Limit of the intersection of events

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Discussion Overview

The discussion revolves around proving the limit of the intersection of an infinite decreasing sequence of events in probability theory. Participants explore the relationship between the probabilities of these events and their complements, focusing on foundational principles and mathematical reasoning.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant suggests that since the events form a decreasing sequence, the intersection of the events equals the last event in the sequence, leading to the conclusion that the probability of the intersection equals the limit of the probabilities of the events.
  • Another participant introduces a dual version involving increasing sequences of events and discusses the necessity of disjoint events for applying σ-additivity.
  • A different approach is proposed using complements of the events, with participants discussing how to express the probability of the intersection in terms of the probabilities of the complements.
  • There is a discussion about the relationship between the union of complements and the intersection of the original events, with some participants questioning how to connect these concepts mathematically.
  • Participants explore the implications of the complements being subsets of each other and how this affects the overall probability calculations.
  • One participant confirms the correctness of a derived formula relating the union of complements to the limit of the probabilities of the original events.

Areas of Agreement / Disagreement

Participants engage in a collaborative exploration of the topic, with some agreeing on the validity of certain mathematical transformations while others raise questions about the assumptions and implications of their approaches. The discussion remains unresolved regarding the most straightforward proof method.

Contextual Notes

Participants acknowledge the need for careful consideration of the properties of complements and the conditions under which certain probability formulas apply. There is an ongoing exploration of how to properly relate different expressions without reaching a definitive conclusion.

stukbv
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Hi, I keep seeing this come up
A1 ⊇ A2 ⊇ A3 ... is an infinite decreasing sequence of events. Prove from first principles
that
P(intersection of Ai from i=1 to infinity) = Lim P(An) as n--> infinity

All i can think of is that since each is a subset of the preceding, then A1 ∩ A2...∩An = An
So clearly P(A1 ∩ A2...∩An) = P(An) and thus the same for limits.
I think this is too simplistic though, is it or isn't it ?

Thanks a lot
 
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Hi stukbv! :smile:

Did you already encounter the dual version of this? That is, if

[tex]A_0\subseteq A_1\subseteq A_2\subseteq ...[/tex]

then

[tex]P\left(\bigcup_{i=0}^{+\infty}{A_i}\right)=\lim_{n\rightarrow +\infty}{P(A_n)}[/tex]

It is the easiest if we prove this first. Now, this statement actually follows from the [itex]\sigma[/itex]-additivity. However, the [itex]\sigma[/itex]-additivity requires the events to be disjoint, which is not the case here. Is there a way to make the events disjoint, though?
 
Yeah I've done it that way I think
Let B1 = A1
B2 = A2∩A1C
B3=A3∩A2C
And so on,
Then the Bi are disjoint and the union Bi (up to n) = An, but also U Bi up to ∞ = Union of Ai upto ∞ (since preceding ones are subsets!) so P(An) = P(U Bi) (to n) = Sum P(Bi (to n) as all bi are disjoint, then let RHS and LHS n tend to infinity and we get sum to infinity of P(Bi) = P(U Bi) to ∞ = P(U Ai) to ∞.
according to my lecturer!
 
Ah yes, that's very good! So now for your question. You need to prove that

[tex]P\left(\bigcap{A_n}\right)=\lim_{n\rightarrow +\infty}{P(A_n)}[/tex]

Now what happens if you take the complements of these events?
 
what like B1 = A1
B2 = A2\A1
B2 = A3\A2
and so on ?
 
No, take the complements. What is

[tex]P\left(\left(\bigcap{A_n}\right)^c\right)[/tex]
 
P((U An^c)?
 
stukbv said:
P((U An^c)?

Yes, carry on...
 
SO P(U An^c) is equal to the sum since they are disjoint?
But i don't see how you can relate the 2 like we did before
 
  • #10
stukbv said:
SO P(U An^c) is equal to the sum since they are disjoint?
But i don't see how you can relate the 2 like we did before

No, they are not disjoint. But you have a formula for

[tex]P\left(\bigcup{A_n^c}\right)[/tex]

Right? You've proven the formula above...
 
  • #11
ok so the limit = p(An^c)
 
  • #12
stukbv said:
ok so the limit = p(An^c)

And what the probability of a complement?
 
  • #13
1-p(An).
I see , but then how do i relate it to my initial statement?
 
  • #14
Well, you know already that the statement

[tex]P\left(\bigcup{A_n^c}\right)=1-\lim_{n\rightarrow +\infty}{P(A_n)}[/tex]

is true. Now try to evaluate the left side.
 
  • #15
surely if they are all subsets of their preceding ones then the union of the complements is just a1^c ?
 
  • #16
We have

[tex]A_1\supseteq A_2\supseteq ...[/tex]

and thus

[tex]A_1^c\subseteq A_2^c\subseteq ...[/tex]

So that doesn't really work.
 
  • #17
ohh so is the LHS 1 - P( intersection of Ai)
 
  • #18
Then you can get rid of the ones and then take limits ?
 
  • #19
so we can say lhs = 1-P(intersection Ai) ?
 
  • #20
Yes, that is correct! Just eliminate the ones and you have the result you wanted to prove!
 
  • #21
Perfect! just in time... thanks again!
 

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