# Does this hold? $P(A \cap B|C) = P(A|C) P(B|C) \implies P(A\cap B|C^\complement) = P(A|C^\complement).P(B|C^\complement)$

If two events $A$ and $B$ are conditionally independent given a third event $C$. Are they conditionally independent on $C^\complement$ as well?

i.e Does this hold? $$P(A \cap B|C) = P(A|C) P(B|C) \implies P(A\cap B|C^\complement) = P(A|C^\complement).P(B|C^\complement)$$

I think that this is not the case. We know that for independence, the two events MUST intersect. Now I can draw a venn diagram in which $A$ and $B$ intersect only when $C$ occurs and outside $C$ they don't. Hence they won't be independent in the universe $C^\complement$. Is this a valid argument? If my argument is incorrect and the above actually holds, provide can you provide a hint for the proof? Also can you explain it intuitively(why it holds or why it doesn't hold in general)?

• Would like to add that intersection of 2 events is a Necessary but not Sufficient condition for independence. – Kartik Mahajan Aug 2 '18 at 3:46
• About this necessity of intersection, if one of the events is empty (or, more generally, has probability zero), then they are trivially independent, right? Or is independence left undefined in this case? – fonini Aug 2 '18 at 4:15
• Sorry that I missed the trivial case. Yes, trivial case can be ignored. – Kartik Mahajan Aug 2 '18 at 4:50

It is a constructive argument.

It is possible to have three events $A,B,C$ where $A\perp B\mid C$ holds, but $A\cap C^\complement$ and $B\cap C^\complement$ are disjoint (so $A\not\perp B\mid C^\complement$).   Therefore conditional independence with respect to an event does not logically entail conditional independence with respect to the complement.

Consider $A=\{1,2,5\} , B=\{2,3,6\}, C=\{1,2,3,4\}, C^\complement=\{5,6\}$ representing events from a fair toss of a die, as a demonstration one such counterexample.

Your two events $A$ and $B$ may be very well-behaved when $C$ occurs, but that tells you nothing about what they may do when $C$ doesn't occur. Examples are very easy to construct.

Let $\Omega=\{1, 2, 3,4\}$ and $\mathsf{P}(\{j\})=1/4$. Define the events $A=\{1,2\}$, $B=\{1\}$, and $C=\{2,3\}$. Then $$0=\mathsf{P}(A\cap B\mid C)=\mathsf{P}(A\mid C)\mathsf{P}(B\mid C)=\frac{1}{2}\times 0.$$ However,

$$\frac{1}{2}=\mathsf{P}(A\cap B\mid C^c)\ne \mathsf{P}(A\mid C^c)\mathsf{P}(B\mid C^c) =\frac{1}{2}\times\frac{1}{2}.$$

Yes, your argument is fine, at least intuitively. You can formalize it if you want:

Picture the following experiment. You start by flipping a fair coin. If it lands heads, you flip another one. But if the first coin landed tails, then you flip two more fair coins, making for a total of three coin flips in this case. Let $C$ be the event "the first coin landed tails", $A$ be the event "the second coin landed tails", and $B$ be the event "the last coin landed tails". Assuming $C$ holds, then $A\cap C$ and $B\cap C$ are independent. But assuming $C$ does not hold, then the second coin is the last coin, so that $A\cap C$ and $B\cap C$ are the same event. This is an explicit counterexample.

• Thanks a lot for this example. – Kartik Mahajan Aug 2 '18 at 5:03