Proof that $P(A \mid C) = 1$ implies $P(A \mid B∩ C) = 1$

Question

This isn't homework. I'm reading Probabilistic Graphical Models by Koller et al.$\space$, and an easy problem in chapter $3$ made me think of a more general problem (which I'm now stuck on). I have everything in place except for this:

Let $P(B \cap C) \neq 0$. Then $P(A \space|\space B ∩ C) = 1$ if $P(A \mid C) = 1$. It's intuitively obvious, but I haven't been able to formally prove it. I get

$$P(A \mid B∩C) = \dfrac{P(A∩B∩C)}{P(B∩C)} = \dfrac{P(C) P(A\mid C) P(B\mid A∩C)}{P(B∩C)} = \dfrac{P(B\mid A∩C)}{P(B|C)}$$

I don't see why that ratio on the right hand side is $1$. Is it? What am I not seeing?

Answer 1

Assume that $\mathbb P(\ \mid C)$ and $\mathbb P(\ \mid B\cap C)$ both exist, that is, that $$\mathbb P(B\cap C)\ne0. $$ Then note that $\mathbb P(A\mid C)=1$ if and only if $\mathbb P(A\cap C)=\mathbb P(C)$ if and only if $\mathbb P(C\setminus A)=0$. Likewise, $\mathbb P(A\mid B\cap C)=1$ if and only if $\mathbb P((B\cap C)\setminus A)=0$. But $(B\cap C)\setminus A\subseteq C\setminus A$, hence the latter implies the former, that is, $$ \mathbb P(A\mid C)=1\implies \mathbb P(A\mid B\cap C)=1. $$

Answer 2

Mmmm ... I don't think it's obvious, really. In fact, I think there are a couple of ready counter-examples.

First, let's say $A\cap C =C$. That is, anytime we have $C$, we have $A$. But let's say $B\cap A=\emptyset$. Then, $P(A|C)=1$. But $P(A|B)=0$, and therefore $P(A|B,C)=0$.

Also, we could say $A\cap C =C$ and $A\cap B=B$, but $B\cap C=\emptyset$. Then the probability of anything given $B$ and $C$ is zero.

AFAICT, your first statement implies the second only in the special case where 1) $A\cap C =C$ and 2) $A\cap (B\cap C) = B\cap C$.

Answer 3

$\mathbb{P}\{A|C\}=1$ implies $\mathbb{P}\{A,C\}=\mathbb{P}\{C\}$, which implies $A\subset C$. Therefore $\mathbb{P}\{A|B,C\}=1$.

Answer 4

For any three events $A,B,C$ ,$P(A∩B∩C)=P[(A ∩C) ∩(B∩C)]=P(A ∩C)+ P(B∩C)-P[(A ∩C) ∪(B∩C)]$$=P(A ∩C)+ P(B∩C)-P[(A∪B)∩C]$$=P(A ∩C)+ P(B∩C)-P(C)-P(A∪B)+P(A∪B∪C)$ . Now $\space$$\space$

$(A ∪ B) ⊆ (A ∪ B ∪C)$

$\implies$ $ P(A∪B) ≤ P(A∪B∪C)$

$\implies$$P(A ∩C)+ P(B∩C)-P(C)$

$\space$ $\space$ $\space$ $\space$$≤P(A ∩C)+ P(B∩C)-P(C)-P(A∪B)+P(A∪B∪C)$

$\implies$$P(A ∩C)+ P(B∩C)-P(C)≤P(A∩B∩C)$....(i) Moreover,

$(A∩B∩C)⊆(B∩C)$$\implies$$P(A∩B∩C)≤P(B∩C)$ , so this along with (i) gives,

$P(A ∩C)-P(C)≤P(A∩B∩C)-P(B∩C)≤0$ . Hence ,

$P(A|C)=1 \implies P(A ∩C)=P(C) \implies$$0≤P(A∩B∩C)-P(B∩C)≤0$

$\implies P(A∩B∩C)=P(B∩C)$$\implies P\big(A|(B∩C)\big)=1$