# Limit of the intersection of events

1. Jun 15, 2011

### stukbv

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 isnt it ?

Thanks a lot

2. Jun 15, 2011

### micromass

Staff Emeritus
Hi stukbv!

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

$$A_0\subseteq A_1\subseteq A_2\subseteq ...$$

then

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

It is the easiest if we prove this first. Now, this statement actually follows from the $\sigma$-additivity. However, the $\sigma$-additivity requires the events to be disjoint, which is not the case here. Is there a way to make the events disjoint, though?

3. Jun 15, 2011

### stukbv

Yeah ive 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!

4. Jun 15, 2011

### micromass

Staff Emeritus
Ah yes, that's very good!! So now for your question. You need to prove that

$$P\left(\bigcap{A_n}\right)=\lim_{n\rightarrow +\infty}{P(A_n)}$$

Now what happens if you take the complements of these events?

5. Jun 15, 2011

### stukbv

what like B1 = A1
B2 = A2\A1
B2 = A3\A2
and so on ?

6. Jun 15, 2011

### micromass

Staff Emeritus
No, take the complements. What is

$$P\left(\left(\bigcap{A_n}\right)^c\right)$$

7. Jun 15, 2011

### stukbv

P((U An^c)?

8. Jun 15, 2011

### micromass

Staff Emeritus
Yes, carry on...

9. Jun 15, 2011

### stukbv

SO P(U An^c) is equal to the sum since they are disjoint?
But i dont see how you can relate the 2 like we did before

10. Jun 15, 2011

### micromass

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

$$P\left(\bigcup{A_n^c}\right)$$

Right? You've proven the formula above...

11. Jun 15, 2011

### stukbv

ok so the limit = p(An^c)

12. Jun 15, 2011

### micromass

Staff Emeritus
And what the probability of a complement?

13. Jun 15, 2011

### stukbv

1-p(An).
I see , but then how do i relate it to my initial statement?

14. Jun 15, 2011

### micromass

Staff Emeritus
Well, you know already that the statement

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

is true. Now try to evaluate the left side.

15. Jun 16, 2011

### stukbv

surely if they are all subsets of their preceding ones then the union of the complements is just a1^c ?

16. Jun 16, 2011

### micromass

Staff Emeritus
We have

$$A_1\supseteq A_2\supseteq ...$$

and thus

$$A_1^c\subseteq A_2^c\subseteq ...$$

So that doesn't really work.

17. Jun 16, 2011

### stukbv

ohh so is the LHS 1 - P( intersection of Ai)

18. Jun 16, 2011

### stukbv

Then you can get rid of the ones and then take limits ?

19. Jun 16, 2011

### stukbv

so we can say lhs = 1-P(intersection Ai) ?

20. Jun 16, 2011

### micromass

Staff Emeritus
Yes, that is correct! Just eliminate the ones and you have the result you wanted to prove!