- #1

mathmari

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I want to prove or disprove that for a probability measure $Q$ over a $\sigma$-algebra $\mathcal{B}$ with $A,B\in \mathcal{B}$ the following hold:

- $Q(A\cup B)=1-Q(\overline{A}\cup \overline{B})$
- $1-Q(\overline{A}\cap \overline{B})=Q(\overline{A})+Q(\overline{B})+Q(\overline{A}\cup \overline{B})$
- $1-Q(\overline{A})-Q(\overline{B})=Q(A)-1+Q(B)$
- $Q(A)\neq \emptyset\Rightarrow A\neq \emptyset$

- Let $\Omega$ be the universal set.

We suppose that the statement is true.

For $A=\emptyset$ and $B=\Omega$ be we get the following:

\begin{align*}&Q(\emptyset\cup \Omega)=1-Q(\overline{\emptyset}\cup \overline{\Omega}) \\ & \Rightarrow Q(\Omega)=1-Q(\Omega\cup \emptyset) \\ & \Rightarrow Q(\Omega)=1-Q(\Omega) \\ & \Rightarrow 2Q(\Omega)=1 \\ & \Rightarrow Q(\Omega)=\frac{1}{2} \\ & \Rightarrow 1=\frac{1}{2}\end{align*}

So, the statement is in general not true.

Is everything correct?

$$$$

- Let $S\in \mathcal{B}$.

We have that $\Omega = S\cup \overline{S}$. Then $Q(\Omega)=Q(S\cup \overline{S})$. Since $Q$ is a probability measure, we get that $Q(\Omega)=1$ and $Q(S\cup \overline{S})=Q(S)+Q(\overline{S})$.

Therefore, we get that:

$$Q(\Omega)=Q(S\cup \overline{S}) \Rightarrow 1=Q(S)+Q(\overline{S}) \Rightarrow Q(\overline{S})=1-Q(S)$$

By De Morgan’s laws we have that $\overline{A\cup B}=\overline{A}\cap \overline{B}$.

We have that $Q(\overline{A\cup B})=1-Q(A\cup B) \Rightarrow Q(\overline{A}\cap \overline{B})=1-Q(A\cup B) \Rightarrow 1-Q(\overline{A}\cap \overline{B})=Q(A\cup B)$.

So, we have to check if $Q(A\cup B)=Q(\overline{A})+Q(\overline{B})+Q(\overline{A}\cup \overline{B})$, right? How could we check that?

$$$$

- We have that $Q(\overline{A})=1-Q(A)$ and $Q(\overline{B})=1-Q(B)$.

So, we get:

$$1-Q(\overline{A})-Q(\overline{B})=1-(1-Q(A))-(1-Q(B))=1-1+Q(A)-1+Q(B)=Q(A)-1+Q(B)$$

So, the statement is true.

Is everything correct?

$$$$

- Could you give me a hint for this statement?