How would you go about doing this? is it really obvious?

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- May 15th 2009, 07:15 AMn7615rProbability Axioms
How would you go about doing this? is it really obvious?

- May 15th 2009, 07:53 AMIsomorphism
So where are you having trouble? Is it in:

The first part : Reading set notations(about the infinitely often)

The second part : The union bound axiom? the one that says probability of union of countable number of events is upperbounded by the sum of probability of those events

In case of basic application problems like these, its beneficial to both (us and you), if you tell us where you are having trouble. - May 15th 2009, 08:22 AMn7615r
i'm having trouble understanding the set notation and how 'infintely often can be written in that way.

and also showing how the P(A)≤P(Bn). is it because A is a subset of Bn?

and how does this relate to the third part? - May 15th 2009, 10:33 AMMoo
Hello,

If $\displaystyle B_n$ happens, it means that there is at least one $\displaystyle A_m$ in its union that happens.

This can be translated as "$\displaystyle \exists m\geq n$, such that $\displaystyle A_m$ happens"

Then, for $\displaystyle A=\bigcap_{n\geq 1} B_n$ to happen, $\displaystyle B_n$ has to happen for any $\displaystyle n\geq 1$

This can finally be translated as "for any n, there exists m>n such that $\displaystyle A_m$ happens".

In other words, there are infinitely many $\displaystyle A_m$ events that happen.

You can read this for more things : http://www.mathhelpforum.com/math-he...up-liminf.html in particular post #4, by**Laurent**(Wink)

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and also showing how the P(A)≤P(Bn). is it because A is a subset of Bn?

Since A is, as you noticed, a subset of $\displaystyle B_n$, then there exists a set $\displaystyle E$,*disjoint*from A, such that $\displaystyle A\cup E=B_n$

By the third axiom, we have (the red equality) :

$\displaystyle \mathbb{P}(B_n)=\mathbb{P}(A\cup E){\color{red}=}\mathbb{P}(A)+\mathbb{P}(E)$

Now, by the first axiom, we know that $\displaystyle \mathbb{P}(E)\geq 0$

And it follows that $\displaystyle \mathbb{P}(B_n)\geq \mathbb{P}(A)$ (Wink)

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and how does this relate to the third part?

And note that if $\displaystyle \sum_{m\geq 1} a_m<\infty$, this implies that the partial reminder (I don't know the exact name) $\displaystyle \sum_{m\geq n} a_m \xrightarrow[]{n\to\infty} 0$

It should be okay from now :)

Quote:

The second part : The union bound axiom? the one that says probability of union of countable number of events is upperbounded by the sum of probability of those events

- May 15th 2009, 10:43 AMn7615r
Thanks so much! this was a great help :)

- May 15th 2009, 11:45 AMIsomorphism