Show that for any $\displaystyle n$ events $\displaystyle A_1 ,A_2 ,...,A_n$ in a finite probability space, we have $\displaystyle P(A_1 \cup A_2 \cup \cdot \cdot \cdot \cup A_n) \leq P(A_1) + ...+P(A_n)$

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- Apr 5th 2010, 04:44 PMqweslProbability question
Show that for any $\displaystyle n$ events $\displaystyle A_1 ,A_2 ,...,A_n$ in a finite probability space, we have $\displaystyle P(A_1 \cup A_2 \cup \cdot \cdot \cdot \cup A_n) \leq P(A_1) + ...+P(A_n)$

- Apr 5th 2010, 04:51 PMAnonymous1
Have you ever heard of Boole's inequality?

http://www.andrew.cmu.edu/course/21-228/lec7.pdf - Apr 5th 2010, 05:00 PMDrexel28
I really know nothing about probability and so this is more of a question then an answer. Doesn't this follow since the probability measure is by definition countably additive?

- Apr 5th 2010, 05:03 PMqwesl
Yes but how is this done using only the following axioms (first principles)

1) $\displaystyle P(\emptyset)=0$

2)$\displaystyle P(\Omega)=1$

3)$\displaystyle P(A_1 \cup \cdot \cdot \cdot \cup A_n)=P(A_1) + ...+ P(A_n)$ for $\displaystyle n$ disjoint sets $\displaystyle A_1 , ...,A_n \subseteq \Omega$ - Apr 5th 2010, 05:12 PMAnonymous1
Use google, proofs this common are usually all over the internet.

http://www.math.ntu.edu.tw/~hchen/te...s/lecture2.pdf

If you don't like that one there are several others...

Yes, I've seen a proof using this argument, wikipedias in fact. - Apr 5th 2010, 05:19 PMDrexel28