If P(a|b) = P(a), then P(a|b,c) = P(a|c)
I think this is a correct statement. If a and b are independent, then it makes sense that the probability of a given b and c is the same as the probability of a given c. But how do I prove that this is correct
On
If P(a|b) = P(a), then P(a|b,c) = P(a|c)
This statement does not always hold, even though it could be correct. Here's an intuitive example from Wiki:
Let events A and B be defined as the probability that person A and person B will be home in time for dinner where both people are randomly sampled from the entire world. Events A and B can be assumed to be independent i.e. knowledge that A is late has minimal to no change on the probability that B will be late. However, if a third event is introduced, person A and person B live in the same neighborhood, the two events are now considered not conditionally independent. Traffic conditions and weather-related events that might delay person A, might delay person B as well. Given the third event and knowledge that person A was late, the probability that person B will be late does meaningfully change.
In the example above, if we are not given the information that person A and person B live in the neighborhood, then the probability of person B arriving on time is (to a good approximation) independent of the person A arriving on time, hence $P(B|A) = P(B)$.But if we know that they live in the neighborhood, we have to condition it on this new information (call it N), then $P(B|A,N) ≠ P(B|N)$ .
In short: new information could change your sample space and in that case the conditional probability could change.
Consider the uniform distribution on $\{1,2,3,4\}$ and consider $A=\{1,2\},B=\{1,3\}$ and $C=\{1,4\}$. (This is a handy example to keep in your back pocket for several counterexamples).
Verify that $A$ is independent of $B$ and $A$ is independent of $C$. However, here we have $\Pr(A\mid B\cap C) = 1\neq \Pr(A\mid C) = \dfrac{1}{2}$