Conditional independence for 3 events

Question

I have the following statement:

Prove or disprove if $P(A_1 | A_2) = P(A_1)$ then $P(A_1 | A_2, A_3) = P(A_1 | A_3)$

I had tried things like:

$P(A_1 | A_{2,3})=\frac{P(A_{1,3})}{P(A_{2,3})} = \frac{P(A_3 | A_{1,2})P(A_{1,2})}{P(A_{2,3})}$

unsuccessfully.

Any hints?

What was your first attempt? Can you add it to the post and we can then correct from there. — Annika, Aug 30 '23 at 18:29
This doesn't parse. "If P(A_1|A_2)" needs something more. A probability is not a boolean. — wnoise, Aug 30 '23 at 18:33

score 4 · Accepted Answer · answered Aug 30 '23 at 18:43

The first example you should typically try for a problem like this is the following:

Uniform distribution over $\{1,2,3,4\}$... $A_1 = \{1,2\},A_2=\{1,3\},A_3=\{1,4\}$. (Equivalently worded with coins: Flip two coins. $A_1$ is that the first flip is heads, $A_2$ the second flip is heads, $A_3$ is that the two flips match).

This is a famous counterexample for many basic problems involving independence.

Specifically, note that $A_1,A_2,A_3$ are each pairwise independent but the three are not mutually independent.

$~$

Here, we have $\Pr(A_1\mid A_2)=\Pr(A_1\mid A_3)=\Pr(A_1)=0.5$

$~$

We have that $A_2\cap A_3 = \{1\}$. We have that $\Pr(A_1\mid A_2,A_3)=\Pr(A_1\mid \{1\}) = 1\neq \Pr(A_1\mid A_3)=0.5$

Conditional independence for 3 events

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