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From Blackwell's theorem, it is possible to draw an equivalence between one signal being a garbling of another and the statistical notion of a Mean Preserving Spread.

Consequently, at least in the context of Blackwell's Theorem, the following is an alternative definition of a MPS:

For a pair of distributions $F$ and $F'$ induced by two signals $\sigma' : \Theta \to \Delta(S')$ and $\sigma : \Theta \to \Delta(S')$, $F$ is Mean Preserving Spread of $F'$ if there exists a Markov Kernel $\gamma : S \to \Delta(S')$ such that:

$$\sigma'(s' | \theta) = \int_{s \in S} \gamma(s' |s) \sigma(s | \theta) ds$$

I know this is true due to the theorem, but I'm trying to prove the result directly, using the following (standard) definition of a Mean Preserving Spread: For a pair of random variables from the pair of distributions $Z \sim F$ and $\tilde{Z} \sim \tilde{F}$, $F$ is a MPS of $\tilde{F}$ if: $$\mathbb{E}(Z | \tilde{Z}) = \tilde{Z}$$

I have made the following partial attempt.

First, I note that in general: $$\mathbb{E}(X|Y) = \int_A \dfrac{f_{X,Y}(x,y)}{f_Y(y)}$$

Where $A$ is the $\sigma$-algebra generated by $Y$.

In this case, I think (though am far from sure) that: $$f_{X,Y}(x,y) = \gamma(s' |s) \sigma(s|\theta)$$

$$f_Y(y) = \gamma(s|s')$$

Consequently:

$$\begin{align} \mathbb{E}(Z|\tilde{Z}) &= \int_{s' \in S'} \dfrac{\sigma'(s'|\theta)}{\gamma(s|s')} .ds'\\ &= \int_{s' \in S'} \dfrac{\sigma(s|\theta) \gamma(s|s')}{\gamma(s|s')} .ds' \\ &= \sigma(s|\theta) \end{align}$$

Would someone be able to confirm that I've specified the densities correctly here, and if I have not, suggest what they should be?

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  • $\begingroup$ The notation $\int_A$ for $A$ a $\sigma$-algebra is rarher unusual. More importantly, where do the densities come from? $\endgroup$ Commented May 21 at 6:38

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