0
$\begingroup$

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

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

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?

$\endgroup$
  • 1
    $\begingroup$ 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 $0$. $\endgroup$ – jbowman Aug 29 '16 at 19:19
  • $\begingroup$ 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 $Q(p)$ (?)... $\endgroup$ – them Aug 29 '16 at 19:20
  • 1
    $\begingroup$ 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. $\endgroup$ – jbowman Aug 29 '16 at 19:24
  • $\begingroup$ @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 ? $\endgroup$ – them Aug 29 '16 at 19:28
  • $\begingroup$ In the Bernoulli case, the proportion of ones converges to $p$ (which is a random variable) $\endgroup$ – Juho Kokkala Aug 29 '16 at 19:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.