Do conditionals determine joint distribution? I read some questions on this matter, but I found a tricky example of conditionals, and I don't really know how to approach this problem.
Let $X,Y$ be random variables such that $X,Y\in\{0,1,2,...\}$. Also we have the following conditionals:
$$P(X=i+1|Y=i)=P(X=i-1|Y=i)=1/2$$
$$P(Y=i+1|X=i)=1-P(Y=i-1|X=i)=1/3$$
Also, let the boundary conditions be:
$$P(X=1|Y=0)=P(X=0|Y=0)=1/2$$
$$P(Y=1|X=0)=1-P(Y=0|X=0)=1/3$$
The tricky thing about this example is the support of the random variables is infinite countable and the way the conditionals are defined. In this particular case of support, can the conditionals determine the joint distribution? If someone can help me on how to approach this example, I would be very grateful, thanks!
 A: [The following is an excerpt from our book Monte Carlo Statistical Methods on the reason why the joint distribution is determined by its conditional distributions, a result known as the Hammersley-Clifford theorem.]
A most surprising feature of the Gibbs sampler is that the conditional
distributions contain sufficient information to
produce a sample from the joint distribution.  (This is the case for both
two-stage and multi-stage Gibbs samplers.) By comparison
with maximization problems, this
approach is akin to maximizing an objective function successively in every direction
of a given basis. It is well known that this optimization method does not
necessarily lead to the global maximum, but may end up in a saddlepoint.
It is, therefore, somewhat remarkable that the
full conditional distributions perfectly summarize the joint density,
although the set of marginal distributions obviously fails to do so.
The following result then shows that the joint density can be
directly and constructively derived from the conditional densities.

Theorem The joint distribution associated with the conditional densities $f_{Y|X}(y|x)$ and $f_{X|Y}(x|y)$ has the joint density $$
  f(x,y) = \frac{f_{Y|X}(y|x)}{\int
 \left[f_{Y|X}(y|x)/f_{X|Y}(x|y)\right] \text dy} . $$

Proof:
Since $f(x,y) = f_{Y|X}(y|x)f_X(x)= f_{X|Y}(x|y)f_Y(y)$ we have
$$
\int \frac{f_{Y|X}(y|x)}{f_{X|Y}(x|y)}\text dy = \int \frac{f_Y(y)}{f_X(x)}\text dy = \frac{1}{f_X(x)},
$$
and the result follows.
This derivation of $f(x,y)$ obviously requires the existence and computation of the integral.  However, this result clearly demonstrates the fundamental feature that the two conditionals are
sufficiently informative to recover the joint density.
Note, also, that this theorem makes the implicit assumption that the joint density $f(x,y)$ exists.
