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Let $(\Omega, \mathscr{A}, \mathbb{R})$ be a probability space.

We define an atom of probability as an event $A \in \mathscr{A}$ such that $\mathbb{P}(A)>0$ and for all events $D \subset A$, $\mathbb{P}(D) = 0$ or $\mathbb{P}(D)=\mathbb{P}(A)$.

What my textbook states about atoms is (and I understand the proof of both of these statements): if $A$ and $B$ are atoms, then either $\mathbb{P}(A \cap B)=\mathbb{P}(A)=\mathbb{P}(B)$ or $\mathbb{P}(A \cap B)=0$, and there can be at most countably many disjoint (up to an event of zero probability) atoms. I'm not sure if this is useful for my question, but I stated it here anyway.

Now, for my question: prove that $\Omega$ can be written as a union $$ \Omega = (\bigcup_{n=0}^{+\infty}A_{n}) \cup X,$$ where $\{A_{n}\}_{n \in \mathbb{N}}$ is a sequence of atoms (does not specify what their intersections are), and $X$ is an event with the following property: for any $p \in (0, P(X))$, there exists an event $B \subset X$ such that $P(B)=p$.

My idea was to take a maximal family of disjoint (again, up to an event of zero probability) atoms, which is at most countable, $\{A_{n}\}_{n \in \mathbb{N}}$, and consider $X := \Omega \setminus \cup_{n=0}^{+\infty}A_{n}$, and prove the required property for that set.

If $Y \subset X$ is an atom, then, seeing as how my family is maximal, $Y$ must non-trivially intersect some $A_{i}$ from my sequence, and therefore $\mathbb{P}(Y\cap A_{i})=\mathbb{P}(A_{i})>0$, but also since $Y \subset X$, $Y \cap A_{i}= \emptyset$, so $\mathbb{P}(Y \cap A_{i})=0$, which is a contradiction, so $X$ cannot contain any atoms. Is this useful for what I'm supposed to prove? I can't seem to see where to go from here, or if this is even the right first step.

EDIT: I'm not even 100% sure if I can take a maximal family without proving it exists in some way, so this possibly isn't the right way to go.

EDIT 2: I've managed to work out that because $X$ cannot contain any atoms, for every $p \in (0, \mathbb{P}(X))$, there must exist an event $B$ such that $\mathbb{P}(B)<p$, therefore $0$ is a limit point of the set of $p \in (0, \mathbb{P}(X))$ for which there exists an event $B$ for which $\mathbb{P}(B) = p$, but this still doesn't give me the whole solution.

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    $\begingroup$ Please clarify : your notation $AB$ is what everyone else denotes by $A\cap B$, i.e. the intersection ? $\endgroup$ – Ewan Delanoy Apr 24 '17 at 7:19
  • $\begingroup$ Yes, in my environment (and I think probability theory as well) $AB = A \cap B$. $\endgroup$ – Matija Sreckovic Apr 24 '17 at 9:39
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    $\begingroup$ Thank you for clarifying this point. Personally I've never seen the notation $AB$ used to mean intersection in any probability textbook, and I would advise you to change all your $AB$'s to $A\cap B$ in your text. $\endgroup$ – Ewan Delanoy Apr 24 '17 at 9:50
  • $\begingroup$ @EwanDelanoy: For what it's worth, there are popular books that use $AB$; Sheldon Ross's books, for instance. But I don't much like it either. $\endgroup$ – Nate Eldredge Apr 26 '17 at 15:57
  • $\begingroup$ @NateEldredge Indeed, $AB$ is already used to mean the product of two random variables $A$ and $B$. $\endgroup$ – Ewan Delanoy Apr 26 '17 at 16:29
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Everything you said is OK, you just need a few complements and simplifications.

To justify the existence of a maximal disjoint set of atoms, you can use Zorn's lemma : let $P$ be the poset whose elements are the sets of disjoint atoms, and whose order is defined by inclusion. Then every chain in $P$ has an upper bound in $P$ (the union of the chain), so the hypothesis of Zorn's lemma is satisfied.

So there is a maximal disjoint set of atoms by Zorn's lemma, the set is necessarily countable as explained in the OP, call it $(A_n)_{n\in{\mathbb N}}$ and let $X=\Omega \setminus \bigcup_{n\in{\mathbb N}}A_n$. Then $X$ is measurable, and atomless as explained in the OP.

Now, let $p\in (0,{\mathbb{P}}(X))$. Let $${\mathscr{A}}_{p,X}=\lbrace U\in {\mathscr{A}} | U \subseteq X, {\mathbb{P}}(U) \leq p \rbrace$$.

Say that two elements $U_1$ and $U_2$ in ${\mathscr{A}}_{p,X}$ are linked iff $U_1\subseteq U_2$. I call $L=U_2\setminus U_1$ the link between $U_1$ and $U_2$. If $L$ is either empty or has positive measure, we say that $U_1$ and $U_2$ are strongly linked, and denote this by $U_1 \leq_{p,X} U_2$.

It is easily checked that $\leq_{p,X}$ defines a partial ordering on ${\mathscr{A}}_{p,X}$. Also, if $C$ is any $\leq_{p,X}$-chain in ${\mathscr{A}}_{p,X}$, since the links between two (distinct) successive elements are all disjoint, $C$ is necessarily countable. The union of $C$ will be in ${\mathscr{A}}_{p,X}$, so it will be an upper bound for $C$. By Zorn's lemma, there is a $\leq_{p,X}$-maximal element, denote it by $M$. We know that $\delta=p-{\mathbb{P}}(M)$ is nonnegative. If $\delta>0$, by the property called "Edit 2" in the OP, there is a $V_1\subseteq X\setminus U$ with $0<{\mathbb{P}}(V_1)<\delta$, but then the set $M'=M\cup V_1$ would contradict the maximality of $M$, so $\delta=0$ and hence $P(M)=p$, which finishes the proof.

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  • $\begingroup$ I'm confused about the definition of linked elements. If $X \in L$, that is $X \in U_{2} \setminus U_{1}$, then $X \in U_{2}$, therefore $X \in V_{2}$. But since $U_{2} \setminus U_{1} = V_{1} \setminus V_{2}$, then $X \in V_{1} \setminus V_{2}$, therefore $X \notin V_{2}$, which is a contradiction. So by this if I'm not mistaken $L$ can't be anything other than $\emptyset$. Did you mean to write $L = U_{1} \setminus U_{2}$ or is the relation meant to be trivial in this way? $\endgroup$ – Matija Sreckovic Apr 26 '17 at 19:55
  • $\begingroup$ @MatijaSreckovic Answer to your first question : my initial answer was indeed incorrect as stated. Please check my corrected version. Answer to your 2nd question ; it is not true than $P(L)$ is always $0$. Consider for example, $\Omega=[0,1]$, $({\mathscr A},{\mathbb P})=$ Lebesgue measure, $p=\frac{2}{3},U_1=[0,\frac{1}{3}], U_2=[0,\frac{2}{3}],$. Then $L$ has probability $\frac{1}{3}$ not zero. $\endgroup$ – Ewan Delanoy Apr 27 '17 at 7:49
  • $\begingroup$ The second comment was meant in accordance with your first definition, but yeah, it's pointless with the fixed definition of linked sets. $\endgroup$ – Matija Sreckovic Apr 27 '17 at 19:25

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