# Does $p(a | b,c) = p(a|c)$ necessarily imply $p(a|b) = p(a)$?

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?

• Your intuition is wrong; it does not imply that $a$ is independent of $b$.With this knowledge, can you find a (dis)proof? Oct 27 '20 at 10:47
• 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. Oct 27 '20 at 10:59
• 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. Oct 27 '20 at 12:34

## 2 Answers

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

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

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

Now consider the following sets

... 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 $$b$$

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