A ⊂ B ⇒ P (A) ≤ P (B) using result of theAxiomsofProbability How can I proof

P(B) ≤ P (AUB ) and

P (AB)≤ P(A) ?

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- Feb 16th 2010, 11:56 AMwadasA ⊂ b ⇒ p (a) ≤ p (b)
A ⊂ B ⇒ P (A) ≤ P (B) using result of the

*Axioms*of*Probability How can I proof*

P(B) ≤ P (AUB ) and

P (AB)*≤ P(A) ?* - Feb 16th 2010, 12:06 PMicemanfan
If A represents a subset of the events corresponding to B, then if A occurs, B occurs.

- Feb 16th 2010, 12:25 PMMoo
Hello,

It's not a matter of events, since we want to use the axioms, what you said is not valid ^^'

To prove that A ⊂ B ⇒ P (A) ≤ P (B), just consider the disjoint sets BnA and BnA', where A' denotes the complement of A.

Their union makes B, and by the third axiom of probability, you can conclude...

P(B) ≤ P(AUB) comes from the fact that B ⊂ AUB

P (AB) ≤ P(A) comes from the fact that AB ⊂ A

That's all... - Feb 16th 2010, 12:32 PMjohanS
If you want to prove in a more formal way (instead of using the suggestion above) how can you write$\displaystyle A \cup B$ as a union of disjoint sets? And more easily, what is $\displaystyle A \cup B$ if $\displaystyle A \subset B$ ?

Because if you have X and Y disjoint, i.e. $\displaystyle X \cap Y = \emptyset$ ten by the axioms of probability you know that $\displaystyle P(X \cup Y) = P(X) + P(Y)$ (You will also need to use the fact that for any event $\displaystyle E, P(E) \geq 0$.) If your not used to work with sets you can try by using venn diagrams. - Feb 16th 2010, 12:34 PMjohanS
sorry, when I started writing your reply wasn't here

- Feb 16th 2010, 12:55 PMicemanfan
- Feb 16th 2010, 01:01 PMMoo
Yes, but it can't be taken as a formal proof of what's being said here (Nod)

Why would A be a subset of B, though they're events ? There's a degree of abstract that can't be found if we talk about "if event A occurs, then so does B". It's a mere translation of the inequality between the probabilities that there are in the first message.

I don't know if I explained well what I thought, I hope you understand what I want to say :( - Feb 16th 2010, 01:18 PMicemanfan