Prove or disprove two simple statements on conditional probability

STATEMENT 1:

if $$P[A|B]=P[B]$$ then $$A$$ and $$B$$ are independent.

My attempt:

Assuming that $$A$$ and $$B$$ are independent then, since $$P[A|B]= P[B]$$ is possible only if $$P[A]=P[B]$$, the statement is not true in general.

STATEMENT 2:

if $$P[A∩B|C]=P[A|C]P[B|C]$$, then A and B are independent events.

I've tried many ways, but I cannot figure out how to prove or disprove the statement. Could you help me? Thanks.

HINTS

Stmt 1: you have proven the reverse direction. You need to show a counter-example where $$P(A\mid B) = P(B)$$ but $$A, B$$ are dependent. Instead your only showed an example where $$A, B$$ independent but $$P(A\mid B) \neq P(B)$$. It's the wrong direction.

The correct counter-example is not difficult. You can easily construct examples with coins, dice (or even just purely abstractly).

Stmt 2: This is more subtle. $$P(A\cap B\mid C) = P(A\mid C) P(B \mid C)$$ means that $$A,B$$ are independent when conditioned on $$C$$. The question is does that make $$A,B$$ independent when unconditioned? I would suggest using a 6-sided die and assigning various subsets of results as $$A,B,C$$ and see if you can first find an example where $$P(A\cap B\mid C) = P(A\mid C) P(B \mid C)$$.

Hope this helps. If you're still stuck after a while lemme know your progress and I'll see if I can think of further hints.

• I really appreciate your help, but I found already tons of examples on the web, the problem is that I can't understand their logic. Here on math exchange for example I find this example for statement 2: math.stackexchange.com/questions/3002222/… the rationale is clear, I do not understand the number attached to those probabilities. If you could provide two easy examples explaining where the number come from it would be of great help. – Kolmogorovwannabe Sep 22 at 8:38
• what do you not understand about that example? that's a very simple example already. which number do you not understand? – antkam Sep 22 at 15:10
• I cannot understand what is the intersection between A and B (which are independent by the way) and why conditioning the intersection to C we get 1/2. $A∩B$ is "get H in both coins", conditioned to "the two coins match", If C happens and the two coin matches, then I have on both tails or head, then 1/2 as prob., is this the right interpretation? What does it mean $P(A|C)$? If C happens, then the two coins match what information gives about the happening of A? Finally, in the example did they prove the wrong direction? – Kolmogorovwannabe Sep 22 at 21:21
• you interpretation of $P(A\cap B \mid C)$ is entirely correctly, and now you understand why that value is $1/2$. As for $P(A \mid C)$, you are also correct that knowing $C$ does not give you any clue about $A$. that's why $P(A \mid C)$ remains $1/2$. in that thread, the question asked in the reverse of your stmt 2. so that answer correctly answered that question, but it is not the proper counter-example for your stmt 2. you need a different example. – antkam Sep 23 at 3:38