probability measure $Q$ on $\left(\Theta,\mathcal{F}\right)$ determines the derived Bayesian experiment $S_Q:=\left(\Theta\times\Omega,\mathcal{F}\otimes\mathcal{A},\Pi\right)$, where $\Pi$ is the unique probability measure on $\left(\Theta\times\Omega,\mathcal{F}\otimes\mathcal{A}\right)$, guaranteed by the product measure theorem (cf. [ASH] 2.6.2), such that $\Pi\left(F\times A\right)=\int_F P\left(\theta, A\right)\space Q\left(d\theta\right)$ for all $F\in\mathcal{F}$, $A\in\mathcal{A}$. $Q$ (or, equivalently, the restriction of $\Pi$ to the parameter space) is a prior distribution, the restriction of $\Pi$ to the sample space is the predictive distribution, and any regular version of the conditional distribution $\Pi\left(\pi_\Theta\in\cdot\mid\pi_\Omega=\omega\right)$, if such exists, is called the family of posterior distributions (indexed by $\omega\in\Omega$); here $\pi_\Theta$ is the natural projection from $\Theta\times\Omega$ to $\Theta$ and $\pi_\Omega$ - the natural projection to $\Omega$. Note that the family of sampling distributions is a regular version of $\Pi\left(\pi_\Omega\in\cdot\mid\pi_\Theta=\theta\right)$.

A $\left(\Gamma,\mathcal{G}\right)$-statistic, where $\left(\Gamma,\mathcal{G}\right)$ is a measurable space, is a measurable function $T:\left(\Omega,\mathcal{A}\right)\rightarrow\left(\Gamma,\mathcal{G}\right)$. Set $\mathcal{T}:=\sigma(T)$.

Bayesian Sufficiency

  1. According to Schervish ([SCH], Lemma 2.6), $\mathcal{T}$ is Bayes sufficient iff for every prior $Q$ there exists a (not necessarily regular) version $\mu:\Omega\times\mathcal{F}\rightarrow\left[0,1\right]$ of the posterior, such that for all $F\in\mathcal{F}$, $\mu\left(\cdot,F\right)$ is $\mathcal{T}/\mathfrak{B}$-measurable ($\mathfrak{B}$ being the standard Borel field on the real line).
  2. According to Nogales, Oyola & Perez ([NOG] at the end of page 76), $\mathcal{T}$ is Bayes sufficient for a given prior $Q$ iff $\mathcal{F}\perp_\mathcal{T}\mathcal{A}$ in $S_Q$ ("$\cdot\perp_\square\cdot$" being the symbol for conditional stochastic independence). Here $\mathcal{F}$, $\mathcal{A}$ and $\mathcal{T}$ are considered as sub-$\sigma$-fields of $\mathcal{F}\otimes\mathcal{A}$ through their identification with $\left\{F\times\Omega:F\in\mathcal{F}\right\}$, $\left\{\Theta\times A:A\in\mathcal{A}\right\}$ and $\left\{\Theta\times A:A\in\mathcal{T}\right\}$, respectively.


  1. [ASH] Ash, Robert B., Doléans-Dade, Catherine A. "Probability and Measure Theory". 2nd Edition, 2000. Academic Press.
  2. [NOG] Nogales, Agustín G., Oyola, José A., Pérez, Paloma. "On Conditional Independence and the Relationship between Sufficiency and Invariance under the Bayesian Point of View". Statistics & Probability Letters 46 (2000) 75-84.
  3. [SCH] Schervish, Mark J. "Theory of Statistics". 1st printing, 1995. Springer-Verlag.

1 Answer 1


No, these definitions are not equivalent. 2 is implied by 1, but likely not vice versa. That 2 is implied by 1 is an immediate consequence of [FUR] Theorem 3 that states that $\mathcal{T}$ is Bayes sufficient for $Q$ (per definition 2) iff definition 1 applies w.r.t. this particular prior (rather than "for every prior $Q$").


[FUR] Furmanczyk, K., Niemiro, W. "Sufficiency in Bayesian Models". Applicationes Mathematicae. 25, 1 (1998), pp. 113-120.


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