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I was browsing a well known technology board and someone posted a pretty standard question about conditional probabilities. As this happens to be the current subject of my lectures (just starting), I decided to take a swing at them and it turns out I have no idea what I'm doing, or what kind of proof (contradiction?) I should be using in these kinds of questions. I can show these are correct/incorrect with arbitrary numbers, but I have a gut feeling there are more succinct ways of showing truth.

If P(a|b,c) = P(a), then P(b|c) = P(b)

I get the feeling that this one is wrong, only because we can't assume c has no bearing on b with only the information that neither affects a. From my limited knowledge of conditional probability, I'm reading the first statement that a is independent from both b and c. I did this, where I assumed the implication was true, and got a contradiction. Nevermind, I made the assumption that P(a|b,c) = P(a) ==> P(a|b) = P(a) and P(a|c) = P(a), and I apparently can't do that?

If P(a | b) = P(a), then P(a | b, c) = P(a | c)

I believe this one is true, because if a and b are independent, then the probability of a, given two conditions - one of which doesn't affect it, and one that might - should just be the probability of a given the other condition (c). I also tried proving this one, but I kept going in a circle.

I've tried contradiction approaches, but I'm not familiar enough with conditional probability to say something is for sure a contradiction (For example, before I realized the incorrect assumption I made on the first one, I assumed the second part was true and arrived at P(b|a,c) = P(b|c), which looks false, but because of the stuff about independence I can't be certain). Most of my 'technique' has revolved around using the definitions

P(a|b) = P(a,b)/p(b) (expanded to) P(a|b,c) = P(a,b,c)/P(b,c)
P(a,b) = P(a|b)P(b) = P(b|a)P(a)

Any help is appreciated, especially if you focus on the "why" of choosing a certain proof.

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  • $\begingroup$ Wellcome to MSE $\endgroup$ – HK Lee Apr 14 '15 at 4:15
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\begin{equation} \begin{split} P(a|b,c) &= P(a)\hspace{30pt}\text{(It means that $a$ is independent of $b$ and $c$)}\\ \Rightarrow \frac{P(a,b,c)}{P(b,c)}&=P(a)\\ \Rightarrow P(a,b,c)&=P(a)P(b,c) = P(a)P(b|c)P(c)\\ \Rightarrow P(a,b,c)&=P(a)P(c)P(b|c) \end{split} \end{equation} If $P(b|c) = P(b)$, then \begin{equation} \begin{split} P(a,b,c)&=P(a)P(c)P(b) \end{split} \end{equation} which means that $a,b,c$ are independent of each other, which means that $b$ is independent of $c$.

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    $\begingroup$ Well, not exactly. What if A is null and B and C are not independent? What if B and C are independent and A=B? $\endgroup$ – BruceET Apr 14 '15 at 16:21
  • $\begingroup$ If $b$ and $c$ are not independent, then we cannot say that $P(b|c) = P(b)$. I wanted to show that $P(b|c) = P(b)$ holds when $b$ and $c$ are independent of each other. If $a=b$ and $b$ and $c$ are independent, then $a,b,c$ are, of course, not independent of each other. My calculations were based only on the information provided by the OP. Thanks for giving the insight. $\endgroup$ – user146290 Apr 14 '15 at 22:05

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