I have been assigned a homework after my first lecture in probability, and I have this question that I couldn't really solve/understand, it has like more 2 weeks time, and I'm wondering if it's a challenging question or I just still didn't learn the material needed.
Here's the question:
Assume that for every sequence $B_n$ of events that satisfy:
$…\subseteq B_4\subseteq B_3\subseteq B_2\subseteq B_1$ and $\bigcap_{n=1}^{\infty}B_n = B$
$\Rightarrow$ $lim_{n\to\infty}P(B_n)=P(B)$.Let $A_n$ be a sequence of events such that:
$lim_{n\to\infty}P(A_n)=0$ and $\sum^\infty_{n=1}P(A^c_n\cap A_{n+1})<\infty$Prove that: $P(A_n i.o)=P(\bigcap^\infty_{k=1}\bigcup_{n\ge k} A_n)=0$
$\mathbf{Me\space Trying \space To \space Understand/Solve:}$
So I have never seen before the meaning of limits in probability, but I have learnt it in Calculus, so I tried to take the given information as a fact and without deeply understanding it ($\lim_{n\to\infty}P(B_n)=P(B))$.
Moving on to proving $P(A_n i.o)=P(\bigcap^\infty_{k=1}\bigcup_{n\ge k} A_n)=0$, I tried to open it up to see how things work:
$(A_1\cup A_2 \cup … \cup A_n)\cap (A_2\cup A_3 \cup … \cup A_n) \cap (A_3 \cup A_4 \cup … \cup A_n)\cap …$
Now I can see that every one of the sets (the unions) is a subset of the one the one that came before it. so according to the thing we assumed I can say that $lim_{n\to\infty}P(\bigcup_{n\ge k}A_n)=A_n$ (Idk How to really prove it, but taking the intersection of all of them would leave me with $A_n$ at the end (intuitively)). And here I got stuck and didn't know my next step.
$\mathbf{Questions:}$
1) What does $i.o$ mean in $P(A_n i.o)$?
2) Am I in the right direction?
3) According to what I've did above, am I able to solve the question without extra knowledge about limits?
4) How Can I Proof $lim_{n\to\infty}P(\bigcup_{n\ge k}A_n)=A_n$ or is it needed to be proven in the general proof?
if I'm in the right direction I would appreciate some hints on how to keep going.
Thanks in advance.
Best Answer
"What does 𝑖.𝑜 mean?"
As already mentioned in the comments $i.o.$ stands for infinitely often and so $$\{A_n\,i.o\}=\{\omega\in\Omega:\omega\in A_n \text{ for an infinite number of indices } n\in\{1,2,\dots\}\}.$$
For a more intuitive explanation see the following example:
Consider the probability model for Bernoulli sequences (coin flips). Let $A$ be the event where the pattern $THTH$ occurs infinitely often, i.e., $$A:=\{A_n\,i.o\},$$ where $A_n$ is the event where the pattern $THTH$ occurs beginning at the $n$th coin flip. For example, $$A_2=\{\omega\in\Omega: \eta_2(\omega)=1,\eta_3(\omega)=-1,\eta_4(\omega)=1,\eta_5(\omega)=-1\},$$where we denoted the outcome of the coin flip by $\eta_1,\eta_2,\dots,$ which takes values $+1$ (tails) or $-1$ (heads).
Notice that the set $$\bigcup_{n\ge k} A_n$$ is the set of coin tossing sequences where the pattern occurs starting at some flip $\ge k$. If the pattern only occurs finitely many times in a particular sequence of flips, then that sequence will not be in this set for a large enough $k.$ Therefore, if we take the intersections of all these unions, we get the set of flips where the sequence occurs infinitely many times, i.e., $$A=\{A_n\,i.o\}=\bigcap_{k=1}^\infty\bigcup_{n\ge k} A_n=:\limsup_{n\to\infty} A_n.$$
For the first exercise, if you already proved that $$P\left(\bigcup_{n=1}^\infty C_n\right)=\lim_{n\to\infty}P\left( C_n\right),$$ for expanding sequence of events, i.e., $C_1\subset C_2\subset \cdots,$ you can apply this statement to the complements of $B_n$'s, since $\left(\bigcap_{n=1}^\infty B_n\right)^c=\bigcup_{n=1}^\infty B_n^c$ and $B_n^c$'s are expanding. Using the fact that $P(D^c)=1-P(D)$, we deduce the result.
If you didn't prove the above statement, here is a hint to do so: Write $\bigcup_{n=1}^\infty C_n=C_1\cup C_2\cup \cdots$ as the union of a sequence of disjoint sets: first start with $C_1,$ then add a disjoint set to obtain $C_1\cup C_2$, then again add a disjoint set to obtain $C_1\cup C_2\cup C_3$ and so on. Then use the definition of the probability measure to add those disjoint sets.
For the second exercise, first notice that $$ \bigcup_{n\ge k} A_n=A_k\cup \bigcup_{n\ge k}(A_{n+1}\setminus A_{n})=A_k\cup \bigcup_{n\ge k} (A_n^c\cap A_{n+1}).$$ As you noticed, the sequence $\left\{\bigcup_{n\ge k} A_n\right\}_k$ is a contracting sequence. Therefore, we can use the previous exercise to deduce $$P\left(\bigcap_{k=1}^\infty \bigcup_{n\ge k} A_n\right) =\lim_{k\to\infty}P\left(\bigcup_{n\ge k} A_n\right)=\lim_{k\to\infty}P\left(A_k\cup \bigcup_{n\ge k} (A_n^c\cap A_{n+1}) \right). $$ Now, by subadditivity, $$\lim_{k\to\infty}P\left(\bigcup_{n\ge k} A_n\right)\le \lim_{k\to\infty} P(A_k) + \lim_{k\to\infty}\sum_{n\ge k}P\left( A_n^c\cap A_{n+1} \right).$$ The first term on the RHS goes to zero by assumption and the second term also goes to zero by the convergence of the sum $\sum^\infty_{n=1}P(A^c_n\cap A_{n+1}),$ which is also discussed in the comments. Hence, we have that $$\lim_{k\to\infty}P\left(\bigcup_{n\ge k} A_n\right)=P\left(\bigcap_{k=1}^\infty \bigcup_{n\ge k} A_n\right) = P(A_n\,i.o)=0.$$