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Suppose $A_{1},\; A_{2},\ldots,\; A_{n}$ are independent events. Show that $A_{1}^{c},\ldots,A_{n}^{c}$ are independent.

Solution:

We begin with $n=2$. $$\begin{align} P(A_{1}^{c}\bigcap A_{2}^{c}) &= P\left(\left(A_{1}\bigcup A_{2}\right)^{c}\right) \cr &=1-P\left(A_{1}\bigcup A_{2}\right) \cr &=1-\left(P(A_{1})+P(A_{2})-P(A_{1}\bigcap A_{2})\right) \cr &=1-P(A_{1})-P(A_{2})+P(A_{1})P(A_{2}) \cr &=\left(1-P(A_{1})\right)\left(1-P(A_{2})\right) \cr &=P(A_{1}^{c})P(A_{2}^{c}). \end{align}$$

So, $A_{1}^{c},\, A_{2}^{c}$ are independent. Then by induction, $A_{1}^{c},\ldots,A_{n}^{c}$ are independent.

Is my proof rigorous/complete?

  • 5
    Well, it looks fine, but I'm afraid you'll have to flesh out the "Then by induction $\,A_1^c,...,A_n^c\,$ are independent"2012-10-26
  • 0
    You have shown it holds for one operator $\cap$. Now you can assume it holds for arbitrary $n$ operators and then prove that it holds for $n+1$. This will be proof by induction2012-10-27
  • 0
    Have you managed to solve the problem?2012-10-29
  • 0
    I don't think that is enough. You have to show for an arbitary combination as well.2012-10-30

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