To define two random variables that satisfy a condition First:
Let $X_{1}, X_{2}, X_{3}, X_{4}$ be 4 random variables.Consider the two auxiliary random variables: $\tilde{X_{1}}, \tilde{X_{2}}$ where their corresponding alphabets $\tilde{\chi_{1}}, \tilde{\chi_{2}}$ equal respectively the alphabets of $X_{1}, X_{2}$.

The problem:
The text then define the auxiliary random variables to be 2 random variables such that the joint distribution of the 6 random variables is defined by:

$$p_{1234\tilde{1}\tilde{2}}(x_{1}, x_{2},x_{3}, x_{4},\tilde{x_{1}}, \tilde{x_{2}}) =\left\{\begin{array}{c c}\dfrac{p_{1234}(x_{1}, x_{2}, x_{3}, x_{4})p_{1234}(\tilde{x_{1}},\tilde{x_{2}}, x_{3},x_{4})}{p_{34}(x_{3},x_{4})}, &  p_{34}(x_{3},x_{4})\neq 0\\
       0, & p_{34}(x_{3},x_{4})= 0
     \end{array}\right.$$

My problem/confusion is: How can I be sure that this is indeed possible? How can I be sure that there exist such random variables $\tilde{\chi_{1}}, \tilde{\chi_{2}}$ that satisfy this condition?
 A: Let $(X,Y)$ be defined on some probability space $(\Omega,\mathcal F,P)$. Consider $\bar\Omega=\Omega\times\Omega$ endowed with the product sigma-algebra $\bar{\mathcal F}=\mathcal F\otimes\mathcal F$ and the probability measure $\mathbb P=P\otimes P$. Consider $\bar X$, $\bar Y$, $\bar Z$ and $\bar T$ defined on $\bar\Omega$ by $\bar X(\omega_1,\omega_2)=X(\omega_1)$, $\bar Y(\omega_1,\omega_2)=Y(\omega_1)$, $\bar Z(\omega_1,\omega_2)=X(\omega_2)$, and $\bar T(\omega_1,\omega_2)=Y(\omega_2)$. Now:

If $X$ is distributed like your $(X_1,X_2)$, if $Y$ is distributed like your $(X_3,X_4)$, and if $(X,Y)$ is distributed like your $(X_1,X_2,X_3,X_4)$, then $(\bar X,\bar Y,\bar Z)$ conditioned on $\bar Y=\bar T$ is distributed like your $(X_1,X_2,X_3,X_4,\tilde X_1,\tilde X_2)$.

To sum up, one may have to enlarge the sample space but one can indeed define $(\tilde X_1,\tilde X_2)$ whose joint distribution with $(X_1,X_2,X_3,X_4)$ is as you described.
A: As Did pointed out, you may need to refine your probability space.
One way to think about this problem is as follows. You first sample your random variables $X_1, \dots, X_4$. Then you sample $\tilde X_1$ and $\tilde X_2$ form a distribution with density 
$$p_{1234}(\tilde x_1,\tilde x_2, X_3, X_4) / p_{34}(X_3,X_4)$$
(here, $X_3$ and $X_4$ are the values from your first sample). Note that this is a valid distribution since all numbers $p_{1234}(\tilde x_1,\tilde x_2, X_3, X_4) / p_{34}(X_3,X_4)$ are non-negative, and $\sum_{\tilde x_1,\tilde x_2} p_{1234}(\tilde x_1,\tilde x_2, X_3, X_4) / p_{34}(X_3,X_4) = 1$. We get r.v. $X_1,X_2,X_3, X_4,\tilde X_1, \tilde X_4$. Their joint distribution is 
$$p_{1234}(x_1, x_2,x_3,x_4) \times \frac{p_{1234}(\tilde x_1,\tilde x_2, x_3, x_4)}{p_{34}(x_3,x_4)},$$
as required.
