Suppose we have three events, $A,X,Y$, and suppose we wish to compute $P[A|X\land Y]$, along with its relatives, $P[A|X\land \neg Y], P[A|\neg X \land Y], P[A|\neg X \land\neg Y]$. We’re given the following probabilities:
- $P[A|X]$
- $P[A|\neg X]$
- $P[A| Y]$
- $P[A|\neg Y]$
We also understand how $X$ and $Y$ interact with just each other, and how the act on their own. For example, we know things like $P[X\land Y], P[X\land \neg Y]$, P[X], etc.
My solution involved attempting to set up a system for the four values we wish to find. The proof of one of the equations is shown below, with the others found similarly.
$$P[A\land X]=P[A\land X\land Y]+ P[A\land X\land \neg Y]$$ $$P[X]P[A|X]=P[X\land Y]P[A| X\land Y] + P[A|X \land \neg Y]P[A|X\land \neg Y]$$
We can then construct similar equations for each of the 4 given conditionals involving $A$:
$$ \begin{cases} P[X]P[A|X]=P[X\land Y]P[A| X\land Y] + P[A|X \land \neg Y]P[A|X\land \neg Y] \\ P[\neg X]P[A|\neg X]=P[\neg X\land Y]P[A| \neg X\land Y] + P[A|\neg X \land \neg Y]P[A|\neg X\land \neg Y] \\ P[Y]P[A|Y]=P[X\land Y]P[A| X\land Y] + P[A|\neg X \land Y]P[A|\neg X\land Y] \\ P[\neg Y]P[A|\neg Y]=P[X\land \neg Y]P[A| X\land \neg Y] + P[A|\neg X \land \neg Y]P[A|\neg X\land \neg Y] \\ \end{cases} $$
While we have four equations and four unknowns, the equations are not linearly independent. For example, $\text{eq}4=\text{eq}1+\text{eq}2-\text{eq}3$.
Is there are way to get these probabilities? I’m leaning towards no, but it’s possible there’s an equation I can throw in that would make the system possible, and I just can’t find that equation.
If it is impossible, what information do you need to compute these probabilities normally? What are the formulas typically used to deal with more complicated conditionals?
You can't calculate the probabilities like $P(A \mid X \land Y)$ in general because you can't distinguish between cases like the two below. In case 1, $P(A \mid X \land Y) < 1$, but in case 2, $P(A \mid X \land Y) = 1$.