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Thread: Proving independence

  1. #1
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    Proving independence

    Hi

    I was wondering, how do check whether the following is independent or not (A \cup B) \cap (A' \cap B')?


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  2. #2
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    Re: Proving independence

    I don't know what you mean by sets being independent. However

    $(A \cup B) \cap (A^\prime \cap B^\prime) = (A \cup B) \cap (A \cup B) = A \cup B$
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  3. #3
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    Re: Proving independence

    Quote Originally Posted by cooltowns View Post
    I was wondering, how do check whether the following is independent or not (A \cup B) \cap (A' \cap B')?
    As stated, this is a meaningless question.
    It is true that $(A \cup B)~ \&~ (A^\prime \cap B^\prime)$ are mutually exclusive.


    Quote Originally Posted by romsek View Post
    I don't know what you mean by sets being independent. However
    $(A \cup B) \cap (A^\prime \cap B^\prime) = (A \cup B) \cap (A \cup B) = A \cup B$
    Assuming that the $^\prime$ is a notation for complement, the above is misstaken.
    $ \begin{align*}(A \cup B) \cap (A^\prime \cap B^\prime)&=(A \cup B) \cap (A\cup B)^\prime \\&=\emptyset \end{align*}$
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  4. #4
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    Re: Proving independence

    Thank you for your replies guys

    Basically, I wanted to investigate whether (A' \cap B') \& (A \cup B) are independent and or mutually exclusive.

    I understand that for them to be mutually exclusive, the following condition must be satifised: (A' \cap B') \cap (A \cup B) = \emptyset

    however for them to be independent, it must satisfy P(A' \cap B') \cup P(A \cup B) =P(A' \cap B')*P(A \cup B)

    You guys mention, it's mutually exclusive, but how can I justify that.
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    Re: Proving independence

    Quote Originally Posted by cooltowns View Post
    Thank you for your replies guys

    Basically, I wanted to investigate whether (A' \cap B') \& (A \cup B) are independent and or mutually exclusive.

    I understand that for them to be mutually exclusive, the following condition must be satifised: (A' \cap B') \cap (A \cup B) = \emptyset

    however for them to be independent, it must satisfy P(A' \cap B') \cup P(A \cup B) =P(A' \cap B')*P(A \cup B)

    You guys mention, it's mutually exclusive, but how can I justify that.
    Plato showed you how to see that they are mutually exclusive (my post was in error).

    Just apply DeMorgan's Law to $(A^\prime \cap B^\prime)$
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    Re: Proving independence

    thanks guys, got it, appreciate the help
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  7. #7
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    Re: Proving independence

    Quote Originally Posted by cooltowns View Post
    Thank you for your replies guys
    Basically, I wanted to investigate whether (A' \cap B') \& (A \cup B) are independent and or mutually exclusive.
    I understand that for them to be mutually exclusive, the following condition must be satifised: (A' \cap B') \cap (A \cup B) = \emptyset
    however for them to be independent, it must satisfy P(A' \cap B') \cup P(A \cup B) =P(A' \cap B')*P(A \cup B)
    You guys mention, it's mutually exclusive, but how can I justify that.
    That is what I posted. Two sets are mutually exclusive if their intersection is null.

    You seem to be asking or thinking of events in a probability space? However, you gave no information whatever about the probability sets and/or probability measure.
    The point is that ordinary sets can be said to be mutually exclusive but the concept of independence is not applied in general. In vector space certain vectors can be independent of one another.

    Now suppose we have a probability space. The subsets of the space are the events.
    Events $A~\&~B$ are mutually exclusive provided: $A\cap B=\emptyset,~\mathcal{P}(A\cup B)=\mathcal{P}(A)+\mathcal{P}(B)~\&~\mathcal{P}(A \cap B)=0$

    Events $A~\&~B$ are independent provided: $\mathcal{P}(A\cap B)=\mathcal{P}(A)\cdot\mathcal{P}( B)$
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