# Mutual indepence doubt

#### earthboy

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.

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#### romsek

MHF Helper
how is going from $2^n-n-1$ equations to $2^n$ equations a "reduction" ?

#### earthboy

Yes, I shouldn't have written "reduced to". I will change it to "can be written as"

#### Plato

MHF Helper
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*}