Is this prior distribution exchangeable? I am reading the book Bayesian data analysis written by Gelman. And I do not confirm the answer of this question:
suppose it is known a priori that the $2J$ parameters $\theta_1,\ldots,\theta_{2J}$ are clustered into two groups, with exactly half being drawn from a $N(1,1)$ distribution, and the other half being drawn from a $N(-1,1)$ distribution, but we have not observed which parameters come from which distribution.
$(a)$ Are $\theta_1,\ldots,\theta_{2J}$ exchangeable under this prior distribution?
$(b)$ Show that this distribution cannot be written as a mixture of $iid$ components.
$(c)$ Why can we not simply take the limit as $J\rightarrow \infty$ and get a couterexample to $de \space Finetti's$ theorem?  
IMO, we have no any information about ordering or grouping of the parameters can be made, so  the parameters $\theta_1,\ldots,\theta_{2J}$ are exchangeable in this prior distribution?
Besides, not all parameters are identity distributed, can we simply say this distribution cannot be written as a mixture of $iid$ components?
We cannot take the limit as $J\rightarrow \infty$ and get a couterexample to $de \space Finetti's$ theorem since this distribution is not exchangeable. Right?  
If you have any idea please write your comment. Let me know I do a right answer or not. Thanks :)
 A: (a) The question doesn't seem to  provide you with the distribution of $\theta_1, \ldots, \theta_{2J}$, so you can't say for sure whether or not the distribution is exchangeable.  E.g. the possibility that $E(\theta_1) = \ldots = E(\theta_{J}) = -1$ seems to be consistent with the wording of the question, but if this possibility obtains then $\theta_j$'s are not exchangeable.  However, it seems likely that Gelman intends the following:
  1) let $(R_1, \ldots, R_{2J})$ have uniform distribution on the set of $\{1, -1\}$-valued sequences having sum 0. 
  2) let $(W_1, \ldots, W_{2J})$ be IID N(0,1) random variables independent of $(R_1, \ldots, R_{2J})$.
  3) let $X_j := W_j + R_j$ for $1 \leq j \leq 2J$.
In this setup, $(X_1, \ldots, X_{2J})$ is an exchangeable sequence because the pairs $(R_1, W_1),  \ldots, (R_{2J}, W_{2J}) $ are exchangeable (because .... [you fill in the blanks] ).  
(b) Try computing $E(X_1X_2)$ for $(X_j)$ defined as above by conditioning on $(R_1, R_2)$. You'll find that the expectation is negative.  Now assume that these $X_j$'s are derived by taking IID draws from a random distribution $\mu$ as in the classical de Finetti theorem, and try to say something about the sign of $E[X_1X_2]$ by conditioning on $\mu$.  You should get an easy contradiction showing that $(X_1, \ldots, X_{2J})$ are not derived by conditional IID sampling.  (Now how do you argue that \emph{Gelman's} $X_j's$ are not derived by conditional IID sampling?)
(c) Drawing n things u.a.r. without replacement from N things is essentially the same as drawing n things u.a.r. with replacement from N things provided n is small compared to N, and this becomes more and more true if n is fixed and N gets large.  If you properly formalize this you may be able to show that when you take an appropriate limit as $J \to \infty$ in the setup above, the limit is exchangeable and satisfies de Finetti's theorem. 
A: I don't know my explanation is correct or not but I try to explain it in detail. Here is the story, there is $\frac{1}{\binom{2J}{J}}$ probability to find the $J$ parameters that follow the distribution $N(1,1)$. Hence the remaining $J$ parameters must be sampled from $N(-1,1)$. Use Bayes theorem, $p(\theta_1,\ldots,\theta_{2J})=p(\theta_{p(1)},\ldots,\theta_{p(J}))p(\theta_{p(J+1)},\ldots,\theta_{p(2J})|\theta_{p(1)},\ldots,\theta_{p(J}))$
$=\sum_{p}\left(\left(\dfrac{1}{\binom{2J}{J}}\prod_{j=1}^{J}N(\theta_{p(j)}|1,1)\right)\prod_{j=J+1}^{2J}N(\theta_{p(j)}|-1,1)\right)$
feel free to write your comment if you have any idea. :D
