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I'm in a ML course, and we had this math refresher quiz. We were asked to prove (or disprove) the following: $$p(a | b,c) = p(a|c) \to p(a|b) = p(a).$$

It is clear that $a$ is not dependent on $b$, when both $b$ and $c$ occur. However, it is not fully clear wether this implies that $a$ itself is independent of $b$. Intuitively, yes. But I don't know how to prove that mathematically. Any suggestions?

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  • $\begingroup$ Your intuition is wrong; it does not imply that $a$ is independent of $b$.With this knowledge, can you find a (dis)proof? $\endgroup$
    – Servaes
    Oct 27 '20 at 10:47
  • $\begingroup$ Thanks for the hint. Unfortunately, I still can't see how to prove it. So far I got $\frac{p(a,b,c)}{p(b,c)} = \frac{p(a,c)}{p(c)}$. Now because of initial condition (i.e. $p(a|b,c) = p(a|c)$) we can discern that $p(b,c) = \frac{p(c)}{p(b)}$. But I don't know how proceed further from here. $\endgroup$
    – User1396
    Oct 27 '20 at 10:59
  • $\begingroup$ Because the statement is false, you can disprove it by giving a counterexample. I won't give one because I don't know in what framework you are thinking about probabilities; you coud describe such an example in plain english at a high school level, or you could construct a probability space and measure in which this occurs. $\endgroup$
    – Servaes
    Oct 27 '20 at 12:34
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$$p(a|b,c)=\frac{p(a,b,c)}{p(b,c)}=\frac{p(a,c)}{p(c)}\frac{p(b|a,c)}{p(b|c)}=$$

$$p(a|b,c)=p(a|c)\cdot P$$

$$P=\frac{p(a,b,c)}{p(a,c)}\cdot\frac{p(c)}{p(b,c)}$$

Now consider the following sets

enter image description here

... it is self evident that $P=1$ and thus

$p(a|b,c)=p(a|c)$ but $a,b$ are not independent as $b \subset a$

In other words, $p(a|b,c)=p(a|c)$ is NOT SUFFICIENT for independence between $a$ and $b$

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Hint: Consider events $a$, $b$ and $c$ such that $p(b|\,c)=1$.

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