# Zero-one law for an independence-like structure

I am a number theorist by profession, so apologies if the answer to this question is "trivially true" or "trivially false".

Let $(\Omega, \mathcal{A}, P)$ be a (non-atomic) probability space. Let $(\mathcal{F}_n)_{n \geq 1}$ be a filtration such $\mathcal{F}_\infty = \mathcal{A}$. Let $(X_n)_{n \geq 1}$ be a sequence of non-negative and bounded (but not uniformly bounded) random variables, such that $X_n$ is $\mathcal{F}_n$-measurable and $E(X_n | \mathcal{F}_{n-1})=E(X_{n})$. Let $B$ denote the set where $\sum_{n=1}^\infty X_n = \infty$.

Question: is it true that $P(B)$ can only be 0 or 1?

(To me, this seems to be a sort of Kolmogorov's zero-one law, where the independence is replaced by some quasi-independence property. However, I cannot see if the answer should be "yes" here, or if some additional integrability conditions or something of that sort would be necessary.)

• In fact, as written the condition is something closer to "uncorrelated". Mar 9, 2018 at 18:41
• No, I don't want a martingale, I need the answer to the question as stated and not the answer to a completely different question. This is no typo, this is just the situation that I have in my application. Mar 9, 2018 at 20:37
• PS: I realize that the structure here is not a martingale, but it is a sequence of random variables which are adapted to a filtration, and that reminded me of a martingale. As I remember it, it is possible to prove Kolmogorov's zero-one law by martingale theory. Maybe a similar thing works here? Mar 9, 2018 at 20:48
• I'm confused - people keep changing the key assumption. What exactly is $E(X_n \mid \mathcal{F}_{n-1})$ supposed to equal? Is it $X_n$? $E[X_n]$? $X_{n-1}$? or what? Mar 9, 2018 at 21:10
• The key assumption ís $E(X_n | \mathcal{F}_{n-1})=E(X_n)$. Consider the special case when $X_n$ is the indicator function of an event $A_n$ for all $n$. Then the assumption says that $E(X_n|\mathcal{F}_{n-1})=P(A_n)$, which means that $A_n$ is independent of the $\sigma$-field generated by $A_1, \dots, A_{n-1}$. So if I am not mistaken, then in the case when $X_n$ is {0,1}-valued, the question reduces to Kolmogorov's zero-one law (not fully sure about that, though). However, in my question $X_n$ are allowed to have more different values. Mar 9, 2018 at 21:16

Here is a counterexample. It's roughly based on the following useful example: if $X$ is the amount you decide to bet on a fair game, and $Y$ is your resulting profit, then $X,Y$ are uncorrelated but not independent.

Let $\xi_n$ be independent with $P(\xi_n = n) = n^{-2}$, $P(\xi_n = 0) = 1-n^{-2}$. Note that $E[\xi_n] = n^{-1}$ and that, by Borel–Cantelli, we have $\sum_n \xi_n < \infty$ a.s.

Let $\theta$ be a fair coin flip (Bernoulli 1/2) independent of all the $\xi_n$. Let $\mathcal{F}_n = \sigma(\theta, \xi_1, \dots, \xi_n)$.

Set $X_n = n^{-1} \theta + \xi_n (1-\theta)$, which is $\mathcal{F}_n$-measurable. Notice that $$E[X_n \mid \theta = 1] = n^{-1} = E[\xi_n] = E[X_n \mid \theta = 0]$$ and that $(X_n, \theta)$ is independent of $(\xi_1, \dots, \xi_{n-1})$. Hence $E[X_n \mid \mathcal{F}_{n-1}] = E[X_n \mid \theta] = n^{-1} = E[X_n]$, so we have the desired condition.

But on the event $\theta = 1$ we have $X_n = n^{-1}$ for all $n$, and then $\sum X_n = \infty$. And if $\theta = 0$ then $X_n = \xi_n$ for all $n$, then $\sum X_n < \infty$. So we have $P(B) = 1/2$.

(Incidentally, the sequence $M_n = \sum_{k=1}^n (X_k - k^{-1})$ is a martingale which converges with probability $1/2$. In fact, in retrospect, I think the desired condition is equivalent to "$X_n$ is a martingale difference sequence plus constants.")

If you require the $X_n$ to be uniformly bounded then the answer is yes; $P(B)$ is 0 or 1 according to whether $\sum_n E[X_n]$ converges or diverges.

Let's say $0 \le X_n \le 1$ for all $n$. Let $c_n = E[X_n]$. If $\sum_n c_n < \infty$ then by monotone convergence we have $E\left[\sum_n X_n\right] < \infty$ and thus $P(B) = 0$. So suppose that $\sum_n c_n = \infty$. Let $Y_n = X_n - c_n$ (so $-1 \le Y_n \le 1$) and $M_n = \sum_{k=1}^n Y_k$. As noted above, $M_n$ is a martingale, which now has bounded increments. On the event $B^c$ we have $M_n = \sum_{k=1}^n X_k - \sum_{k=1}^n c_k \to -\infty$ since the first sum converges and the second diverges. However, according to this exercise, there is probability zero to have $M_n \to -\infty$. (It says that almost surely, $\limsup M_n$ is either finite or $+\infty$.) So in this case $P(B)=1$.

• Thank you! Do you think there also exist a counterexample if one has the additional assumption that the $X_n$ are uniformly bounded? Mar 10, 2018 at 18:42
• @KurisutoAsutora: There does not. See edit. Mar 10, 2018 at 19:12
• Very interesting. Thank you very much! Mar 10, 2018 at 19:19