Is it necessarily true that$P(A\lvert B) = 1 \implies P(A\lvert C)=1$ where $C\subseteq B$

Question

I've been trying my luck at the seemingly trivial and intuitive statement that $P(A\lvert B) = 1 \implies P(A\lvert C)=1$ where $C\subseteq B$ but I have struggled to prove it.

My attempts:

$$P(A\lvert C)= \frac{P(A\cap C)}{P(C)}\geq\frac{P(A\cap C)}{P(C)}\times \frac{P(C)}{P(B)}=\frac{P(A\cap C)}{P(C)}\times\frac{P(C\cap B)}{P(B)}=\frac{P(A\cap C)}{P(B)}\times\frac{P(C\cap B)}{P(C)}\\ $$

$$=\frac{P(A\cap C)}{P(B)}\times P(B\lvert C)$$

I do not know where I am going with this... any hints?

Answer 1

$P(A\mid B)=1$ means that $B\setminus A$ is a set of (probability) measure zero. Then so is $C\setminus A$.

Answer 2

If $P(A\mid B)=1$ then $P(AB) = P(B)$. But also $P(B)=P(AB)+P(A^cB)$, so this means $P(A^cB)=0$. Now, since $C\subset B$, we have $P(C)=P(BC)$, which we can write as $$ P(C)=P(BC)=P(ABC) + P(A^cBC).$$ Now argue that the first term on the RHS equals $P(AC)$, while the second term is zero.

Answer 3

Remember that when we say $P(A) = 1$, it means that $A$ happens "almost surely" (or in other words, as far as the probability is concerned, $A$ always happens). So $P(A|B)$ means that whenever $B$ happens, $A$ happens. Saying that $C \subseteq B$ means that if $C$ happens, $B$ surely happens (not even almost surely, this time, surely!). So if $C$ happens, $B$ surely happens, and if $B$ happens, $A$ happens almost surely... so it's a bit like $C \implies B \implies A$ but with some probability jargon around it.

More formally, $1 = P(A|B) = \frac{P(A \cap B)}{P(B)}$, so $$ P(A \cap B) = P(B) = P(A \cap B) + P(A^c \cap B) $$ The second equality follows from the measurability of $B$, which is required to consider probabilities. It follows that $P(A^c \cap B) = 0$. Since $C \subseteq B$, we also have $$ 0 \le P(A^c \cap C) \le P(A^c \cap B) = 0 \implies P(A^c \cap C) = 0. $$ Running our above equalities in reverse, we get $$ P(C) = P(A \cap C) + P(A^c \cap C) = P(A \cap C) \implies P(A|C) = 1. $$ Hope that helps,

Answer 4

Algebraically, consider two conditional probabilities:

$$\begin{align*} P(A\mid C) &= \frac{P(A\cap C)}{P(C)}\\ P(A\mid(B\setminus C)) &= \frac{P(A\cap (B\setminus C))}{P(B\setminus C)} \end{align*}$$

The mediant of $P(A\mid C)$ and $P(A\mid(B\setminus C))$ (in the above fractional forms) is

$$\frac{P(A\cap C) + P(A\cap(B\setminus C))}{P(C) + P(B\setminus C)} = \frac{P(A\cap B)}{P(B)} = P(A\mid B)$$

using $C\subseteq B$ and therefore $C \cup (B\setminus C) = B$.

If (assuming the contrary) that $P(A\mid C) \ne P(A\mid (B\setminus C))$, then by the mediant inequality,

$$\begin{align*} \min(P(A\mid C), P(A\mid (B\setminus C))) < P(A\mid B) &< \max(P(A\mid C), P(A\mid (B\setminus C)))\\ 1 &< \max(P(A\mid C), P(A\mid (B\setminus C))) \end{align*}$$

which contradicts that $P(A\mid C)$ and $P(A\mid (B\setminus C))$ are probabilities.

Therefore the opposite holds,

$$P(A\mid C) = P(A\mid (B\setminus C)) = P(A\mid B) = 1$$

(unless $C=\emptyset$)