Mutual indepence doubt

Feb 2010
163
15
in the 4th dimension
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:

romsek

MHF Helper
Nov 2013
6,647
2,994
California
how is going from $2^n-n-1$ equations to $2^n$ equations a "reduction" ?
 
Feb 2010
163
15
in the 4th dimension
Yes, I shouldn't have written "reduced to". I will change it to "can be written as"
 

Plato

MHF Helper
Aug 2006
22,462
8,634
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
\(\displaystyle \begin{align*}\mathcal{P}(\mathit{A} \mathit{B'})&=\mathcal{P}(\mathit{A})-\mathcal{P}(\mathit{A} \mathit{B}) \\&=\mathcal{P}(\mathit{A})-\mathcal{P}(\mathit{A})\mathcal{P}(\mathit{B}) \\&=\mathcal{P}(\mathit{A})(1-\mathcal{P}(\mathit{B})) \\&=\mathcal{P}(\mathit{A})(\mathcal{P}(\mathit{B'})) \end{align*}\)