Sort-of converse of Kolmogorov zero-one theorem

Let $$(\Omega, \mathscr F, \mathbb P)$$ be a probability space. The Kolmogorov zero-one theorem states that

Suppose we have independent random variables $$X_1, X_2, ...$$. Then $$\forall \ A \in \bigcap_n \sigma(X_n, X_{n+1}, ...)$$, $$P(A) = 0$$ or $$1$$.

If we choose $$X_k = 1_{A_k}$$ for events $$A_1, A_2, ...$$, then we have:

Suppose we have independent events $$A_1, A_2, ...$$. Then $$\forall \ A \in \bigcap_n \sigma(A_n, A_{n+1}, ...)$$, $$P(A) = 0$$ or $$1$$.

Now, is this following conjecture true? If not, can it be modified slightly to be true?

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.

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

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 Theorem (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

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

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

Based on what @FedorPetrov pointed out, it seems like we need $$f(n)$$'s s.t.

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

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: There's of course showing that any sequence that satisfies $$(*)$$ will not be independent, but I have a feeling it's more of showing that any independent sequence will never satisfy $$(*)$$.

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 $$(*)$$ 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.

• Did you check this in some simple situation where the tail sigma-algebra is still trivial, e.g. $X_n$ is a Markov chain on $\{0,1\}$, and $A_n=1_{X_n=0}$? How would you construct your events $B_n$ then? Dec 27 '15 at 13:33
• Well, existence of such independent sequence is something that should be proved... In general, I'm rather sceptical, since there are a lot of examples of nonindependent sequences with trivial tail sigma-algebra. Dec 27 '15 at 16:39
• Even if such function $f(n)$ exists, how does it imply equality of sigma-algebras? Dec 28 '15 at 10:27
• @BCLC: please see Durrett's book, chapter on Ergodic Theory Dec 28 '15 at 10:37
• I would not call that a converse of Kolmogorov theorem. Because this theorem is not really a statement about the "set-theoretic" $\sigma$-fields $\sigma(A_n, A_{n+1}, \ldots)$, but rather about their completions $\bmod \mathbb{P}$ (the independance assumption is a "$\bmod P$" property). Dec 28 '15 at 21:48

I don't think that there can be such a selection function $f$.

Define $A_i$ through independent events $C_i$ and $D_i$, each of which occurs with $50\%$ chance, as follows. $A_1=D_1$ and for $i\ge 2$, if $C_i$, then $A_i=D_i$, otherwise $A_i=A_{i-1}$. If $A \in \bigcap_n \sigma(A_n, A_{n+1}, ...)$, then $P(A) = 0$ or $1$ should hold, with a similar proof as for the Kolmogorov Zero-One Theorem. On the other hand, no $A_i$ and $A_j$ are independent.

• what do you mean by 'if $C_i$' ? You mean if $\omega \in C_i$, then $A_i = D_i$ ?
– BCLC
Apr 17 '16 at 10:29
• $C_i$ is an event, so I mean if $C_i$ occurs. Each of $C_i$ and $D_i$ are like independent coin tosses - if $C_i$ comes up heads, $A_i$ is unchanged, if $C_i$ comes up tails, $A_i$ gets the value of $D_i$. Apr 17 '16 at 12:37
• Right. Anyway, how does that prove or suggest that such $f$ does not exist?
– BCLC
Apr 17 '16 at 17:12
• None of the $A_i$'s are independent, so whatever your $f$ is, already $A_{f(1)}$ and $A_{f(2)}$ won't be independent. Apr 17 '16 at 20:39
• Ah thanks, domotorp. Dumb question: If $C_i$ and $D_i$ are independent, how can the choice of $A_i$ depend on whether or not $\omega \in C_i$?
– BCLC
Apr 18 '16 at 8:35