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:


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

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


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


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 $A_{f(n)}$.


  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.

Based on: https://math.stackexchange.com/questions/605301

  • 7
    $\begingroup$ 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? $\endgroup$ Dec 27 '15 at 13:33
  • 3
    $\begingroup$ 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. $\endgroup$ Dec 27 '15 at 16:39
  • 2
    $\begingroup$ Even if such function $f(n)$ exists, how does it imply equality of sigma-algebras? $\endgroup$ Dec 28 '15 at 10:27
  • 1
    $\begingroup$ @BCLC: please see Durrett's book, chapter on Ergodic Theory $\endgroup$ Dec 28 '15 at 10:37
  • 1
    $\begingroup$ 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). $\endgroup$ 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.

  • $\begingroup$ what do you mean by 'if $C_i$' ? You mean if $\omega \in C_i$, then $A_i = D_i$ ? $\endgroup$
    – BCLC
    Apr 17 '16 at 10:29
  • $\begingroup$ $C_i$ 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 $D_i$. $\endgroup$
    – domotorp
    Apr 17 '16 at 12:37
  • $\begingroup$ Right. Anyway, how does that prove or suggest that such $f$ does not exist? $\endgroup$
    – BCLC
    Apr 17 '16 at 17:12
  • $\begingroup$ 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. $\endgroup$
    – domotorp
    Apr 17 '16 at 20:39
  • $\begingroup$ 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 $\omega \in C_i$? $\endgroup$
    – BCLC
    Apr 18 '16 at 8:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.