About the equation $\mathbb{P}(A\cap B | C\cap D) =\mathbb{P}(A|C)\mathbb{P}(B|D) $

Question

If $(A,C)$ and $(B,D)$ are $\mathbb{P}$-independent, then it holds true that: $$\mathbb{P}(A\cap B | C\cap D) = \frac{\mathbb{P}(A\cap B \cap C\cap D)}{\mathbb{P}(C\cap D)} = \frac{\mathbb{P}(A\cap C)\mathbb{P}(B\cap D)}{\mathbb{P}(C)\mathbb{P}(D)} =\mathbb{P}(A|C)\mathbb{P}(B|D).$$ Now, I want to investigate this relation more deeply without assuming that $(A,C)$ and $(B,D)$ are $\mathbb{P}$-independent.

Suppose only that $A$ and $B$ are $\mathbb{P}(\cdot|C\cap D)$-independent. Then: $$\mathbb{P}(A\cap B | C\cap D)=\mathbb{P}(A | C\cap D)\mathbb{P}(B | C\cap D)=\frac{\mathbb{P}(A \cap C\cap D)}{\mathbb{P}(C\cap D)}\frac{\mathbb{P}(B \cap C\cap D)}{\mathbb{P}(C\cap D)}=(*).$$ Now, suppose also that $A\cap C$ and $D$ are $\mathbb{P}$-independent and $B\cap D$ and $C$ are $\mathbb{P}$-independent. Then: $$(*)=\frac{\mathbb{P}(A\cap C)\mathbb{P}(D)}{\mathbb{P}(C\cap D)}\frac{\mathbb{P}(B \cap D)\mathbb{P}(C)}{\mathbb{P}(C\cap D)}= \mathbb{P}(A|C)\mathbb{P}(B|D)\left(\frac{\mathbb{P}(C)\mathbb{P}(D)}{\mathbb{P}(C\cap D)}\right)^2,$$ and so we see that what it is missing to obtain the initial equation is requiring that $C$ and $D$ are $\mathbb{P}$-independent.

Now, I'm wondering if the assumed conditions alone imply that the initial equation holds, i.e. if:

1. $A$ and $B$ are $\mathbb{P}(\cdot|C\cap D)$-independent;
2. $A\cap C$ and $D$ are $\mathbb{P}$-independent;
3. $B\cap D$ and $C$ are $\mathbb{P} $-independent;
4. $\mathbb{P}(A\cap B | C\cap D)>0$;

is it true that $$\mathbb{P}(A\cap B | C\cap D) =\mathbb{P}(A|C)\mathbb{P}(B|D)?$$

I think that the answer is no, since if this is the case, the last equation implies the $\mathbb{P}$-indipendence of $C$ and $D$ also, and I suspect that this would be too much... I'm a bit lost finding a counter-example. Any help?

Answer 1

UPDATE Here is a "100% non-trivial" counter-example where every probability is neither $0$ nor $1$. :)

Sorry this is kinda tedious but it is the "smallest" such example I can think of. (If you allow some prob to be $1$ then smaller counter-examples exist.)

Uniformly draw a number from $\{1, 2, \dots, 24\}$ and define:

• $C = \{1, 2, \dots, 12\}$ and $P(C) = 12/24 = 1/2$

• $D = \{9, 10, \dots, 20\}$ and $P(D) = 12/24 = 1/2$

• $C \cap D = \{9, 10, 11, 12\}$ and $P(C\cap D) = 4/24 = 1/6 \neq P(C)P(D)$, so they are dependent.

• $A = \{1, 2, 9, 11\} = A \cap C, P(A\cap C) = 4/24 = 1/6$

• $P(A\cap C \cap D) = P(\{9, 11\}) = 2/24 = 1/12 = P(A \cap C) P(D)$, so $(A\cap C) \perp D$

• $B = \{9, 10, 19, 20\} = B \cap D, P(B \cap D) = 4/24 = 1/6$

• $P(B\cap D \cap C) = P(\{9, 10\}) = 2/24 = 1/12 = P(B\cap D) P(C)$, so $(B \cap D) \perp C$

Finally, when conditioned on $C\cap D = \{9, 10, 11, 12\}$, we have (excuse my abuse of notation):

• $P(A|C \cap D) = {\{9, 11\} \over \{9, 10, 11, 12\}} = 1/2$

• $P(B|C \cap D) = {\{9, 10\} \over \{9, 10, 11, 12\}} = 1/2$

• $P(A \cap B | C \cap D) = {\{9\} \over \{9,10,11,12\}} = 1/4$, so $A \perp B$ when conditioned on $(C \cap D)$.

Whew!

This "pictorial view" might be easier (pick one of $24$ squares uniformly randomly):

+----+----+----+----+----+----+
|  C |A C |A CD|   D|   D|    |
+----+----+----+----+----+----+
|  C |A C |ABCD| B D|   D|    |
+----+----+----+----+----+----+
|  C |  C | BCD| B D|   D|    |
+----+----+----+----+----+----+
|  C |  C |  CD|   D|   D|    |
+----+----+----+----+----+----+

BTW your big equation becomes:

$$\frac14 = \frac13 \times \frac13 \times ({\frac12 \times \frac12 \over \frac16})^2$$

P.S. if you allow some prob to be $1$, specifically $P(A|C\cap D) = P(B | C\cap D) = P(A\cap B | C \cap D) = 1$, then a smaller example is the 2nd row above, i.e.

+----+----+----+----+----+----+
|  C |A C |ABCD| B D|   D|    |
+----+----+----+----+----+----+

---

As the question is written, the answer is No, but the counter-example I have below is "trivial" and may not be what you have in mind:

• $A=B=\emptyset$, i.e. the empty set (the impossible event)

• $C=D$ and $P(C) = P(D) > 0$

Your big equation reduces to $0 = 0 \times ({P(C)P(D) \over P(C \cap D)})^2$ which is trivially true even if $P(C)P(D) \neq P(C\cap D)$, i.e. even if $C, D$ are dependent.

Answer 2

Assuming 1.,2.,3.and 4., the equation in your question is true if and only if $C$ and $D$ are independent.

In order to get a counter-example, it's enough to build an example in which 1.,2.,3. and 4. hold and in which $C$ and $D$ are dependent.

Take a 52-cards French deck and a die. Pick a card at random and roll the die.

Calling $v$ the value on the die, which ranges from $1$ to $6$, set:

$C:=\{ v \geq 3\}$

$D:=\{ v \leq 4\}$

Both $C$ and $D$ have probability $2/3$, $C \cap D$ has probability $1/3$, so $C$ and $D$ are dependent.

Now set:

$A:=\{ v\in\{3,4\} $and the card is red$ \} \cup \{ v\in\{5,6\} $and the card is a heart$ \}$

$B:=\{ v\in\{3,4\} $and the card is a Queen or a King$ \} \cup \{ v\in\{1,2\} $and the card is a Jack$ \}$

We only have to show that the four condition 1., 2., 3., and 4. hold.

1.) $C \cap D =\{ v\in\{3,4\}\}$, and given $C \cap D$ the event $A$ concerns only the suit of the card while $B$ concerns only the value of the card, so they are independent.

2.)Since $A \subset C$, you need to show that $A$ and $D$ are independent. \begin{align} P(A)&=(1/3)(2/4)+(1/3)(1/4)=1/4\\ P(D)&=2/3\\ P(A \cap D)&=(1/3)(2/4)=2/12=P(A)P(D) \end{align}

3.)Here also, since $B \subset D$, you need to show that $B$ and $C$ are independent. \begin{align} P(B)&=(1/3)(2/13)+(1/3)(1/13)=1/13\\ P(C)&=2/3\\ P(B \cap C)&=(1/3)(2/13)=2/13=P(B)P(C) \end{align}

4.) Since the events $A,B,C,D$ have non-empty intersection, 4. holds.