# Conjecture related to Kolmogorov 0-1 Law (for events)

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.

• Perhaps a proof by construction, where $B_1 = A_1, B_2 = A_2 - A_1, \dots$? Dec 24, 2015 at 1:55
• 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. Dec 25, 2015 at 14:20
• 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 $\mathcal{G}$? Dec 27, 2015 at 10:51
• 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. Dec 27, 2015 at 11:03
• 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 Dec 27, 2015 at 11:06

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.

• 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
– BCLC
Dec 30, 2015 at 22:30
• I read the exercise. $\mathcal H = \tau_{B_n}$ ?
– BCLC
Dec 30, 2015 at 22:49
• 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. Dec 31, 2015 at 20:05
• I'm not sure I follow. Just from the assumptions, the $A_i$'s ARE independent?
– BCLC
Jan 1, 2016 at 2:36
• 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? Jan 1, 2016 at 10:31