# Is my proof rigorous/complete?

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.

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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" –  DonAntonio Oct 26 '12 at 18:53
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 induction –  Alex Oct 27 '12 at 1:37
Have you managed to solve the problem? –  Davide Giraudo Oct 29 '12 at 10:32
I don't think that is enough. You have to show for an arbitary combination as well. –  Pk.yd Oct 30 '12 at 2:47