Probability question

• Apr 5th 2010, 04:44 PM
qwesl
Probability 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 PM
Anonymous1
Quote:

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

Have you ever heard of Boole's inequality?

http://www.andrew.cmu.edu/course/21-228/lec7.pdf
• Apr 5th 2010, 05:00 PM
Drexel28
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 PM
qwesl
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 PM
Anonymous1
Quote:

Originally Posted by qwesl
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 $\displaystyle A_1 , ...,A_n \subseteq \Omega$

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...

Quote:

Originally Posted by Drexel28
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?

Yes, I've seen a proof using this argument, wikipedias in fact.
• Apr 5th 2010, 05:19 PM
Drexel28
Quote:

Originally Posted by Anonymous1

Yes, I've seen a proof using this argument, wikipedias in fact.

Oops, probably should have wikid it first! (Giggle) God bless Wikipedia!