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I am reading Statistical Theory: A Concise Introduction, by Felix Abramovich and Ya'acov Ritov. In the the appendix, the authors provide a primer on basic probability theory. In discussing the probability function, the authors write the following:

The probability function assigns to each event $A \in \mathcal{A}$ a real number $P(A)$ called the probability of $A$ satisfying the following conditions:

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  1. Since $(A_1 \cup A_2)^c = A_1^c \cap A_2^c$, then $P(A_1 \cap A_2) = 1 - P((A_1 \cap A_2)^c) = 1 - P(A_1^c \cup A_2^c)$.

Note that $A$ denotes an event.

Intuitively, I can tell that the proposition is true. However, I wanted to prove it in my notes. I unsuccessfully attempted to do this by using De Morgan's laws and the fact that $P(A^c) = 1 - P(A) \ \forall \ A$.

I was wondering if people could please take the time to show me the correct proof for this.

It seems to me that we have an if/then statement of the form, "Since (If) $(A_1 \cup A_2)^c = A_1^c \cap A_2^c$, then $P(A_1 \cap A_2) = 1 - P((A_1 \cap A_2)^c) = 1 - P(A_1^c \cup A_2^c)$.

A (Hypothesis): $(A_1 \cup A_2)^c = A_1^c \cap A_2^c$.

B (Conclusion): $P(A_1 \cap A_2) = 1 - P((A_1 \cap A_2)^c) = 1 - P(A_1^c \cup A_2^c)$.

So for a proof, we need to either work forwards from A to B or backwards from B to A.

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  • $\begingroup$ This is DeMorgan's Law, pure and simple. Nothing else is needed except the substitutions shown in step B. What more could be said?? $\endgroup$ – whuber Jan 12 '18 at 17:45
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$(A_1 \cup A_2)^c = A_1^c \cap A_2^c \rightarrow (A_1^c \cup A_2^c)^c = A_1 \cap A_2 \rightarrow P(A_1 \cap A_2) = P((A_1^c \cup A_2^c)^c) \rightarrow P(A_1 \cap A_2) = 1-P(A_1^c \cup A_2^c).$

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  • $\begingroup$ Thanks for the response. How did you get from $(A_1 \cup A_2)^c = A_1^c \cap A_2^c \rightarrow (A_1^c \cup A_2^c)^c = A_1 \cap A_2$? Specifically, I don't understand how you went from $(A_1 \cup A_2)^c$ to $(A_1^c \cup A_2^c)^c$? Isn't the complement of $(A_1 \cup A_2)^c = A_1^c \cap A_2^c$? So you would get $(A_1^c \cap A_2^c)^c$? And the complement of $A_1^c \cap A_2^c$ would be $(A_1 \cup A_2)^c$ $\endgroup$ – The Pointer Jan 12 '18 at 12:47
  • $\begingroup$ What is true for $A_1$ and $A_2$ must also be true for $A_1^c$ and $A_2^c$, given it is something that holds generally (I mean: not for specific $A_1$ and $A_2$, but for any couple of events). $\endgroup$ – Federico Tedeschi Jan 12 '18 at 14:55
  • $\begingroup$ A more specific way of putting it might be to say "take the complements of all the sets in the first statement, remembering that $(A^c)^c = A$, and see what you get." $\endgroup$ – jbowman Jan 12 '18 at 18:43

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