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Thread: Probability space

  1. #1
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    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!
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  2. #2
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    Re: Probability space

    start by showing (easy)

    $$\displaystyle \mu(A \cap B \cap C) \leq \frac{2}{3}$$
    Thanks from detalosi
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  3. #3
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    Re: Probability space

    We have that :

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

    $\displaystyle \mu(A \cap B \cap C) \leq \mu(A\cap B) \Rightarrow$
    $\displaystyle \mu(A \cap B \cap C) \leq \frac{2}{3}$
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  4. #4
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    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.
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  5. #5
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    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 $\leq \mu (X)$
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  6. #6
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    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} $
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  7. #7
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    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?
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  8. #8
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    Re: Probability space

    we must have

    $$\mu (A)=\mu (B)=\mu (C)=\frac{5}{6}$$
    Thanks from detalosi
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  9. #9
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    Re: Probability space

    I am unsuccessfully trying verify the above hint. Could you provide another hint?
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  10. #10
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    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 $=1$
    Thanks from detalosi
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  11. #11
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    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.
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  12. #12
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    Re: Probability space

    Quote Originally Posted by detalosi View Post
    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.
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  13. #13
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    Re: Probability space

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

    I intended to write:

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