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Thread: Mutual indepence doubt

  1. #1
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    Mutual indepence doubt

    The definition of mutual independence of events $\displaystyle A_{1},..A_{n}$, state that that the events should satisfy equations like $\displaystyle p(A_{i}A_{j})=P(A_{i})P(P_{j}),P(A_{i}A_{j}A_{k})= P(A_{i})P(A_{j})P(A_{k}), .....P(A_{1}...A_{n})=P(A_{1})P(A_{2})....P(A_{n})$
    These comes down to a total of $\displaystyle 2^n-n-1$ equations.I understand this much.

    My book says that it is readily seen that this definition can be written as a set of $\displaystyle 2^n$ equations, obtained from the last equation on replacing an arbitrary number of events $\displaystyle A_{j }$ by their complements $\displaystyle A'_{j}$, using the fact that if $\displaystyle A$ and $\displaystyle B$ are independent then $\displaystyle P(AB')=P(A)P(B')$

    I cant understand what it means or how they arrive at $\displaystyle 2^n$ equations, or even how it gives 4 equations for 2 events, using the complements.
    Please help me.
    Last edited by earthboy; Nov 19th 2018 at 11:10 AM.
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  2. #2
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    Re: Mutual indepence doubt

    how is going from $2^n-n-1$ equations to $2^n$ equations a "reduction" ?
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    Re: Mutual indepence doubt

    Yes, I shouldn't have written "reduced to". I will change it to "can be written as"
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    Re: Mutual indepence doubt

    Quote Originally Posted by earthboy View Post
    My book says that it is readily seen that this definition can be written as a set of $\displaystyle 2^n$ equations, obtained from the last equation on replacing an arbitrary number of events $\displaystyle A_{j }$ by their complements $\displaystyle A'_{j}$, using the fact that if $\displaystyle A$ and $\displaystyle B$ are independent then $\displaystyle P(AB')=P(A)P(B')$ I cant understand what it means or how they arrive at $\displaystyle 2^n$ equations, or even how it gives 4 equations for 2 events, using the complements.
    Suppose that $\displaystyle \mathit{A~\&~B}$ are independent events then: We will write $\displaystyle \mathcal{P}(\mathit{A} \mathit{B})$ for $\displaystyle \mathcal{P}(\mathit{A}\cap \mathit{B})$
    $\displaystyle \left\{ \begin{array}{l}i)~\mathcal{P}(\mathit{A} \mathit{B})=\mathcal{P}(\mathit{A})\cdot \mathcal{P}(\mathit{B})\\ii)~\mathcal{P}(\mathit{A } \mathit{B'})=\mathcal{P}(\mathit{A})\cdot \mathcal{P}(\mathit{B'})\\iii)~\mathcal{P}(\mathit {A'} \mathit{B})=\mathcal{P}(\mathit{A'})\cdot \mathcal{P}(\mathit{B})\\iv)~\mathcal{P}(\mathit{A '} \mathit{B'})=\mathcal{P}(\mathit{A'})\cdot \mathcal{P}(\mathit{B'})\end{array} \right.$
    There. four equation showing independance of $\displaystyle \mathit{A},~\mathit{B},~\mathit{A'},~\&~\mathit{B' }$


    Here is ii) done for you: $\displaystyle \mathcal{P}(\mathit{A})=\mathcal{P}(\mathit{A} \mathit{B})+\mathcal{P}(\mathit{A} \mathit{B'})$ using this we get
    $\displaystyle \begin{align*}\mathcal{P}(\mathit{A} \mathit{B'})&=\mathcal{P}(\mathit{A})-\mathcal{P}(\mathit{A} \mathit{B}) \\&=\mathcal{P}(\mathit{A})-\mathcal{P}(\mathit{A})\mathcal{P}(\mathit{B}) \\&=\mathcal{P}(\mathit{A})(1-\mathcal{P}(\mathit{B})) \\&=\mathcal{P}(\mathit{A})(\mathcal{P}(\mathit{B' })) \end{align*}$



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