# Probability measure: prove or disprove

• MHB
• mathmari
In summary, the conversation discusses the proof or disproof of several statements involving a probability measure over a sigma-algebra. The statements involve sets A and B, with some additional constraints. The conversation includes a step-by-step process of verifying the truth or falsehood of the statements, including checking for specific values of A and B. In the end, it is concluded that the statements are generally true, with a caveat that the sets A and B must be disjoint for one of the statements to hold.
mathmari
Gold Member
MHB
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?

mathmari said:
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)

I like Serena said:
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)
I like Serena said:
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)

mathmari said:
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)

I like Serena said:
We have the additional constraint that $S_1$ and $S_2$ must be disjoint. (Nerd)

Otherwise everything is correct. (Mmm)

Great! Thank you! (Yes)

## What is a probability measure?

A probability measure is a function that assigns a number between 0 and 1 to a set of possible outcomes in a probability space, representing the likelihood of that outcome occurring.

## How is a probability measure used to prove or disprove a statement?

A probability measure can be used to prove or disprove a statement by assigning probabilities to the possible outcomes and calculating the probability of the statement being true based on those probabilities.

## What is the difference between a discrete and a continuous probability measure?

A discrete probability measure is used for events with a finite or countably infinite number of outcomes, while a continuous probability measure is used for events with an uncountable number of outcomes.

## How do you prove that a probability measure is valid?

A probability measure must satisfy three axioms: the probability of an event must be between 0 and 1, the probability of the entire sample space must be 1, and the probability of mutually exclusive events must be additive. To prove validity, these axioms must be satisfied.

## Can a probability measure be used for events with uncertain outcomes?

Yes, a probability measure can be used for events with uncertain outcomes. In fact, that is the main purpose of a probability measure - to assign probabilities to uncertain events and quantify the likelihood of those events occurring.

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