Probability of occurrence of infinitely many events?

Question

I'd like some hep with this prof, please, I'm rather confused.

In a sample space, let E_n be a sequence of events, each one with probability p(E_n). Suppose the series Sum(n =1, oo) p(E_n) converges a let E = {infinitely many events E_n occur}. Show that p(E) = 0.

Thank you

Answer 1

Define the event F_n to be {E_i occurs for some i at least n} = Union(i=n, infty) E_n. Then:

E = Intersection(n=1, infty) F_n

Thus to prove p(E) = 0 it is sufficient to prove that lim_{n->infty} F_n = 0. But consider this:

p(F_n) = p(Union(i=n, infty) E_i ) is less than or equal to Sum(i=n, infty) p(E_i).

Since Sum(n =1, oo) p(E_n) converges, Sum(i=n, infty) p(E_i) must go to zero as n->infty, and so p(F_n) goes to zero, which proves the claim.

Edit: The other answer is not quite correct. It only proves that the probability of a specific infinite subset of the E_n's all occuring is zero, not that the probability of infinitely many E_n's occuring is zero. This mistake is similar to the mistake one might make when considering the probability of picking any random point in the interval [0,1]. The probability of any specific point being picked is zero, but this certainly doesn't mean the probability of picking a point in say [0,1/2] is zero.

Note that the other proof only actually requires the slightly weaker hypothesis that p(E_n)->0 as n ->infty. In fact this weaker hypothesis is not sufficient to give your desired result. You can see this by considering the unit interval with the uniform probability distribution. It is then possible to create a sequence of events {E_n} so that p(E_n) = 1/n for every n, and every point in [0,1] lies in infinitely many {E_n}. Thus for every point in [0,1] there is some infinite subset of {E_n} so that that point lies in each member of the infinite subset. Therefore the entire interval [0,1] is in E.

Answer 2

I'll give you another proof that is quite similar to Sean's proof. Your example is a particular case of this more general case..

Maybe you have already studied measure theory. If you have, you probably know that a sample space can be seen as a measure space and a random variable can be seen as a measurable function. And, since we can identify events with sets (measurable sets), the probability function is a measure. The measure of an event. A finite measure with values in [0, 1].

So, lets reparaphrase your original problem and consider it in a more general context:

Let (X, M, u) be a measure space - X a set, M a sigma-algebra defined on M and u a measure defined on M. Let E_n be a sequence of sets of M (measurable sets) such that Sum (n =1, oo) u(E_n) converges. Let E = {x in X | such that x belongs to infinitely many sets E_n}. Then, u(E) = 0. In measure theory language, this is the same as saying almost every x in X belongs to finitely many sets E_n .

Proof: (very similar to Sean's proof)

We know that E, called the lim sup E_n (limit superior of E_n) is given by E = Intersection (n =1, oo) ((Union (m = n, oo) E_m)). If F_n = Union (m = n, oo) E_m, then E = Intersection (n= 1, oo) F_n.

It's clear that the sets F_n form a decreasing sequence: F_1 > F_2 > F_3, ....here > means "contains". And since Sum (n =1, oo) u(E_n) converges, u(E_1) < oo. Therefore, according to the property known as "continuity from above" of the measure, we have u(E) = lim u(F_n).

In virtue of the definition of F_n and of the sub-additivity of the measure, for each n we have

0 <= u(F_n) <= Sum(m= n, oo) u(E_m) (1).

Since Sum (n =1, oo) u(E_n) converges, it follows (property of convergent series) that lim (n -> oo) Sum(m=n, oo) u(E_m) = 0. So, in virtue of (1), it follows by squeeze that lim u(F_n) = u(E) = 0, which is the desired conclusion in a more general context.

This could also be proved very nicely and simply using the Lebesgue integral with respect to the measure u.

Hope this helps (instead of confusing you.....)

Steiner
[e-mail] (if you want any further explanation that I can give)

EDIT: I agree with Sean, there's a slight mistake in Ksoileau's proof

Answer 3

Suppose an event E is defined as: event E_k_1 occurs and event E_k_2 occurs and ... for some set of infinite set of integers {k_1,k_2,...}.
Fix e>0. Suppose all P(E_k_i)>=e for every i=1,2,3,.... Then Sum(i=1,oo)p(E_k_i)>=e*oo=oo. But this contradicts the assumption that Sum(i=1,oo)p(E_n) converges, hence there must exist some j such that P(E_k_j) Since the event E implies the event E_k_j, P(E)<=P(E_k_j) hence P(E)