$P(A|B) > P(A) \implies P(A^c|B^c) > P(A^c)$ Intuition

Question

I want to better understand the intuition behind this: $P(A|B) > P(A) \implies P(A^c|B^c) > P(A^c)$ i.e.

The proof is fairly strait-forward, but I'm having trouble linking the mechanics of the proof to concrete intuition (I've drawn a diagram attached to help aide any explanation). I.e. I don't want to fell like im just randomly moving symbols around.

I'm particularly interested in a step in the proof which is the third line below:

$P(A^c|B^c) = \frac{P(A^c \cap B^c)}{1 - P(B)}$

$P(A^c|B^c) = \frac{1-(P(A)+P(B)-P(A \cap B))}{1 - P(B)}$

$P(A^c|B^c) > \frac{1-(P(A)+P(B)-P(A)P(B))}{1 - P(B)} = P(A^C)$

This uses an implication from our given fact $P(A|B) > P(A) \implies P(A \cap B) > P(A)P(B)$

If anyone can give some kind of geometric or intuitive explanation what is happening here when we plug this into the formulae. What are we actually saying? What does this mean?

(E.g. it's very easy to see De Morgan's Law visually. I'm searching for some kind of intuition to unwrap the third line of the proof.) Thanks!

Question Summary

1. Intuition in general behind this idea.
2. intuition behind the third line of the proof. Thanks!

Answer 1

I think one of the brilliant things about algebraic manipulation is that while it is often insightful to understand intuitively all the steps, you will still arrive at correct results by just following the rules without thinking too much. I think it is much more important to get intuition for the results rather than the manipulations, so that's where I wish to help.

I can't really see the exact boundaries of $A$ and $B$ on your drawing. But I think you can visualize the following:

Imagine $A$ and $B$ as areas as in your drawing. $P(A\mid B)>P(A)$ means that when you are inside $B$, the part of $A$ contained in $B$ takes up a bigger fraction of the space in $B$ than $A$ does of the entire set $S$. You can imagine it as sets as well: $S=\lbrace 1,2,\dots,10\rbrace$, $A=\lbrace 1,2,3,4\rbrace$, $B=\lbrace 2,3,4,5\rbrace$.

Then it should be intuitive that the area outside $A$ intersecting the area outside $B$ takes up relatively more space than the area outside $A$ does in all of $S$. Indeed, as $A$ and $B$ intersect a lot, we should expect their complements to do so as well. In our set example, $A^c=\lbrace 5,6,7,8,9,10\rbrace$, $B^c=\lbrace 1,6,7,8,9,10\rbrace$, with intersection $A^c\cap B^c = \lbrace 6,7,8,9,10\rbrace$. This exactly means that $P(A^c\mid B^c)>P(A^c)$.

Answer 2

The best intuition for me would focus on "the dog who didn't bark." Suppose $A$ is the event that there is a stranger in the house, and $B$ is the event that the dog barks. Then $A^c$ is the event that there is no stranger, and $B^c$ is the event that the dog doesn't bark. Let's say that we think that $P(A|B)>P(A)$, meaning that the fact that the dog barked makes it more likely that there was a stranger. The implication is then that, if we heard no dog barking, then it becomes more likely that there was no stranger. The basic idea is that we can take the relationship between events and automatically get information between their complements (or duals).

There is a bit of an issue in the first line of your proof, where you write $P(A) \cap P(B)$, which we don't have a definition for. An amended proof might go:

$$ \begin{aligned} P(A^c | B^c) &= 1 - P(A | B^c) \\ &= 1 - \dfrac{P(A\cap B^c)}{P(B^c)}, \text{ definition of conditional prob.}\\ &= 1 - \dfrac{P(A\cap B^c)}{1-P(B)}, \text{ law of total prob.}\\ &= 1 - P(A) + P(A) - \dfrac{P(A\cap B^c)}{1-P(B)}, \text{ adding and subtracting $P(A)$} \\ &= 1 - P(A) + \dfrac{P(A) - P(A)P(B) - P(A\cap B^c)}{1-P(B)}, \text{ algebra} \\ &= 1 - P(A) + \dfrac{(P(A\cap B) + P(A\cap B^c)) - P(A)P(B) - P(A\cap B^c)}{1-P(B)}, \text{ law of total prob.}\\ &= 1 - P(A) + \dfrac{P(A\cap B) - P(A)P(B)}{1-P(B)}, \text{ algebra} \\ &> 1 - P(A), \text{ so long as $P(A\cap B) > P(A)P(B)$} \\ &= P(A^c) \\ \end{aligned} $$

But you might notice that, so long as $P(B)>0$, we have that $P(A\cap B) > P(A)P(B)$ if and only if $\dfrac{P(A\cap B)}{P(B)} > P(A)$. The left hand side is just $P(A|B)$. This was also the third line of your proof. For me, the best intuition for this statement (i.e. $P(A\cap B) > P(A)P(B)$) is that $A$ and $B$ provide us with information about each other. That's the meaning of the original hypothesis that $P(A|B)>P(A)$. If that's the case, then there is a better chance that the two happen together, or $P(A\cap B)$, than there is that they happen independently, $P(A)P(B)$. The latter includes, most notably, the rare possibilities that one happens without the other and vice versa. However, once we know that one of these two events happened, we know that it is more likely that the other happened too.