# Kolmogorov 0-1 Law Converse?

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.

• I'm pretty sure that $A_{f(n)}$ won't work, since it seems pretty easy to make pairwise dependent variables that still satisfy the tail 0-1 property. I think it will be hard to come up with invariably independent events unless you use tail events. So the question is: is there a sequence of tail events whose tail is the original tail? Jan 17, 2016 at 20:59
• I would guess that it is false, but it does seem possible that you could make a sequence of tail events that generated the tail, which would make it true. Jan 18, 2016 at 19:17
• No, I was talking about your conjecture - I just thought it would be very hard to come up with independent events when there is almost no restriction on the $A_n$. The exception would be tail events, which you did assume would have probability 0 or 1 in your conjecture. But we would need the tail of those tail events to be the original tail to satisfy your conjecture (I think that question is also of independent interest.). Jan 18, 2016 at 22:27
• Isn't the assumption already for tail events? Jan 18, 2016 at 23:15
• @BCLC The $A_i$'s are certainly not tail events for the sequence $\lbrace A_i \rbrace$. What I am referring to is the assumption that $\forall A \in \bigcap_n \sigma (A_n, A_{n+1}, \ldots), P(A) = 0$ or $1$. That's the tail. Jan 20, 2016 at 15:56