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.


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.



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.

  • $\begingroup$ 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. $\endgroup$ – Kolmogorovwannabe Sep 22 at 8:38
  • $\begingroup$ what do you not understand about that example? that's a very simple example already. which number do you not understand? $\endgroup$ – antkam Sep 22 at 15:10
  • $\begingroup$ 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? $\endgroup$ – Kolmogorovwannabe Sep 22 at 21:21
  • $\begingroup$ 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. $\endgroup$ – antkam Sep 23 at 3:38

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