Property of Probability Set Function

Theorem

Let $\mathcal{B}$ : Set of events

Then

1. $P(A)=1-P(A^C),\forall A\in \mathcal{B}$
2. $P(\varphi )=0$
3. $A\subset B⟹P(A)\le P(B)$
4. $0\le P(A)\le 1,\forall A\in \mathcal{B}$
5. $A,B\subset \mathcal{C}⟹P(A\cup B)=P(A)+P(B)-P(A\cup B)$

Proof

Let

• $\mathcal{C}$ : Sample space
• $\mathcal{B}$ : Set of events

1. $P(A)=1-P(A^C),\forall A\in \mathcal{B}$

Let $A$ be any subset of $\mathcal{B}$

We have $\mathcal{C}=A\cup A^C$ and $A\cap A^C=\varphi$. Thus from axiom 2 and 3 of probability set function, it follows that

$$P(\mathcal{C})=P(A\cup A^C)=P(A)+P(A^C)=1$$

2. $P(\varphi )=0$

Based on point 1, we have that $P(A)=1-P(A^C)$.

Choose $A=\varphi$. Because ${\varphi }^C=\mathcal{C}$, we have

$$\begin{array}{rl}P(\varphi )& =1-P({\varphi }^C)\\ & =1-P(\mathcal{C})\\ & =1-1\\ & =0\end{array}$$

3. $A\subset \mathcal{B}⟹P(A)\le P(\mathcal{B})$

Let $B$ be any subset of $\mathcal{B}$ and let $A$ be any subset of $A$.

Now $B=A\cup (A^C\cap B)$ and $A\cap (A^C\cap B)=\varphi$.

Hence from 3rd axiom of probability set function,

$$P(B)=P(A)+P(A^C\cap B)$$

From 1st axiom of probability set function, we have $P(A^C\cap B)\ge 0$.

Now we have

$$\begin{array}{rlrl}& & P(B)& =P(A)+P(A^C\cap B)\\ ⟺& & P(A^C\cap B)& =P(B)-P(A)\ge 0\\ ⟺& & P(B)& \ge P(A)\end{array}$$

4. $0\le P(A)\le 1,\forall A\subset \mathcal{B}$

Since $\varphi \subset A\subset \mathcal{C}$, by point 3 we have that

$$P(\varphi )\le P(A)\le P(\mathcal{C})⟹0\le P(A)\le 1$$

5. $A,B\subset \mathcal{C}⟹P(A\cup B)=P(A)+P(B)-P(A\cup B)$

Let $A$ and $B$ be any subset of $\mathcal{C}$.

We have

$$\begin{array}{rlr}A\cup B& =U\cap (A\cup B)& \text{Identity law}\\ & =(A\cup A^C)\cap (A\cup B)& \text{Complement law}\\ & =A\cup (A^C\cap B)& \text{Distributive law}\\ & & \\ B& =U\cap B& \text{Identity law}\\ & =(A\cup A^C)\cap B& \text{Complement law}\\ & =(A\cap B)\cup (A^C\cap B)& \text{Distributive law}\end{array}$$

¹²³

Thus, from 3rd axiom of probability set function, $P(A\cup B)=P(A)+P(A^C\cap B)$ and

$$P(B)=P(A\cap B)+P(A^C\cap B)$$

Now we have

$$\begin{array}{rl}P(A\cup B)-P(A)& =P(A^C\cap B)\\ & \\ ⟺P(B)& =P(A\cap B)+P(A\cup B)-P(A)\\ ⟺P(A\cup B)& =P(A)+P(B)-P(A\cap B)\end{array}$$

which completes the proof.

Footnotes

1. Identity Laws of Sets ↩

2. Complement Laws of Sets ↩

3. Distributive Laws of Sets ↩