# In probability, if we have an equality, can we condition on an arbitrary event and guarantee that the equality still holds?

I have a question regarding equalities and conditioning probabilities. I was reading through this proof for conditional independence, which can be found here. In the proof, they first state that $$p(A|B)=\frac{p(A\cap B)}{P(B)}$$, which comes from the definition of conditional probability. This I understand. But, they then go on to state that $$p(A|B,C)=\frac{p(A\cap B|C)}{P(B|C)}$$ by conditioning on $$C$$. This seems like a strong claim. Is it is case that when you have an equality of probabilities, you may condition on any arbitrary event given that it occurs with a non-zero probability? For example, if I know that $$p(A\cap B) + p(A^c\cap B)=p(B)$$, may I also state that $$p(A\cap B | C) + p(A^c\cap B | C)=p(B | C)$$ by conditioning on $$C$$?

Further, if this is true, how might I go about proving it? And if it's false, is there a counter-example?

• If $P(C)>0$ and $P(B \mid C)> 0$ then $P(A\mid B,C)=\frac{P(A\cap B\mid C)}{P(B\mid C)}$ does not seem to me any stronger a claim than $P(A\mid B)=\frac{P(A\cap B}{P(B)}$. Commented Feb 6, 2023 at 1:36

## 1 Answer

If you understand that $$P(A|B, C)$$ really means $$P(A|B \cap C)$$ (in probability, comma "," is frequently interpreted as the synonym for "intersection"), then it is still a matter of applying definition: \begin{align} & P(A|B \cap C) \\ =& \frac{P(A \cap B \cap C)}{P(B \cap C)} \\ =& \frac{P(A \cap B | C) P(C)}{P(B | C)P(C)} \\ =& \frac{P(A \cap B | C)}{P(B | C)}. \end{align} This completes the proof.