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Math Help - Characteristic Function

  1. #1
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    Characteristic Function

    Given  A \in \mathcal{P}(X) define the characteristic function  \chi_{A}: X \to \{0,1\} by  \chi_{A}(x) = \begin{cases} 0&\text{if}\ x \not \in A\\ 1&\text{if}\ x \in A \end{cases} .

    (i) Prove that the function  x \mapsto \chi_{A}(x) \chi_{B}(x) is the characteristic function of the intersection  A \cap B .

    So  x \in A \cap B \Rightarrow x \in A \ \text{and} \ x \in B . By definition, the characteristic function always has values of  0 or  1 . Therefore the characteristic function is  \chi_{A} \chi_{B} .

    (ii) Find the subset  C whose characteristic function is given by  \chi_{C}(x) = \chi_{A}(x) + \chi_{B}(x) - \chi_{A}(x) \chi_{B}(x) . Isn't this just  A \Delta B ?


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    (i) Prove that the function  x \mapsto \chi_{A}(x) \chi_{B}(x) is the characteristic function of the intersection  A \cap B .

    So suppose that  x \in A \cap B then  x \in A and  x \in B ; thus by definition,
     \chi_{A}(x)=1 ,  \chi_{B}(x)=1 , & <br />
\chi _{A \cap B}  = 1 therefore \left( {\forall x \in A \cap B} \right)\left[ {\chi _{A \cap B} (x) = \chi _A (x)\chi _B (x)} \right]

    Now if x \notin A \cap B\quad  \Rightarrow \quad x \notin A \vee x \notin B which means  \chi_{A}(x)=0 or  \chi_{B}(x)=0 , & \chi _{A \cap B}  = 0 therefore \left( {\forall x \notin A \cap B} \right)\left[ {\chi _{A \cap B} (x) = \chi _A (x)\chi _B (x)} \right].



    You have this one correct!:
    (ii) Find the subset  C whose characteristic function is given by  \chi_{C}(x) = \chi_{A}(x) + \chi_{B}(x) - {\color{red}2}\chi_{A}(x) \chi_{B}(x) . Isn't this just  A \Delta B ?
    Last edited by Plato; August 19th 2007 at 09:31 AM.
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  3. #3
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    why did you add the 2?
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  4. #4
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    Quote Originally Posted by shilz222 View Post
    why did you add the 2?
    Because that is the characteristic function for the symmetric difference operator.
    I thought that is what you wanted.
    But upon reading the question again, I realize that the answer is C = A \cup B that is <br />
{\chi _{A \cup B} (x) = \chi _A (x) + \chi _B (x) - \chi _A (x)\chi _B (x)}<br />
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  5. #5
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     \chi_{A}(x) corresponds to set  A

     \chi_{B}(x) corresponds to set  B

     \chi_{A}(x) \chi_{B}(x) corresponds to set  A \cap B

    So  (A + B) - (A \cap B) = (A \cup B) - (A \cap B) = A \Delta B ?

    How does this correspond to  A \cup B ?

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  6. #6
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    Suppose that x \in A \cup B then exactly one of these is true:
    \left( {x \in A - B} \right) \vee \left( {x \in B - A} \right) \vee \left( {x \in A \cap B} \right).
    Consider them one at a time.
    1) \left( {x \in A - B} \right)\quad  \Rightarrow \quad \chi _{A \cup B} (x) = \chi _A (x) + \chi _B (x) - \chi _{A \cap B} (x) = 1 + 0 - 0 = 1.

    2) \left( {x \in B - A} \right)\quad  \Rightarrow \quad \chi _{A \cup B} (x) = \chi _A (x) + \chi _B (x) - \chi _{A \cap B} (x) = 0 + 1 - 0 = 1.

    3) \left( {x \in A \cap B} \right)\quad  \Rightarrow \quad \chi _{A \cup B} (x) = \chi _A (x) + \chi _B (x) - \chi _{A \cap B} (x) = 1 + 1 - 1 = 1.

    Moreover, this is clear \left( {x \notin A \cup B} \right)\quad  \Rightarrow \quad \chi _{A \cup B} (x) = 0.


    This is a very well known problem. What you posted above is incorrect.
    This is correct: \chi _{A\Delta B} (x) = \chi _A (x) + \chi _B (x) - 2\chi _{A \cap B} (x).
    Again this is well known.
    Last edited by Plato; August 19th 2007 at 01:43 PM.
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  7. #7
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    Thanks.


    I learned something new today. I see the characteristic function has many applications in other mathematical fields (i.e. probability theory).
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  8. #8
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    Quote Originally Posted by shilz222 View Post
    I learned something new today. I see the characteristic function has many applications in other mathematical fields (i.e. probability theory).
    The characteristic function in probability theory is completely unrelated: it is \phi_X(t) = E(e^{itX}) and is just another name for the Fourier transform.

    Probability theorists often rename standard mathematical objects this way. E.g., convergence with probability 1 is another name for convergence almost everywhere.
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