Probability measure: prove or disprove

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.
  • #1
mathmari
Gold Member
MHB
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Hey! :eek:

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?
 
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  • #2
mathmari said:
Hey! :eek:

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)
 
  • #3
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)
 
  • #4
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)
 
  • #5
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)
 

Related to Probability measure: prove or disprove

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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