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My solution seems too long winded. This should take less writing, no? Here it is.

$B^c$ means complement of $B$ (i.e. $P(B) = 1 - P(B^c)$

If $A$ and $B$ are independent events:

$$ P(A \cap B) = P(A)P(B) $$

So I do:

$$ P(A^c \cap B^c) = P(A^c | B^c)P(B^c) = (1 - P(A|B^c))P(B^c) = (1 - \frac{P(A\cap B^c)}{P(B^c)})P(B^c) = (1 - \frac{P(B^c|A)P(A)}{P(B^c)})P(B^c) $$

Now, I can use the result that since $A$ and $B$ are independent $P(B^c|A) = 1 - P(B|A) = 1 - P(B) = P(B^c)$. Then:

$$ (1 - \frac{P(B^c|A)P(A)}{P(B^c)})P(B^c) = (1 - \frac{P(B^c)P(A)}{P(B^c)})P(B^c) = (1 - P(A))P(B^c) = P(A^c)P(B^c) $$

Therefore, since $P((A^c \cap B^c) = P(A^c)P(B^c)$ then $A^c$ and $B^c$ are independent.

s5s
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  • One approach: use indicator functions. Since the indicator for the complement of a set is a (very simple) function of the indicator of the set, https://stats.stackexchange.com/questions/94872 provides a demonstration. – whuber Feb 27 '19 at 18:54

1 Answers1

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Here is another way:

$$\begin{align} P(A^c\cap B^c) &=1-P(A\cup B)=1-P(A)-P(B)+P(A\cap B)\\ &= 1-P(A)-P(B)+P(A)P(B) \\ &=1-P(A)-P(B)(1-P(A))=(1-P(A))(1-P(B)) \\ &= P(A^c)P(B^c)\end{align}$$

gunes
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  • You need to be careful with your basic arithmetic. $P(A\cup B)$ equals $P(A)+P(B)-P(A\cap B)$ whereas you equate it to $P(A)+P(B)+P(A\cap B)$ in your first line, and then you reverse the mistake when you replace $-P(B)-P(A)P(B)$ on the second line with $-P(B)(1-P(A))$ on the third line; the latter expression actually equals $-P(B)+P(A)P(B)$ which is what you _should_ have gotten on the second line.Since the final result is correct, I guess this is an instance of two wrongs making a right.... – Dilip Sarwate Feb 27 '19 at 20:25
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    @DilipSarwate, corrected it, much appreciated. this is also an instance where one should be extra careful when writing answers on mobile :) – gunes Feb 27 '19 at 20:54