MHB Probability measure: prove or disprove

[mathmari]

Hey!

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:

1. $Q(A\cup B)=1-Q(\overline{A}\cup \overline{B})$
2. $1-Q(\overline{A}\cap \overline{B})=Q(\overline{A})+Q(\overline{B})+Q(\overline{A}\cup \overline{B})$
3. $1-Q(\overline{A})-Q(\overline{B})=Q(A)-1+Q(B)$
4. $Q(A)\neq \emptyset\Rightarrow A\neq \emptyset$

I have done the following :

1. 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?
   $$$$
   
2. 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?
   
   $$$$
   
3. 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?
   
   $$$$
   
4. Could you give me a hint for this statement?

 

[I like Serena]

Hey!

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:

1. $Q(A\cup B)=1-Q(\overline{A}\cup \overline{B})$
2. $1-Q(\overline{A}\cap \overline{B})=Q(\overline{A})+Q(\overline{B})+Q(\overline{A}\cup \overline{B})$
3. $1-Q(\overline{A})-Q(\overline{B})=Q(A)-1+Q(B)$
4. $Q(A)\neq \emptyset\Rightarrow A\neq \emptyset$

Hey mathmari!

1. All correct. (Nod)

2. How about checking for, say, $A=B=\Omega$?

3. Good.

4. The statement $S\Rightarrow T$ is true if and only if $\lnot T \Rightarrow \lnot S$.
Can we prove the latter? (Wondering)

 

[mathmari]

2. How about checking for, say, $A=B=\Omega$?

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

The left side is equal to $1-Q(\overline{\Omega}\cap \overline{\Omega})=1-Q(\emptyset\cap \emptyset)=1-Q(\emptyset)=1-0=1$.

The right side is equal to $Q(\overline{\Omega})+Q(\overline{\Omega})+Q(\overline{\Omega}\cup \overline{\Omega}) =Q(\emptyset)+Q(\emptyset)+Q(\emptyset\cup \emptyset)=0+0+Q(\emptyset)=0$.

Therefore, the statement does not hold in general, right? (Wondering)

4. The statement $S\Rightarrow T$ is true if and only if $\lnot T \Rightarrow \lnot S$.
Can we prove the latter? (Wondering)

So, we have to prove that $\lnot A\neq \emptyset \Rightarrow \lnot Q(A)\neq 0$, i.e. that $A= \emptyset \Rightarrow Q(A)=0$, i.e. $Q(\emptyset )=0$, right? (Wondering)

Since $Q$ is a probability measure, we have that $Q(\Omega)=1$ and that $Q(S_1\cup S_2)=Q(S_1)+Q(S_2)$, for $S_1, S_2\in \mathcal{B}$.
So, we get the following:
$$Q(\Omega)=Q(\Omega\cup \emptyset)=Q(\Omega)+Q(\emptyset) \Rightarrow 1=1+Q(\emptyset) \Rightarrow Q(\emptyset)=0$$
right?

So, since it holds that $\lnot A\neq \emptyset \Rightarrow \lnot Q(A)\neq 0$, it follows that $Q(A)\neq 0\Rightarrow A\neq \emptyset$, right? (Wondering)

 

[I like Serena]

Since $Q$ is a probability measure, we have that $Q(\Omega)=1$ and that $Q(S_1\cup S_2)=Q(S_1)+Q(S_2)$, for $S_1, S_2\in \mathcal{B}$.

We have the additional constraint that $S_1$ and $S_2$ must be disjoint. (Nerd)

Otherwise everything is correct. (Mmm)

 

[mathmari]

We have the additional constraint that $S_1$ and $S_2$ must be disjoint. (Nerd)

Otherwise everything is correct. (Mmm)

Great! Thank you! (Yes)