How is $P(A|B') = P(A|B)$ the definition for independence of events $A,B$?

Question

Notations: $P(XY) := P(X \cap Y)$ and $X' := X^{\small \complement}$.

I read from a comment on MSE that $P(A | B) = P(A)$ and $P(A | B) = P(A | B')$ are equivalent definitions of independence of two events and they both lead to $P(AB) = P(A)P(B)$. The first definition now seems rather natural to me, thanks to @lulu and @Ryan G.

We assume that the probabilities $0 < P(A), P(B) < 1$.

Derivation from the first def.: $P(A|B) = P(A) \iff \frac{P(AB)}{P(B)} = P(A) \iff P(AB) = P(A)P(B)$.

Derivation from the second def.: $P(A|B) = P(A|B') \iff \frac{P(AB)}{P(B)} = \frac{P(AB')}{P(B')}$.

Now if they are equivalent, then $\frac{P(AB')}{P(B')} = P(A) \iff P(A | B') = P(A)$ which is not necessarily the case.

Can anyone please confirm this or tell me what exactly is the statement/assumption that I am missing?

Answer 1

Assuming $P(A|B)=P(A|B')$, we have \begin{align*}& & \frac{P(AB)}{P(B)} &= \frac{P(AB')}{P(B')} \\ &\Leftrightarrow &P(B')P(AB) &= P(B)P(AB') \\ &\Leftrightarrow & (1-P(B))P(AB) &= P(B)P(AB') \\ &\Leftrightarrow & P(AB) &= P(B)(P(AB')+P(AB)), \end{align*} and since $P(AB') + P(AB) = P(A)$ we have shown $P(A|B) = P(A|B')$ is equivalent to $P(AB) = P(B)P(A)$.

Answer 2

We will prove the equivalence of the definitions.

The case for $P(A|B)=P(A)\implies P(A|B)=P(A|B')$ is trivial.

The other case, for $P(A|B)=P(A|B')\implies P(A|B)=P(A)$ can be proved using the law of total probability:

$$P(A)=\\P(B)P(A|B)+P(B')P(A|B')=\\P(B)P(A|B)+ (1-P(B))P(A|B)=\\P(A|B)$$

Where in the second equality we used the assumption about $P(A|B)=P(A|B')$ and that $P(B')=1-P(B)$

Answer 3

If $A$ is independent of $B$ then $P(AB')=P(A)-P(AB)=P(A)-P(A)P(B)=P(A)(1-P(B))=P(A)P(B')$, so $A$ is also independent of $B'$. Hence $P(A\mid B)=P(A\mid B')=P(A)$.

Conversely, if $P(A\mid B)=P(A\mid B')$ then multiplying both sides by $P(B)P(B')=P(B)(1-P(B))$ gives $P(AB)(1-P(B))=P(AB')P(B)$, that is $P(B)(P(AB')+P(AB))=P(AB)$, which is the same as $P(B)P(A)=P(AB)$ since $P(AB')+P(AB)=P(A)$.

Answer 4

1. Let $0<P(B)<1.$ Then $$\text{the probability of event }A \textbf{ does not depend on } \;\textbf{whether or not}\; \textbf{ event $B$ occurs} \\\iff P(A|B)=P(A|B^c) \\\iff \frac{P(A\cap B)}{P(B)}=\frac{P(A\cap B^c)}{P(B^c)} \\ =\frac{\text n(A\cap B^c)}{\text n(B^c)} \\ =\frac{\text n(A)-\text n(A\cap B)}{\text n(B^c)} \\ =\frac{P(A)-P(A\cap B)}{P(B^c)} \\ =\frac{P(A)-P(A\cap B)}{1-P(B)} \\\iff P(A\cap B)=P(A)P(B).$$ Notice that this intuitive characterisation of pairwise independence excludes the cases where the subjects' probabilities are $(0,0),(1,1),(0,1)$ or $(1,0).$

2. On the other hand, the more common intuitive characterisation $$\text{the probability of event }A \textbf{ is unaffected by the knowledge that event $B$ occurs},$$ i.e., $P(A|B)=P(A),$ excludes only the $(0,0)$ case.

3. In contrast, the formal definition—by sidestepping division by $0$—applies with no restriction on the subjects' probabilities.

   Apart from the restrictions, all three are equivalent to one another.

Answer 5

We can use Bayes theorem too for proving equivalence. Let's show that from second definition we can readily obtain the first definition. Going from the other way will be similar.

$P(B|A)$

$=\frac{P(A|B)P(B)}{P(A|B)P(B)+P(A|B')P(B')}$, by Bayes

$=\frac{P(A|B)P(B)}{P(A|B)P(B)+P(A|B)P(B')}$, since $P(A|B)=P(A|B')$, by second definition of independence

$=\frac{P(A|B)P(B)}{P(A|B)(P(B)+P(B'))}$

$=\frac{P(A|B)P(B)}{P(A|B)}$, since $P(B)+P(B')=1$, by law of countable unions

$=P(B)$, assuming $P(A|B) \neq 0$.

Hence, we have, $P(B|A)=P(B)$, and this is precisely another way of writing the first defination of independence (swap A and B).

Answer 6

There are only so many ways to prove the equivalence of the two forms of independence. However, as with most somewhat difficult concepts, a particular approach may resonate better with someone than other approaches. So at the risk of possibly duplicating someone else's formulas, I will present my approach along with my explanations.

---

Equivalence of the Two Forms

Suppose that $$ P(A\mid B)=P(A\mid B')\tag1 $$ Then $$ \begin{align} P(A) &=P(A\cap B)+P(A\cap B')\tag{2a}\\[3pt] &=P(A\mid B)P(B)+P(A\mid B')P(B')\tag{2b}\\[3pt] &=P(A\mid B)P(B)+P(A\mid B)(1-P(B))\tag{2c}\\[3pt] &=P(A\mid B)\tag{2d} \end{align} $$ Explanation:
$\text{(2a)}$: Disjoint Union
$\text{(2b)}$: Bayes' Theorem
$\text{(2c)}$: apply $(1)$ and Disjoint Union
$\text{(2d)}$: simplify

Suppose that $$ P(A\mid B)=P(A)\tag3 $$ Then $$ \begin{align} P(A\mid B') &=\frac{P(A\cap B')}{P(B')}\tag{4a}\\ &=\frac{P(A)-P(A\cap B)}{P(B')}\tag{4b}\\ &=\frac{P(A\mid B)-P(A\mid B)P(B)}{P(B')}\tag{4c}\\[5pt] &=P(A\mid B)\tag{4d} \end{align} $$ Explanation:
$\text{(4a)}$: Bayes' Theorem
$\text{(4b)}$: Disjoint Union
$\text{(4c)}$: apply $(3)$ and Bayes' Theorem
$\text{(4d)}$: Disjoint Union and simplify

---

Disjoint Union

By Disjoint Union, we mean that when $P(A\cap B)=0$, we have $P(A\cup B)=P(A)+P(B)$. This can be viewed as an application of the Inclusion-Exclusion Principle. We use it here to say that $P(A\cap B)+P(A\cap B')=P(A)$ and $P(B)+P(B')=1$.

Answer 7

How is $P(A|B′)=P(A|B)$ the definition for independence of events $A,B$?

The equality says that whether given that $B$ occurs or not, we shall have the same expectation that $A$ occurs. IE: Our probability measure for $A$ is not dependent on the occurrence of $B$.

That means exactly that: event $A$ is independent of event $B$.