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(1) For every ¯ µ ∈ Δ( Q ) , ¯ µ ( { q ∈ Q | t (¯ µ, q ) = PASS } ) = 1 . 

(2) For every ¯ ζ ∈ Δ (Δ( Q )) , one has 

for co-meager many ¯ µ ∈ Δ( Q ) . 

<img src='content_image/321662.jpg'>

This proposition has the following nice interpretation: Assume that an expert claims to know the distribution from which an element q ∈ Q is drawn. Say that the expert passes the test if t (¯ µ, q ) = PASS, where ¯ µ is the distribution proclaimed by the expert and q is the actual realization. The first property says that if the expert provides the correct distribution of q then he passes with probability 1. The second property ensures that, if the expert randomizes his theory according to some distribution ¯ ζ , then he will fail the test with probability 1 for a large (co-meager) set of truths. 

Proposition 3 was proved by Olszewski and Sandroni (2009b). Their theorem is formulated for the underlying space of realization Q = { 0 , 1 } $^{N}$but their theorem and proof carry through for every compact metric space. We need the case Q = [0 , 1]. 

4.2. Proof. We first prove that when forecasts are made using an exchangeable theory, then the forecasts over a single realization already determines the entire theory. 

Lemma 1. Let x = ( s$_{0}$, s$_{1}$, . . . ) ∈ { 0 , 1 } N be a realization and let µ ′ = ε$_{λ}$ ′ and µ ′′ = ε$_{λ}$ ′′ be two exchangeable theories. If µ $^{′}$(1 | s$_{0}$, . . . , s$_{n}$$_{−}$$_{1}$ ) = µ $^{′′}$(1 | s$_{0}$, . . . , s$_{n}$$_{−}$$_{1}$ ) for every n ∈ N then λ ′ = λ ′′ (and, in particular, µ ′ = µ $^{′′}$). 

Proof. Under the assumption of the lemma, it follows by induction from (1) that 

<img src='content_image/321659.jpg'>

for every n ∈ N . From (6) and (4) it follows that 

<img src='content_image/321660.jpg'>

