Given three discrete random variables A,B,C, does P(A)P(C) = P(B)P(C) imply P(A)=P(B)?

Question

We know that there may exist some values of C that have probability of 0, but the entire sample space must have some values of C that have nonzero probability. If $P(A=a)P(C=c) = P(B=b)P(C=c)$ for all 3-tuples $(a,b,c)\in \Omega_A \times \Omega_B \times \Omega_C$, then does this mean that $P(A)=P(B)$ for all 2-tuples $(a,b)\in \Omega_A \times \Omega_B$?

Answer 1

Your notation is seriously off. I think what you mean is: if $P(A=a, C=c) = P(B=b, C=c)$ for all $(a,b,c) \in \Omega_A \times \Omega_B \times \Omega_C$, then is $P(A=a) = P(B=b)$ for all $(a,b) \in \Omega_A \times \Omega_B$?

The answer is yes, because (for a discrete random variable $C$) there is a countable set $S = \{c_n: n \in \mathbb N\}$ such that $C \in S$ with probability $1$, and then by countable additivity $P(A=a) = \sum_{n \in \mathbb N} P(A=a, C=c_n)$ and $P(B=b) = \sum_{n \in \mathbb N} P(B=b, C=c_n)$.

Answer 2

I just realized that the answer is really simple: Choose $c'\in\Omega_C$ such that $P(C=c')\ne 0$. Then, for any $(a,b)\in\Omega_A\times\Omega_B$, $$P(A=a)P(C=c')=P(B=b)P(C=c') \rightarrow P(A=a) \frac{P(C=c')}{P(C=c')}=P(B=b)\frac{P(C=c')}{P(C=c')} \rightarrow P(A=a) = P(B=b)$$

Sorry, I was up late and missed the obvious answer!