How to show P(A|A∪B) ≥ P(A|B)?

Question

How would one show $P(A\,|\,A\cup B)\ge P(A\,|\,B)$?

I first tried to expand the LHS using the definition of conditional probability and therefore it becomes:

$\frac{P(A \cap (A \cup B))}{P(A \cup B)}$

Then I tried to use distribution law to further expand the numerator and get:

$P((A \cap A) \cup (B \cap A))$

Since $A \cap A$ is just A and it becomes $P(A \cup (B \cap A))$.

However, if I use distribution law again, it looks like it will go back to the previous step and I'm just repeating doing the same thing. In the solution, it mentioned the total probability law, but I'm wondering how to link these two together and what should be the correct way to prove this.

Answer 1

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

On the other hand,

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

Now, let $x=P(A\cap B), y=P(B), z=P(A\setminus B)$. The original inequality now boils down to:

$$\frac{x+z}{y+z}\ge\frac{x}{y}$$

which is equivalent to $x\le y$ i.e. $P(A\cap B)\le P(B)$ - which is certainly true.

Answer 2

Break things down into the disjoint components $P(A\setminus B)$, $P(A\cap B)$, and $P(B\setminus A)$. That is, $$ \begin{align} P(A)P(B) &=\overbrace{(P(A\setminus B)+P(A\cap B))}^{P(A)}\overbrace{(P(B\setminus A)+P(A\cap B))}^{P(B)}\\ &=\underbrace{(P(A\setminus B)+P(A\cap B)+P(B\setminus A))}_{P(A\cup B)}P(A\cap B)+P(A\setminus B)P(B\setminus A)\\ \end{align} $$ Thus, $$ \begin{align} P(A)P(B) &=P(A\cup B)P(A\cap B)+\overbrace{P(A\setminus B)P(B\setminus A)}^{\ge0}\\ &\ge P(A\cup B)P(A\cap B) \end{align} $$ Therefore, since $A\cap(A\cup B)=A$, we get by cross-multiplication $$ \underbrace{\frac{P(A)}{P(A\cup B)}}_{P(A|A\cup B)}\ge\underbrace{\frac{P(A\cap B)}{P(B)}}_{P(A|B)} $$

Answer 3

A variational proof.

AFFIRMATION 01 If there are four numbers $L$, $A$, $a$ and $\ell$ such that $$ L+\ell = A+a \qquad \mbox{ and } \qquad L \geq A \geq a \geq\ell \geq 0 \qquad (\,\ast \, ) $$ then necessarily $$ A\cdot a \geq L\cdot \ell \geq 0 \qquad (\,\ast \,\ast \, ). $$ If the inequalities in $(\,\ast \, )$ are strict then the inequalities in $(\,\ast \ast, )$ are strict. Equality in $(\,\ast\, \ast \,)$ is valid if, and only if, $L=\ell $.

The proof is immediate. By $(L-\ell)\geq (A-a) $ and $$ A\cdot a = \frac{1}{4}\left( A+a \right)^2 - \frac{1}{4}\left( A-a \right)^2 \\ L\cdot \ell = \frac{1}{4}\left( L+\ell \right)^2 - \frac{1}{4}\left( L-\ell\right)^2 $$ we have $A\cdot a- L\cdot \ell \geq 0$.

AFFIRMATION 02 By Inclusion-exclusion principle we have for all events $A$ and $B$ $ P(A\cup B) + P(A \cap B ) = P(A) + P(B). $ Note that $P(A \cap B ) \leq P(A) \leq P(A\cup B)$ and $P(A \cap B ) \leq P(B) \leq P(A\cup B)$. By AFFIRMATION 01 we have $$ P(A\cup B) \cdot P(A \cap B ) \leq P(A) \cdot P(B) \qquad (\,\ast \,\ast \,\ast \,) $$

Commonly the conditional probability $P(A \mid B)$ is defined as $P(A \mid B)=\dfrac{P(A \cap B)}{P(B)}$. But not to lose the definition in the case where $P(B)=0$ we can define the conditional probability $P(A \mid B)$ as the single number that multiplies $P(B)$ to make it equal to $P(A \cap B)$. In other words, $$ P(A \cap B) = P(A \mid B) \cdot P(B) \qquad (\,\ast \,\ast \,\ast \,\ast \,) $$ Thus, depending on the context of the problem in the case of $P(B)=0$ one can consider $P(A \mid B)=1$ or $P(A \mid B)=0$ or some other convenient value.

Let's use the definition $(\,\ast \,\ast \,\ast \,\ast \,)$ above to prove the inequality $P(A \mid A \cup B) \geq P(A \mid B)$. Assuming otherwise there would be events $A$ and $B$ such that
$$ P(A \mid A \cup B) < P(A \mid B). $$ Multiplying both sides of the above inequality by $P(A\cup B) \cdot P(B)\geq 0$ $$ P(A \mid A \cup B)P(A\cup B) \cdot P(B) \leq P(A \mid B) \cdot P(B) \cdot P(A\cup B). $$ By $P(A)=P(A \cap ( A \cup B)) = P(A \mid A \cup B)P(A\cap B)$ and $P(A \cap B) = P(A \mid B) \cdot P(B)$ $$ P(A) \cdot P(B) \leq P(A \cap B) \cdot P(A\cup B). $$ But this contradicts Affirmation 02. So we can only conclude that there are no events $A$ and $B$ such that $ P(A \mid A \cup B) < P(A \mid B)$. Thus, for every event $A$ and $B$ the opposite inequality holds $$ P(A \mid A \cup B) \geq P(A \mid B) $$

Answer 4

Absorption.

$\begin{align}&\quad A\cap(A\cup B) \\&= (A\cap A)\cup(A\cap B)&&\text{Distribution}\\&=(A)\cup(A\cap B)&&\text{Idempotence}\\&=(A\cap(B^\complement\cup B))\cup(A\cap B)&&{\cap}\text{ Identity}\\&=((A\cap B^\complement)\cup(A\cap B))\cup (A\cap B)&&\text{Distribution}\\&=(A\cap B^\complement)\cup((A\cap B)\cup(A\cap B))&&\text{Association}\\&=(A\cap B^\complement)\cup(A\cap B)&&\text{Idempotence}\\&=A\cap(B^\complement\cup B)&&\text{Distribution}\\&=A&&{\cap}\text{ Identity}\end{align}$