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.


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.



$$\def\Pr{\mathop{\mathsf P}} \Pr(A\mid B)~{=\Pr(A\mid B\cap C)\Pr(C\mid B)+\Pr(A\mid B\cap C^\complement)\Pr(C^\complement\mid B)\\ =\Pr(A\mid B\cap C)\Pr(C\mid B)+\Pr(A\mid B\cap C^\complement)(1-\Pr(C\mid B))\\=(\Pr(A\mid B\cap C)-\Pr(A\mid B\cap C^\complement))\Pr(C\mid B)+\Pr(A\mid B\cap C^\complement)}\tag 1$$

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:

  • $(\Pr(C\mid B)-\Pr(C))=0$ or,
  • $(\Pr(A\mid B\cap C)+\Pr(A\mid B\cap C^\complement))=0$ .

Now the first case occurse exactly when $B,C$ are indepencent, by definition.

Exactly when does the second case happen?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.