Probability with a joined condition

Question

I want to know the probability $P(A|X,Y)$, given that I know $P(A|X)$, $P(A|Y)$, $P(A)$, $P(X)$, $P(Y)$ and given, that $X$ and $Y$ are independent. I'm also going to assume that $X$ and $Y$ are conditionaly independent.

My shot at the solution: $$P(A|X,Y)=\frac{P(X,Y|A)P(A)}{P(X,Y)}=\frac{P(X|A)P(Y|A)P(A)}{P(X)P(Y)}$$ $$P(X|A)=\frac{P(A|X)P(X)}{P(A)}$$ $$P(A|X,Y)=\frac{P(A|X)P(A|Y)}{P(A)}$$

My problem with this solution, that if $P(A)$ is sufficiently small, the final probability is greater than 1, which should not be possible.

What am I doing wrong?

edit

My guess is that if I assume independence and conditional independence together I have some restrictions to what values for $P(A)$ I can choose. But I can't prove or imagine this.

Answer 1

You cannot assume two events are conditionally independent (given a third) simply because they are pairwise independent.   Witness the following counter example:

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Take the sample space $\Omega = \{1,2,3,4\}$ with unbiased probability for all outcomes, $\Bbb P(\{\omega\})=\tfrac 14~\mathbf 1_{\omega\in\Omega}$, and consider the events: $$X=\{1,2\}, Y=\{2,3\}, A=\{1,2,3\}$$

Then $\Bbb P(X)=\tfrac 1 2, \Bbb P(Y)=\tfrac 12, \Bbb P(X,Y)=\tfrac 14$ so $X$ and $Y$ are pairwise independent.

However $\Bbb P(X\mid A)=\tfrac 23, \Bbb P(Y\mid A)=\tfrac 23, \Bbb P(X,Y\mid A)=\tfrac 1 3$, and $\tfrac 13\neq \tfrac 49$, so they are not conditionally independent given $A$.

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Now if we are discussing events which are also conditionally independent then:

Let $P_A=\Bbb P(A), P_{A^\complement}=\Bbb P(A^\complement), P_X=\Bbb P(X), P_{X\mid A}=\Bbb P(X\mid A)$ and so on

$$\begin{align} P_{X,Y\mid A} P_A + P_{X,Y\mid A^\complement} P_{A^\complement} ~=~& P_{X,Y} \\ P_{X\mid A} P_{Y\mid A}P_A + P_{X\mid A^\complement} P_{Y\mid A^\complement} P_{A^\complement} ~=~& P_X P_Y \\ ~=~& (P_{X\mid A} P_A + P_{X\mid A^\complement} P_{A^\complement}) (P_{Y\mid A} P_A + P_{Y\mid A^\complement} P_{A^\complement}) \\ ~=~& P_{X\mid A} P_{Y\mid A} P_A^2 + (P_{X\mid A} P_{Y\mid A^\complement} + P_{X\mid A^\complement} P_{Y\mid A}) P_A P_{A^\complement} + P_{X\mid A^\complement} P_{Y\mid A^\complement} P_{A^\complement}^2 \\ P_{X\mid A} P_{Y\mid A} P_A P_{A^\complement} + P_{X\mid A^\complement} P_{Y\mid A^\complement} P_A P_{A^\complement} ~=~& (P_{X\mid A} P_{Y\mid A^\complement} + P_{X\mid A^\complement} P_{Y\mid A}) P_A P_{A^\complement} \\ P_{X\mid A} (P_{Y\mid A}-P_{Y\mid A^\complement}) P_A P_{A^\complement} ~=~& P_{X\mid A^\complement} (P_{Y\mid A}-P_{Y\mid A^\complement}) P_A P_{A^\complement} \end{align}$$

So either $\Bbb P(X\mid A)=\Bbb P(X\mid A^\complement)$ , $\Bbb P(Y\mid A)=\Bbb P(Y\mid A^\complement)$ , $\Bbb P(A)=0$ , or $\Bbb P(A)=1$

However, if $\Bbb P(X\mid A)=\Bbb P(X\mid A^\complement)$ , then $\Bbb P(X)=\Bbb P(X\mid A)\Bbb P(A)+\Bbb P(X\mid A)\Bbb P(A^\complement) \\ =\Bbb P(X\mid A)$

Thus if $X$ and $Y$ are both pairwise independent and conditionally independent given $A$, then either: $X$ is independent of $A$, $Y$ is independent of $A$, or $A$ is either almost surely certain or almost surely impossible.

My guess is that if I assume independence and conditional independence together I have some restrictions to what values for $\Bbb P(A)$ I can choose. But I can't prove or imagine this.

$$\Bbb P(A)\in\{0, \Bbb P(A\mid X), \Bbb P(A\mid Y), 1\}$$