# How the De Finetti's Representation Theorem works in this case?

For the special case of infinite sequence of $$\{0,1\}$$ valued random variables the theorem is stated as $$Pr(x_1, \ldots, x_n) = \int_0^1 p^{(\sum_{i=1}^n x_i)}[1-p]^{(n - \sum_{i=1}^n x_i)} dQ(p)\,.$$ Where $$Q(p)$$ is a measure on probability for success $$p \in [0,1]$$ (that is $$Q(\cdot)$$ is a measure on space $$[0,1]$$)

So that any infinite exchangeable sequence of $$\{0,1\}$$ valued random variables can be though of as a realization of a two step process

1. choose the value $$p$$ of the success probabilities according to $$Q$$.

2. condition on $$p$$ the $$X_i$$ are i.i.d. Bernoulli$$(p)$$.

Consider now a sequence such that $$X_1 = X_2 = \cdots = X_n = 1$$ with probability $$\frac{1}{2}$$ and $$X_1 = X_2 = \cdots = X_n = 0$$ with probability $$\frac{1}{2}$$. This is an exchangeable sequence of $$\{0,1\}$$ valued variables.

The "two step process" for this sequence works as the following: with probability $$\frac{1}{2}$$ pick $$p=1$$otherwise pick $$p = 0$$. This probability measure can be described as $$Q(p) = \frac{1}{2}\delta_{1}(p) + \frac{1}{2}\delta_{0}(p)\,.$$ If I plug this in the theorem I get

$$\int_0^1 p^{(\sum_{i=1}^n x_i)}^{(n - \sum_{i=1}^n x_i)} dQ(p) = \frac{1}{2} \left( 0^{(\sum_{i=1}^n x_i)} [1-p]^{(n - \sum_{i=1}^n x_i)}\right) + \frac{1}{2} \left( 1^{(\sum_{i=1}^n x_i)} ^{(n - \sum_{i=1}^n x_i)}\right) = 0 \quad (?)\,.$$

It's late, what am I doing wrong here?

Additionally, the other claim is that the value for $$p$$ should equal to the limiting long run proportion of successes, how does it works in this case?

• A hint... $0^0 = 1$. Look at the exponents in your final equation; they will either equal $n$ or $0$ if the probability is $1$ or $0$. – jbowman Aug 29 '16 at 19:19
• Oh I see, so in the only case where $\sum_i x_i = 0$ or $\sum_i x_i = n$ I get $0^0$, and that would recover the probability, thanks! I still however can not reconcile the claim that the long run proportion of successes i.e. $n^{-1}\sum x_i \to p$, should (somehow) also converge to the distribution $Q(p)$ (?)... – them Aug 29 '16 at 19:20
• You won't get $Q$ back. What happens is that $p$ is equal to $1$ or $0$, and the long run proportion of successes does equal $1$ or $0$ respectively. – jbowman Aug 29 '16 at 19:24
• @jbowman in general settings of this theorem, what does the empirical distribution function of $x_i$ (distribution that places mass $n^{-1}$ on $x_i$ converges to ? – them Aug 29 '16 at 19:28
• In the Bernoulli case, the proportion of ones converges to $p$ (which is a random variable) – Juho Kokkala Aug 29 '16 at 19:52