I recently stumbled upon de Finetti`s (pretty cool) representation theorem (What is so cool about de Finetti's representation theorem?). I wondered whether the RV $\Theta$ that arises in this context follows the rules of the probability calculus. Or framed differently: I am pretty sure that it does and I am looking for a reference where I can look up this topic.

An example for what I have in mind: if we have an exchangeable sequence of throws with a die, we can create the following (exchangeble sequences):

$$X:=(X_1,X_2, ...): (\Omega,A)\to \in ({\{0,1\}^\infty}, 2^{\{0,1\}^\infty})$$, where $X_i(\omega)= 1$ if the outcome is even and $X_i(\omega)=0 $ otherwise.

Furthermore we could define

$$Y:=(Y_1, Y_2, ...): (\Omega,A)\to ({\{0,1\}^\infty}, 2^{\{0,1\}^\infty})$$ where $Y_i(\omega)=1$ if the outcome is 3, and $Y_i(\omega)=0$ otherwise.

Now we could form the new sequence:

$$Z:=(Z_1, Z_2, ...):=(Y_{\phi(1)}, Y_{\phi(2)}, ...)$$ where $\phi(n)$ indicates whether the $n$th die-throw was uneven.

Now it seems to me that it should hold that $\Theta_Z=\frac{\Theta_Y}{\Theta_X} \quad a.s.$

Now as I already said I am looking for a reference that covers the relationship between this parameter that arises in Bayesian exchangeability modelling and the rules of the probability calculus


1 Answer 1


Infinite additivity:

Let $(X_{i,k})_{i=1...\infty}$, $k=1...\infty$ be Random Variables, where $X_{i,k}=1$ if the event $E_k$ happens at the i-th toss (0 otherwise) and we assume that $E_k \cap E_j = \emptyset \quad k \neq j $. Furthermore let all $(X_{i,k})_{i=1...\infty} $and $(Z_i)_{i=1...\infty}=(\sum_{k=1}^\infty X_{i,k})_{i=1...\infty}$ be infinitely exchangeable. Therefore $Z_i$ indicates whether one of the events E happens (as they are mutually exclusive)

We know by De Finetti`s law of large numbers that for every set of infinitely exchangeable Random Variables that our beliefs about the outcomes can be represented via our beliefs about the limiting relative frequency. Where I denote the parameter for the sequence $(X_{i,k})_{i=1... \infty}$ with $\Theta_k$ and the parameter for $(Z_i)_{i=1...\infty}=(\sum_{k=1}^\infty X_{i,k})_{i=1...\infty}$ with $\Theta_Z$, What I want to show is that $$\Theta_Z= \sum_{k=1}^\infty \Theta_k$$ By De Finetti's law of large numbers we know that all exchangeable sequences converge towards their parameter with probability 1: $$P( \lim_{n \to \infty} 1/n \sum_{i=1}^n X_{i,k})=\Theta_k \quad a.s. $$ $\forall k =1...\infty$ and $$P( \lim_{n \to \infty} 1/n \sum_{i=1}^n \sum_{k=1}^\infty X_{i,k})=\Theta_k \quad a.s.$$. As the events are disjoint we know that the Random variable $Z_i=(\sum_{k=1}^\infty X_{i,k})$ indicates that one of these events $E_K$ happens at the $i-th$ trial. Therefore we have that $$\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n \sum_{k=1}^K X_{i,k}=\lim_{n \to \infty} \sum_{k=1}^K \frac{1}{n} \sum_{i=1}^n X_{i,k} = \sum_{k=1}^K \lim_{n \to \infty}\frac{1}{n} \sum_{i=1}^n X_{i,k} = \sum_{k=1}^K\Theta_k \quad \forall K \in \mathbb{N}$$ and therefore

letting $K \to \infty $ we obtain that $\Theta_Z = \sum_{k=1}^\infty \Theta_k \quad a.s.$

Conditional Probability:

Assumptions: $(X_i)_{i \in \mathbb{N}}, (Y_i)_{i \in \mathbb{N}}, (X_iY_i)_{i=1...\infty}$ are infinitely exchangeable and $P(\sum_{i=1}^\infty Y_i<\infty)=0$. By De Finetti`s representation theorem we can express our beliefs about the oucomes via our beliefs about the limiting relative frequency, I denote with $\Theta_{X}, \Theta_{Y} ,\Theta_{XY}$ the respective Parameters.

Now we define $$Z:(\Omega, A)\to (\mathscr{M}, 2^{\mathscr{M}}), \quad Z(\omega)=(X_{\phi_2(\omega)}(\omega), X_{\phi_2(\omega)}(\omega), ...) $$

where $\mathscr{M}=\{\{0,1\}^n|n=1...\infty\}$ and $\phi_k(\omega) =n$ if $Y_n(\omega)=1, \sum_{i=n}^NY_n(\omega)=k$ .Note that $Z(\omega)$ can also be finitely dimensional if one obtains only a finite number of successes. We can apply De Finetti's representation theorem again and obtain:

the parameter $\Theta_Z$, such that $$P((Z_1, ..., Z_n)=(z_1, ..., z_n))= \int_\Omega \Pi_{i=1}^n \theta^{z_i}(1-\theta)^{1-z_i}dP(\Theta_Z)$$

What we want show is that: $\Theta_Z= \frac{\Theta_{XY}}{\Theta_{Y}}$.

This follows again from De Finetti's law of large numbers, as

$$\Theta_Z= \lim_{n \to \infty}\frac{\sum_{i=1}^nX_i(\omega)Y_i(\omega)}{\sum_{i=1}^nY_i({\omega)}} =\frac{\lim_{n \to \infty}1/n\sum_{i=1}^nX_i(\omega)Y_i(\omega)}{\lim_{n \to \infty}1/n\sum_{i=1}^nY_i({\omega)}}= \frac{\Theta_{Y X}}{\Theta_{Y}} \quad a.s.$$

Note that $\Theta_Y\neq0 \quad a.s.$ as I assumed that $P(\sum_{i=1}^\infty Y_i<\infty)=0$.


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