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In Zen`s answer at What is so cool about de Finetti's representation theorem? that is concerned with De Finetti's 0 -1 representation theorem, he says that "De Finetti's law of large numbers" states that it holds that for an exchangeable sequence of RVs $\{X_i\}_{i \in \mathbb{N}}$

$$\lim_{n \to \infty }\overline{X}_n=\Theta \quad \quad \text{almost surely}$$

where $\Theta$ is the Random Variable that arises out of De Finetti`s representation theorem. However it is not clear to me for which measure this equivalence holds almost surely. Is the relevant measure the measure $\mu$ such that $\mu[\Theta]$ is the prior probability measure?

Sebastian
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The analysis ought to be taking place in a general probability space $(\Omega, \mathscr{U}, \mathbb{P})$ with a sample space that encompasses both the observable exchangeable sequence $\mathbf{X}:\Omega \rightarrow \mathbb{R}^\infty$ and the parameter $\Theta:\Omega \rightarrow \mathbb{R}$. The statement of interest is then taken with respect to the probability measure $\mathbb{P}$, so it can be stated more formally by defining an event positing equivalence, and then saying that this event occurs with probability one under the measure $\mathbb{P}$. Formally, this is:

$$\mathbb{P} ( \mathcal{E}) = 1 \quad \quad \text{where} \quad \quad\mathcal{E} \equiv \left\{ \omega \in \Omega \Bigg| \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^n X_i(\omega) = \Theta(\omega) \right\} \in \mathscr{U}.$$

O'Neill (2009) argues that the best way to form the parameter $\Theta$ in this analysis is to define it as the Banach limit of the sequence $\mathbf{X}$, which means that it exists for all possible real sequences (i.e., it is well-defined) and it corresponds to the Cesàro limit of the sequence when that exists. That way, the parameter is equal to the Banach limit of the sequence for all $\omega \in \Omega$ (which is stronger than almost sure equivalence), and $\mathcal{E}$ is the event that the Cesàro limit of the observable sequence exists, which occurs with probability one under the condition of exchangeability.

The prior measure $\mu_\Theta$ for the parameter $\Theta$ is formed from the overall measure $\mathbb{P}$ by restricting attention to events relating to the parameter only. If we take $\mathscr{U}$ to be the class of all Borel sets in $\mathbb{R}^\infty$ (in order to describe events pertaining to the whole observable sequence) then the domain of the measure $\mu_\Theta$ would be the subspace of Borel sets in $\mathbb{R}$ (to describe the parameter). The event $\mathcal{E}$ would not be in this domain, and so it would not make sense to try to apply the prior measure $\mu_\Theta$ to describe a probability for this event.

Ben
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  • Thanks again for this clarifying answer. If we define with $X_k: \Omega \to \mathbb{R}^k$ the outcome of the first $k$ experiments, then $\mathbb{P}(X_k)$ (the push-forward measure) is the agent`s believe about the first $k$ outcomes. and $\mathbb{P}(\Theta)= \mu_\Theta$, right? – Sebastian Sep 18 '18 at 07:14
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    Not really. Remember that random variables are mappings from the domain $\Omega$ to some codomain. The probability measure $\mathbb{P}$ is a set-function whose domain is the class $\mathscr{U}$, which is a sigma-field of subsets on $\Omega$. The probability measure operates on *events* in the sample space (which are sets in $\mathscr{U}$), not random variables (which are mappings from the sample space to a codomain). The probability measures for random variables are formed from the overarching measure $\mathbb{P}$ by looking at the sigma-field of events for that random variable. – Ben Sep 18 '18 at 08:27
  • ah i meant $X_k(\mathbb{P})$ and $\Theta(\mathbb{P})$ of course – Sebastian Sep 18 '18 at 08:29
  • That notation also does not make sense to me, since $\mathbb{P}$ is not an element of $\Omega$, so it is not clear how it could be an argument for a random variable. – Ben Sep 18 '18 at 08:30
  • (To understand this well, it is necessary to learn a bit of "measure theory". If you have not already done so, have a read up on probability measures and sigma-fields and have a look at how probability spaces are formed on sub-sigma-fields within a larger probability space.) – Ben Sep 18 '18 at 08:30
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    With $\mathbb{Q}:=X_k(\mathbb{P})$ I refer to the measure that is defined via: $\mathbb{Q}(A)= \mathbb{P}(X_k^{-1}(A))$ – Sebastian Sep 18 '18 at 08:33
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    I was similarily confused when my professor used this notation :D – Sebastian Sep 18 '18 at 08:34
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/83312/discussion-between-0rangetree-and-ben). – Sebastian Sep 18 '18 at 08:36