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Let $(\Omega, \mathscr F, \mathbb P)$ be a probability space. Conjecture:

Suppose we have events $A_1, A_2, ...$ s.t. $\forall \ A \in \bigcap_n \sigma(A_n, A_{n+1}, ...)$, $P(A) = 0$ or $1$. There exists an independent sequence of events $B_1, B_2, ...$ s.t.

$$\tau_{A_n} := \bigcap_n \sigma(A_n, A_{n+1}, ...) = \bigcap_n \sigma(B_n, B_{n+1}, ...) := \tau_{B_n}$$

Is this true?


I think there exists a function $f: \mathbb N \to \mathbb N$ s.t. $A_{f(n)}$'s are independent so we can choose $B_n = A_{f(n)}$. Is that true? Why/Why not? If not, how else can I prove or disprove the conjecture above? If it is true, I think it can be proven by modifying the proof of the Kolmogorov 0-1 Law (for events).


Perhaps one of these subsequences of sets is independent:

$$A_n$$

$$A_{2n}, A_{2n+1}$$

$$A_{3n}, A_{3n+1}, A_{3n+2}$$

$$\vdots$$

$$A_{mn}, A_{mn+1}, A_{mn+2}, ..., A_{mn+(m-1)}$$

$$\vdots$$

I think we have that

$$\tau_{A_n} = \tau_{A_{mn+i}} := \bigcap_n \sigma(A_{mn+i}, A_{m(n+1)+i}, ...)$$

where $m \in \mathbb N$ and $i \in \{0, 1, 2, ..., m-1\}$.


It seems like we need any such $f(n)$, if it exists, to satisfy the following condition:

$$\sigma(A_{f(n)}, A_{f(n+1)}...) \subseteq \sigma(A_n, A_{n+1}, ...) \tag{**}$$

which I guess is true if (and only if?) $f(n) \ge n$.


Other possible candidates for $f(n)$: (assume the variables are s.t. $f: \mathbb N \to \mathbb N$ is satisfied. If need be, $(**)$ or $f(n) \ge n$ too.)

  1. $\sum_{i=0}^{m} a_i n^i$

  2. $2^n, 3^n, ...$

  3. $\sum_{i=1}^{m} b_i c_i^n$

  4. $\lfloor{t^n}\rfloor, \lceil{t^n}\rceil$ (I guess $t > e^{1/e}$)

  5. $\lfloor{\sum_{i=1}^{m} b_i c_i^n}\rfloor, \lceil{\sum_{i=1}^{m} b_i c_i^n}\rceil$

  6. $\lfloor{\text{linear combination of trigonometric functions}}\rfloor, \lceil{\text{linear combination of trigonometric functions}}\rceil$

  7. $\lfloor{\text{Some linear combination of the above}}\rfloor, \lceil{\text{Some linear combination of the above}}\rceil$


Assuming the conjecture is true, I guess it's not necessary to find $f(n)$ that works for all possible sequences of events $A_1, A_2, ...$ because such $f(n)$ may not even exist.


To disprove the conjecture: I guess we must show that such a sequence $B_n$ being independent implies $B_n$ tail will never equal $A_n$ tail since $B_n$ tail will be $\mathbb P-$trivial by Kolmogorov 0-1 Law (for events).

Something that might help: we could show that $\forall \ A \in \bigcap_n \sigma(A_{f(n)}, A_{f(n+1)}, ...), P(A) = 0$ or $1$ and $\forall n \in \mathbb N, A_{f(n)}, A_{f(n+1)}, ...$ is not independent, but I'm not quite sure that the conjecture is disproved because we could construct some $B_n$'s that look like:

  1. $$B_n = A_{n+1} \setminus A_n$$

  2. $$B_n = A_{n} \setminus A_{n-1}, A_0 = \emptyset$$

  3. $$B_n = \bigcap_m A_{mn}$$

  4. $$B_n = \bigcup_m A_{mn}$$

  5. $$B_{2n} = \bigcap_m A_{mn}, B_{2n+1} = \bigcup_m A_{mn}$$

  6. $$B_n = \limsup_m A_{mn}$$

  7. $$B_n = \liminf_m A_{mn}$$

  8. $$B_{2n} = \limsup_m A_{mn}, B_{2n+1} = \liminf_m A_{mn}$$

Not to say of course that any of those $B_n$'s satisfy $\tau_{A_n} = \tau_{B_n}$ but that $B_n$ need not be in the form $A_{f(n)}$.


Borel-Cantelli:

  1. If $\sum_n P(A_n) < \infty \to 0 = P(\limsup A_n) = P(\limsup A_{mn}) \ \forall m \in \mathbb N$. Hence $B_m = \limsup A_{mn}$ is independent.

  2. If $\sum_n P(A_n) = \infty$, then maybe this extension of Borel-Cantelli? Not quite sure I understand it or how it would be helpful. I don't think we can conclude anything if we have $P(\limsup A_n)$.

  3. Then there's the case of $\sum_n P(A_n) = \infty$ but the conditions earlier aren't satisfied.

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    $\begingroup$ Perhaps a proof by construction, where $B_1 = A_1, B_2 = A_2 - A_1, \dots$? $\endgroup$ – jbowman Dec 24 '15 at 1:55
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    $\begingroup$ To me this conjecture seems unlikely to be true unless you add extra conditions, or you mean that the two $\sigma$-algebra's completions agree (which holds almost trivially). However I can't see a counter example. $\endgroup$ – P.Windridge Dec 25 '15 at 14:20
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    $\begingroup$ In any case I think you can start with the (simpler) question: "Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. Suppose $\mathcal{G}\subset\mathcal{F}$ is a countably generated $\sigma$-algebra and that $\mathbb{P}(A)= 0$ or $1$ for any event $A \in \mathcal{G}$. Is there a sequence of independent events $B_1,B_2,\ldots$ in $\mathcal{F}$ with tail $\sigma$-algebra $\mathcal{G}$? $\endgroup$ – P.Windridge Dec 27 '15 at 10:51
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    $\begingroup$ A $\sigma$-algebra $\mathcal{G}$ is countably generated if there exists $F_1,F_2,\ldots$ s.t. $\mathcal{G} = \sigma(F_1,F_2,\ldots)$. It is straightforwards to find examples where the tail $\sigma$-algebra is not countably generated. $\endgroup$ – P.Windridge Dec 27 '15 at 11:03
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    $\begingroup$ More generally, a sub-$\sigma$-algebra of a countably generated $\sigma$-algebra may not itself be countably generated! Actually look at Exercise 1.1.18 in math.mit.edu/~dws/175/prob01.pdf $\endgroup$ – P.Windridge Dec 27 '15 at 11:06
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If you want events $B_n$ that are independent in an interesting manner (not simply because $\mathbb{P}(B_n) = 0$ or $\mathbb{P}(B_n) = 1$) then the conjecture is false.

Here is a pedantic example. Suppose $(\Omega, \mathcal{F},\mathbb{P})$ is a suitably rich probability space.

Let $A \in \mathcal{F}$ be $\mathbb{P}$-null, i.e. $\mathbb{P}(A) =0$. Take $A_i = A$, so that the tail $\sigma$-algebra is $\mathcal{G} = \{\emptyset, A, A^c, \Omega\}$.

Note that in particular $\mathcal{G}$ is finite.

Now, suppose that $B_1, B_2, \ldots$ is an independent sequence of events with $\mathbb{P}(B_n)$ bounded away from $0$ and $1$. Then the tail $\sigma$-algebra $\mathcal{H}$ is not countably generated. (See e.g. Exercise 1.1.18 http://math.mit.edu/~dws/175/prob01.pdf, which uses an argument like I outlined above- any countably generated $\mathbb{P}$-trivial $\sigma$-algebra has an atom of mass $1$, but $\mathcal{H}$ has no such atom).

So, $\mathcal{G}$ is finite but $\mathcal{H}$ is not even countably generated.


Edit 2: if you accept $\mathbb{P}(B_n) = 0$ then you can replicate any countably generated $\mathbb{P}$-trivial $\sigma$-algebra. In more detail, suppose that $\mathcal{G}$ is generated by events $E_1, E_2, \ldots \in \mathcal{G}\subset\mathcal{F}$. If $\mathcal{G}$ is $\mathbb{P}$-trivial then the $E_n$ are all independent, by virtue of being null (or $E_n^c$ being null). Now make a triangular construction for the $B$ events: $B_{1,1} = E_1$, $B_{2,1} = E_1, B_{2,2} = E_2,\ldots,B_{k,j} = E_j$, $1\le j \le k$.

Then $(B_{k,j})$ is a countable sequence (with natural ordering for the indices) of independent events whose tail $\sigma$-algebra is $\mathcal{G}$.

So, here I think is the key question: suppose that $\mathcal{G}$ is a non-countably-generated $\mathbb{P}$-trivial tail $\sigma$-algebra (coming from non-null events which might be dependent). Can $\mathcal{G}$ be realised as the tail $\sigma$-algebra for some null events?

Edit 1: A gray area is what happens if you accept $\mathbb{P}(B_n)\to 0$, although that doesn't seem to be the thrust of the original question.

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  • $\begingroup$ Thanks P.Windridge, but I'm not quite sure I understand. 1 If we include $P(B_n)=0$ or $1$, the conjecture is (trivially?) true? 2 Is what you're trying to prove in Edit 2? If so, is your $\mathcal G$ equal to my $\tau_{A_n}$? I edited OP for shorthand $\endgroup$ – BCLC Dec 30 '15 at 22:30
  • $\begingroup$ I read the exercise. $\mathcal H = \tau_{B_n}$ ? $\endgroup$ – BCLC Dec 30 '15 at 22:49
  • $\begingroup$ Hi BCLC, (1) I'm saying that if we include $P(B_n)=0$ then the conjecture is true for all choices of the events $A_1,A_2,\ldots$ that have a "nice" tail $\sigma$-algebra (where "nice" here means countably generated). (2) Yes $\mathcal{G}=\tau_{(A_n)}$ and $\mathcal{H}$ is your $\tau_{(B_n)}$. N.B. the linked exercise uses "$A_n$" for what-should-be your $B_n$, and "$B_n$" denotes a candidate sequence of generating events (used to get a contradiction. $\endgroup$ – P.Windridge Dec 31 '15 at 20:05
  • $\begingroup$ I'm not sure I follow. Just from the assumptions, the $A_i$'s ARE independent? $\endgroup$ – BCLC Jan 1 '16 at 2:36
  • $\begingroup$ In Exercise 1.1.18 of math.mit.edu/~dws/175/prob01.pdf, the $A_i$'s are independent events, which you should think of as the $B_i$'s in your conjecture. Is that what you were asking? $\endgroup$ – P.Windridge Jan 1 '16 at 10:31

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