Let $A$, $B$ and $C$ be three random variables and $p_{A,B,C}=p_Ap_Bp_{CA,B}$ and $q_{A,B,C}=p_Aq_{BA}p_{CA,B}$ be two distributions on them. Then, we can conclude that? \begin{align*} \lvert I_p(A,B;C)I_q(A,B;C)\rvert\leq\log\lvert\mathcal{B}\rvert. \end{align*} where $\mathcal{B}$ is the alphabet set of $B$ and $I(.;.)$ is the mutual information.

$\begingroup$ By $AB$, do you mean the vector $(A,B)$ or do you mean the product? And there seems to be a typo in your definition of $q$. $\endgroup$ – Pat Devlin Jan 27 '17 at 0:56

$\begingroup$ I mean $(A,B)$. No, only instead of $p_B$ we have $q_{BA}$. $\endgroup$ – Math_Y Jan 27 '17 at 6:30

$\begingroup$ What are $p_A$ and $q_{BA}$? $\endgroup$ – Pat Devlin Jan 27 '17 at 12:26

$\begingroup$ $p_A$ is pmf of random variable $A$ and $q_{BA}$ is conditional pmf of $B$ conditioned on $A$. $\endgroup$ – Math_Y Jan 27 '17 at 14:29

$\begingroup$ How does that differ from $p_{BA}?$ $\endgroup$ – Pat Devlin Jan 27 '17 at 14:36
The setup is the following.
We have random variables $A,B,C$ and $A', B', C'$. We know $A$ and $A'$ have the same distribution, and for all $(a,b)$, the random variable $C  (A,B)=(a,b)$ has the same distribution as $C'  (A', B') = (a,b)$.
Under the above hypothesis, the claim you want is in fact false. Here's a counterexample.
Let $A$ and $A'$ be uniformly distributed over $\{1, 2, \ldots, a\}$. Let $B$ and $B'$ be independently drawn from $\{0,1\}$ such that $\mathbb{P}(B=0) = \mathbb{P}(B'=1) = \varepsilon$. And let $C = AB$ and $C' = A' B'$ be the product of the other two.
(I'm allowing the $\varepsilon$ in there just so that we don't have to condition on any events having probability $0$ in the definitions of $C$ and $C'$, but you should imagine $\varepsilon \approx 0$. And every $\approx$ below becomes equality in the limit as $\varepsilon \to 0$.)
Then we have $H(A,B) = H(A', B') = H(A) + H(B) \approx H(A) = \log(a)$. On the other hand, $H(A,B  C) \approx 0$ since with probability $1\varepsilon$, $(A,B)$ is uniquely determined by $C$. Similarly $H(A', B'  C') \approx H(A')$ since $C'$ usually tells us nothing about $A'$. This give us
$$ I(A, B; C)  I(A', B'; C') = H(A,B)  H(A,BC)  [H(A',B')  H(A',B'C')] = H(A',B'C')  H(A,BC) \approx \log(a), $$ which can be made arbitrarily large.