Limit of the intersection of events

[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 isn't it ?

Thanks a lot

 

[micromass]

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?

 

[stukbv]

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!

 

[micromass]

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?

 

[stukbv]

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

 

[micromass]

No, take the complements. What is

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

 

[stukbv]

P((U An^c)?

 

[micromass]

P((U An^c)?

Yes, carry on...

 

[stukbv]

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

 

[micromass]

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

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

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

 

[stukbv]

ok so the limit = p(An^c)

 

[micromass]

ok so the limit = p(An^c)

And what the probability of a complement?

 

[stukbv]

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

 

[micromass]

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.

 

[stukbv]

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

 

[micromass]

We have

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

and thus

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

So that doesn't really work.

 

[stukbv]

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

 

[stukbv]

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

 

[stukbv]

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

 

[micromass]

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

 

[stukbv]

Perfect! just in time... thanks again!