1. ## Probability space

Hello,

I have a probability space $\displaystyle (X,\mathcal{A},\mu)$. Considering sets $\displaystyle A$, $\displaystyle B$ and $\displaystyle C$ in $\displaystyle \mathcal{A}$ with:

$\displaystyle \mu(A \cap C) = \mu(A\cap C) = \mu(B\cap C)=\frac{2}{3}$.

How do I begin to show that $\displaystyle \frac{1}{2}\leq\mu(A \cap B \cap C) \leq \frac{2}{3}$?

Appreciate the help!

2. ## Re: Probability space

start by showing (easy)

$$\displaystyle \mu(A \cap B \cap C) \leq \frac{2}{3}$$

3. ## Re: Probability space

We have that :

$\displaystyle A \cap B \cap C \subset A\cap B \Rightarrow$

$\displaystyle \mu(A \cap B \cap C) \leq \mu(A\cap B) \Rightarrow$
$\displaystyle \mu(A \cap B \cap C) \leq \frac{2}{3}$

4. ## Re: Probability space

Could you provide hints for the second part: $\displaystyle \mu(A \cap B \cap C) \geq \frac{1}{2}$.

Thanks a lot.

5. ## Re: Probability space

consider the four sets

$S=\displaystyle A \cap B \cap C$

$(A\cap B)\backslash S$

$(B\cap C)\backslash S$

$(A\cap C)\backslash S$

the sum of their measures must be $\leq \mu (X)$

6. ## Re: Probability space

I have done the following:

Let $\displaystyle S=A \cap B \cap C$ for brevity.

Since $\displaystyle S \subset (A\cap B)$ and $\displaystyle \mu(S) <\infty$ we have that $\displaystyle \mu((A\cap B) \setminus S ) = \mu(A\cap B) - \mu(S)$

Since $\displaystyle S \subset (A\cap C)$ and $\displaystyle \mu(S) <\infty$ we have that $\displaystyle \mu((A\cap C) \setminus S ) = \mu(A\cap C) - \mu(S)$

Since $\displaystyle S \subset (B\cap C)$ and $\displaystyle \mu(S) <\infty$ we have that $\displaystyle \mu((B\cap C) \setminus S ) = \mu(B\cap C) - \mu(S)$

Then :

$\displaystyle \mu(A\cap B) - \mu(S) + \mu(A\cap C) - \mu(S) + \mu(B\cap C) - \mu(S) \leq \mu(X) + \mu(S) \leq 1 \Leftrightarrow$

$\displaystyle 3\cdot \mu(A\cap B) - 3\cdot \mu(S) + \mu(S) \leq 1 \Leftrightarrow$

$\displaystyle \mu(S) \geq \frac{1}{2}$

7. ## Re: Probability space

I have another question pertaining to the following two conditions:

$\displaystyle \mu(A \cap C) = \mu(A\cap C) = \mu(B\cap C)=\frac{2}{3}$ and $\displaystyle \mu(A \cap B \cap C) = \frac{1}{2}$.

I am asked to give an example of a probability space with sets $\displaystyle A,B,C \in \mathcal{A}$ which satisfy the above conditions.

I have done the following :

I have chosen the probability space $\displaystyle ([0,1], \mathcal{B}([0,1]),\mu])$.

If I put $\displaystyle A=[0,\frac{2}{3}]$ and $\displaystyle B=[0,\frac{3}{4}]$ then $\displaystyle A \cap B =[0,\frac{2}{3}]$ and $\displaystyle \mu(A \cap B)=\frac{2}{3}$.

If I put $\displaystyle C=[0,\frac{1}{2}]$ then $\displaystyle A \cap B \cap C=[0,\frac{1}{2}]$ and $\displaystyle \mu(A \cap B \cap C)=\frac{1}{2}$.
However $\displaystyle \mu(A \cap C)\neq\frac{2}{3}$ and $\displaystyle \mu(B \cap C)\neq\frac{2}{3}$.

Is there a systematic approach?

8. ## Re: Probability space

we must have

$$\mu (A)=\mu (B)=\mu (C)=\frac{5}{6}$$

9. ## Re: Probability space

I am unsuccessfully trying verify the above hint. Could you provide another hint?

10. ## Re: Probability space

If we consider the four disjoint sets as above

$\mu (S)=\frac{1}{2}$ and each of the other three sets has measure $\frac{1}{6}$

the sum of the measures of these sets $=1$

11. ## Re: Probability space

I have been misreading the question all along, $\displaystyle \mu(S)=\frac{1}{2}$ and $\displaystyle \mu(A \cap B)= \mu(A \cap C) = \mu(B \cap A)=\frac{2}{3}$ may be used as conditions to finder the other measures.

I have found: $\displaystyle A=\left[0,\frac{5}{6} \right]$, $\displaystyle B=\left[\frac{1}{3},\frac{7}{6} \right]$, $\displaystyle C=\left[0,\frac{5}{6} \right]$

Thanks.

12. ## Re: Probability space

Originally Posted by detalosi
I have been misreading the question all along, $\displaystyle \mu(S)=\frac{1}{2}$ and $\displaystyle \mu(A \cap B)= \mu(A \cap C) = \mu(B \cap A)=\frac{2}{3}$ may be used as conditions to finder the other measures.

I have found: $\displaystyle A=\left[0,\frac{5}{6} \right]$, $\displaystyle B=\left[\frac{1}{3},\frac{7}{6} \right]$, $\displaystyle C=\left[0,\frac{5}{6} \right]$

Thanks.
I'm confused why you keeping writing things like

$\mu(A \cap C) = \mu(A \cap C)$

and

$\mu(A \cap B) = \mu(B \cap A)$

I'd think the first example was a typo but you've written it twice.

The second isn't quite a typo but I'd think it was obvious the intersection operation being commutative and all.

13. ## Re: Probability space

Just me going 150 mph but I do see your point.

I intended to write:

$\displaystyle \mu(A\cap B)= \mu(A\cap C) = \mu(B\cap C)$