Calculating probability of (A and B), based on conditional probabilities of A and B

Question

If we have P(A given Z) and also p(B given Z), which of the following two methods is correct if we want to calculate P(A and B), assuming A and B are independent? Most likely method 1 is correct, but can you explain why and even even better, provide an intuitive expalnation?

Method 1:

$$P(A \cap B | Z) = P(A | Z) \cdot P(B | Z) $$

$$ P(A \cap B) = \int P(A | Z) \cdot P(B | Z) f(Z)\,dz$$ Method 2 $$P(A)=\int P(A|Z)f(Z)\,dz$$ $$P(B)=\int P(B|Z)f(Z)\,dz$$ $$P(A \cap B) = P(A)*P(B)= \int P(A|Z)f(Z)\,dz * \int P(B|Z)f(Z)\,dz$$

Answer 1

We are not given that $$P(A \cap B|Z)=P(A |Z)P(B|Z)$$

For example $A$ is the event that we get a head from tossing a fair coin. Let $Z$ be the event of getting a head from tossing another coin. Let $B$ be the event that the outcome of the two coin tosses being different. Then $A$ and $B$ are independent events.

However, given $Z$, $A$ and $B$ are not the same.

$$P(A=h, B=s|Z=h)=P(A=h, Z=h|Z=h)=P(A=h)=\frac12$$

$$P(A|Z)=\frac12$$

$$P(B|Z)=\frac12$$

\begin{align} P(A \cap B)= P(A)P(B) \end{align}

is true from the definition of independence.

Remark: You might like to use $f(z)$ rather than $f(Z)$.