Suppose that the following identity holds. $$P(A|B)=P(A|B \cap C)P(C)+P(A|B \cap C^C)P(C^C)$$ Also assume that, $P(A|B \cap C) \neq P(A|B)$ and $P(A)>0$. Then we have to show that $B$ and $C$ are independent events, i.e. $P(B \cap C)=P(B)P(C)$.
Note : The original problem was deriving the identity, assuming independence of $B$ and $C$, which can be shown easily. I'm stuck at this converse version.
Source : Rohatgi, Saleh - Problem $9$, page $36$.
Any help would be much appreciated. Thank you.
On
Generally:
$$\def\Pr{\mathop{\mathsf P}} \Pr(A\mid B)~{=\Pr(A\mid B\cap C)\Pr(C\mid B)+\Pr(A\mid B\cap C^\complement)\Pr(C^\complement\mid B)\\ =\Pr(A\mid B\cap C)\Pr(C\mid B)+\Pr(A\mid B\cap C^\complement)(1-\Pr(C\mid B))\\=(\Pr(A\mid B\cap C)-\Pr(A\mid B\cap C^\complement))\Pr(C\mid B)+\Pr(A\mid B\cap C^\complement)}\tag 1$$
And so, when $C, B$ are (pairwise) independent:
$$\Pr(A\mid B)~{=\Pr(A\mid B\cap C)\Pr(C)+\Pr(A\mid B\cap C^\complement)\Pr(C^\complement)\\=(\Pr(A\mid B\cap C)-\Pr(A\mid B\cap C^\complement))\Pr(C)+\Pr(A\mid B\cap C^\complement)}\tag 2$$
So when $(2)$ is true and given that $(1)$ generally is, then $$(\Pr(A\mid B\cap C)+\Pr(A\mid B\cap C^\complement))\cdot(\Pr(C\mid B)-\Pr(C))=0\tag 3$$
So $(2)$ is true when either:
Now the first case occurse exactly when $B,C$ are indepencent, by definition.
Exactly when does the second case happen?
It suffices to consider the case in which $P(B)>0$. Provided this, note that $$P(A|B)=\frac{P(A\cap B)}{P(B)},$$ while $$P(A\cap B)=P(A|B\cap C)P(B\cap C)+P(A|B\cap C^c)P(B\cap C^c).$$ Therefore, $$\tag{*}P(A|B)=P(A|B\cap C)\frac{P(B\cap C)}{P(B)}+P(A|B\cap C^c)\frac{P(B\cap C^c)}{P(B)}\\=P(A|B\cap C)P(C|B)+P(A|B\cap C^c)P(C^c|B)$$ Use the identity, we get $$0=[P(A|B\cap C)-P(A|B\cap C^c)][P(C|B)-P(C)].$$ Use the condition that $P(A|B)\ne P(A|B\cap C)$ (indeed, if $P(A|B\cap C)=P(A|B\cap C^c)$, we would have $P(A|B)=P(A|B\cap C)[P(C|B)+P(C^c|B)]=P(A|B\cap C)$ by (*), a contradiction), we conclude that $P(C|B)=P(C)$, which yields independence.